Answer:
816.33 kg
Explanation:
To solve this problem we'll need to keep in mind the formula for the kinetic energy of an object:
E = 0.5 * m * v²Where E is the kinetic energy, m is the object's mass and v its velocity.
We input the data given by the problem:
500000 J = 0.5 * m * (35 m/s)²And solve for m:
m = 816.33 kgAn exothermic reaction releases 325 kJ. How much energy is this in calories
An exothermic reaction releases 325 kJ. 1359.8 kJ energy is this in calories.
An exothermic reaction is a chemical reaction that releases heat energy into the surroundings. During an exothermic reaction, the products of the reaction have less potential energy than the reactants, and the excess energy is released in the form of heat.
One calorie is defined as the amount of energy required to raise the temperature of one gram of water by one degree Celsius.
One joule is defined as the amount of energy required to apply a force of one newton over a distance of one meter.
1 cal = 4.184 J
1 cal = 0.004184 kJ
325000 cal = x kJ
0.004184/ 1 = x / 325000
x = 1359.8 kJ
Thus, 1359.8 kJ energy is this in calories.
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Proton, Neutron and Electron of 23 13 Aluminum
The atomic mass of Aluminum is 23, which means it has a total of 23 particles in its nucleus, including protons and neutrons.
Aluminum has an atomic number of 13, which means it has 13 protons in its nucleus.
To find the number of neutrons, we subtract the atomic number from the atomic mass. So, Aluminum has 23 - 13 = 10 neutrons in its nucleus. Electrons are the negatively charged particles that orbit around the nucleus of an atom.
Aluminum, being a neutral atom, has an equal number of electrons to the number of protons in its nucleus, which is 13. These electrons are distributed in different energy levels or shells around the nucleus.
Aluminum is a widely used metal in different applications due to its unique properties such as low density, high strength, and resistance to corrosion. It is used in the manufacturing of cans, foils, and aircraft parts. The number of protons and electrons determines the atomic number and chemical properties of an element. The number of neutrons affects the stability and isotopes of an element.
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how to name Type 2 ionic compounds. AuCl3
To name Type 2 ionic compounds such as AuCl₃, you need to use the Stock system or Roman numeral system to indicate the oxidation state of the cation. Some steps are; Identify the cation, Determine the charge, Write the name, and combine two names.
Here are the steps to name AuCl₃; Identify the cation and anion. In this case, the cation is Au³⁺ and the anion is Cl⁻.
Determine the charge on the cation by using the anion's charge and balancing the charges to zero. Since Cl⁻ has a charge of -1 and there are three Cl⁻ ions in the compound, the total negative charge is -3. Therefore, the Au³⁺ ion has a charge of +3.
Write the name of the cation first, followed by the name of the anion with an -ide ending. Since the cation is Au³⁺, we use the name "gold(III)" to indicate the oxidation state of +3. The anion is Cl⁻, so we add the -ide ending to get "chloride".
Combine the two names to get the compound's name: "gold(III) chloride".
Therefore, the name of the Type 2 ionic compound AuCl₃ is "gold(III) chloride".
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An ideal gaseous reaction (which is a hypothetical gaseous reaction that conforms to the laws governing gas behavior) occurs at a constant pressure of 35.0 atm and releases 71.8 kJ of heat. Before the reaction, the volume of the system was 7.00 L
After the reaction, the volume of the system was 2.60 L Calculate the total internal energy change, ΔE , in kilojoules.
Express your answer with the appropriate units.
When, the volume of the system was 2.60 L. Then, the total internal energy change is -71.8 kJ.
We can use the first law of thermodynamics to find the total internal energy change;
[tex]Δ_{E}[/tex] = q + w
where q is the heat transferred to or from the system and w is the work done on or by the system. At constant pressure, work done is given by;
w = -P[tex]Δ_{V}[/tex]
where P is pressure and [tex]Δ_{V}[/tex] is change in volume.
Using the given values, we have:
q = -71.8 kJ (since heat is released)
P = 35.0 atm = 3.56×10⁶ Pa (using the conversion factor 1 atm = 101325 Pa)
[tex]Δ_{V}[/tex] = 7.00 L - 2.60 L = 4.40 L = 4.40×10⁻³ m³ (using the conversion factor 1 L = 10⁻³ m³)
Therefore,
w = -P[tex]Δ_{V}[/tex]= -(3.56×10⁶ Pa)(4.40×10⁻³ m³) = -15.7 J
= -1.57×10⁻² kJ
Thus, the total internal energy change is;
[tex]Δ_{E}[/tex] = q + w = (-71.8 kJ) + (-1.57×10⁻² kJ)
= -71.8 kJ - 1.57×10⁻² kJ
= -71.8 kJ
Therefore, the total internal energy change is -71.8 kJ.
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if 3 moes of cl reacts with 3 moles oxygen, then which substance is the limitting reactant and excess reactant
If 3 moles of cl reacts with 3 moles oxygen, there is no limiting reactant or excess reactant because the reactants are in stoichiometric proportions.
To determine the limiting reactant and excess reactant, we need to compare the stoichiometry of the reaction to the given amounts of each reactant.
The balanced chemical equation for the reaction between chlorine (Cl2) and oxygen (O2) can be represented as follows:
2Cl2 + O2 → 2Cl2O
According to the balanced equation, it requires 2 moles of chlorine (Cl2) to react with 1 mole of oxygen (O2) to produce 2 moles of chlorine oxide (Cl2O).
Given that we have 3 moles of chlorine (Cl2) and 3 moles of oxygen (O2), we can determine the limiting reactant by comparing the ratio of moles between the two reactants.
The ratio of Cl2 to O2 required for complete reaction is 2:1. However, since we have equal amounts of Cl2 and O2 (both 3 moles), neither reactant is present in excess.
Therefore, in this scenario, there is no limiting reactant or excess reactant because the reactants are in stoichiometric proportions. All of the chlorine and oxygen will be consumed in the reaction, resulting in the complete conversion to chlorine oxide (Cl2O).
It's important to note that if the amounts of Cl2 and O2 were different, the reactant present in lesser quantity would be the limiting reactant, and the reactant in greater quantity would be the excess reactant.
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9. Which of the following gas laws is calculated with the pressure and
volume variables at a constant temperature?
Formula
4 points
P₁V₁ = P₂V₂
P₁ = first pressure
P2 = second pressure
V₁ = first volume
The gas law that is calculated with the pressure and volume variables at a constant temperature is Boyle's Law. Boyle's Law states that the pressure (P) of a gas is inversely proportional to its volume (V) when temperature (T) is held constant.
Mathematically, it is expressed as P₁V₁ = P₂V₂, where P₁ and V₁ represent the initial pressure and volume, and P₂ and V₂ represent the final pressure and volume.According to Boyle's Law, if the volume of a gas is reduced while keeping the temperature constant, the pressure will increase proportionally.
Similarly, if the volume is increased, the pressure will decrease. This relationship holds as long as the temperature remains constant throughout the process. Boyle's Law is one of the fundamental gas laws and provides insights into the behavior of gases under changing pressure and volume conditions at a constant temperature.
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The bulbs used for fluorescent lights have a mercury gas pressure of 1.06 Pa at 40.°C. How many milligrams of liquid mercury must evaporate at 40.°C to yield this pressure in a 1.62-L fluorescent bulb?
____mg
The ideal gas law can be used to determine the pressure of mercury gas inside a fluorescent bulb:
PV = nRT
where
R is the gas constant,
n is the number of moles,
P is the pressure, V is the volume, and T is the temperature in Kelvin.
As we solve for n, we get:
n = pv / rt
Given the low pressure and moderate temperature, we can assume that mercury gas behaves better.
Therefore, the amount of mercury gas in the bulb is expressed as:
n = (1.06 Pa)(1.62 L)/(8.31 J/K/mol)(40 + 273.15 K) = [tex]7.71 * 10^-^4[/tex] mol
The molar mass of mercury can be used to determine how much liquid mercury is needed to produce this amount of gas:
m = nM
where M is the molar mass of mercury and m is its mass in mass units.
M = 200.59 g/mol
m = (7.71 x 10^-4 mol)(200.59 g/mol) = 0.154 g
Therefore, to achieve a pressure of 1.06 Pa, 154 mg of liquid mercury must evaporate at a temperature of 40 °C for a 1.62-L fluorescent light bulb.
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Energy
4p
3d
4s
3p
3s
2p
2s
1s
Answer:
Energy
4p ⇵ ⇵ ⇵
3d ⇵ ⇵ ⇵ ⇵ ⇵
4s ⇵
3p ⇵ ⇵ ⇵
3s ⇵
2p ⇵ ⇵ ⇵
2s ⇵
1s ⇵
Which best explains why an Al 3+ ion is smaller than an Al atom?
In forming the Al³+ ion, the Al atom loses the electrons in its outermost energy
level, causing a decrease in the atomic radius.
In forming the Al3+ ion, the Al atom gains three protons and the resulting net
positive charge keeps the electrons more strongly attracted to the nucleus,
reducing the radius.
The Al3+ ion contains more electrons than the Al atom, which results in a greater
attraction for the nucleus and a smaller atomic radius.
In forming the A13+ ion, the Al atom adds electrons into a higher energy level,
causing a decrease in the atomic radius.
There are more protons in an Al3+ ion than there are in an Al atom.
In forming the Al³⁺ ion, the Al atom loses the electrons in its outermost energy level, causing a decrease in the atomic radius, hence option A is correct.
The number of protons in the nucleus of AlandAl3+ AlandAl3+ is the same, however there are differing numbers of electrons in the final shell. Al³⁺ is smaller than Al because it has fewer electrons.
The Al atom will become an Al³⁺ ion when it loses its third electron and develops a tri-positive charge on it. In forming the Al³⁺ ion, the Al atom loses the electrons in its outermost energy level, causing a decrease in the atomic radius.
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an isomer of C3H7O undergoes one step oxidation reaction. Answer the following questions due to this reaction.
a) Write a full symbol equation for this reaction b) Name the proper reagent and catalyst for this reaction.
c) Why do you think there is no need to remove the product from the reaction vessel?
The specific equation depends on the isomer and the oxidizing agent used. An example of a general oxidation reaction could be:
C₃H₇OH + [O] → C₃H₆O + H₂O
Common oxidizing agents for organic compounds include potassium permanganate (KMnO₄), potassium dichromate (K₂Cr₂O₇), or hydrogen peroxide (H₂O₂).
Whether or not the product needs to be removed from the reaction vessel depends on the specific reaction and its desired outcome. In some cases, the product may be of interest for further reactions or analysis, and therefore, it would be retained in the reaction vessel.
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7. A single molecule of nitrogen (N₂) will have a velocity of 400.0m/s.
(a) What equation & unit conversions will you use to calculate the
kinetic energy of one molecule?
(b) What is the kinetic energy of one molecule of nitrogen?
The kinetic energy of a single molecule of nitrogen (N₂) with a velocity of 400.0 m/s is approximately 3.72 x 10^-21 joules.
To calculate the kinetic energy of a single molecule of nitrogen (N₂) with a velocity of 400.0 m/s, we can use the equation for kinetic energy: KE = 1/2 * m * v², where KE represents the kinetic energy, m is the mass of the molecule, and v is the velocity.
(a) Unit conversions:
The mass of a nitrogen molecule (N₂) is approximately 28 atomic mass units (u) or 4.65 x 10^-26 kilograms (kg).
The given velocity is already in meters per second (m/s), so no further conversion is needed.
(b) Calculating the kinetic energy:
Plugging the values into the kinetic energy equation, we have:
KE = 1/2 * (4.65 x 10^-26 kg) * (400.0 m/s)²
Evaluating the expression:
KE = 1/2 * 4.65 x 10^-26 kg * 160000 m²/s²
= 3.72 x 10^-21 joules (J)
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Hi
Please help on question asap if the answer is correct I'll rate you five stars a thanks and maybe even brainliest!
What do you notice about the values?
0.9 amps ÷0.03v=30 ohms.
1.9amps÷0.07v=27.149 ohms
3.1. amps ÷0.10v =31 ohms
3.9. amps ÷ 0.12v =032.5 ohms
5. amps ÷0.15v=33.33 ohms
6.1. amps ÷0.19v=32.1053 ohms
Answer: The values you provided show that as the current (measured in amps) and voltage (measured in volts) increase, the resistance (measured in ohms) remains relatively constant.
Looking at the values, it appears that the current (in amps) divided by the voltage (in volts) yields resistance values (in ohms) that are relatively close to each other. The calculated resistance values range from approximately 27 ohms to 33 ohms, with some variation in between.
Write the chemical formula for this molecule from the picture.
I understand there are 4 Hydrogens
2 Carbons, and 1 Sulfur, but I have no clue how to format it
Answer:
[tex]C_{2}H_{2}S[/tex]
What is the relationship between an elements position and it’s atomic mass in the periodic table
Answer:
The chemical elements are arranged from left to right and top to bottom in order of increasing atomic number, or the number of protons in an atom's nucleus.
Explanation:
A rectangular ingot of gold is 23.7 cm long by 75.5 mm wide by 10.9 cm high. If 1.0 cm³ of gold weighs 19.30 g, what is the price of the ingot in SA Rand if the current price of gold is $1629.8 per ounce. (1oz = 28.35g Exchange rate = 9.6 R/$).
The price of the ingot in South African Rand (SAR) is 19,533,171.507 R.
To calculate the price of the ingot in South African Rand (SAR), we need to follow these steps:
Convert the dimensions of the ingot to cm³:
Length = 23.7 cm
Width = 75.5 mm = 7.55 cm
Height = 10.9 cm
The volume of the ingot is calculated by multiplying these dimensions:
Volume = Length × Width × Height
= 23.7 cm × 7.55 cm × 10.9 cm
= 1830.0475 cm³
Calculate the weight of the ingot in grams:
Since 1 cm³ of gold weighs 19.30 g, we can multiply the volume of the ingot by this conversion factor to obtain the weight:
Weight = Volume × 19.30 g
= 1830.0475 cm³ × 19.30 g
= 35,380.1375 g
Convert the weight of the ingot to ounces:
Since 1 ounce is equal to 28.35 g, we can divide the weight of the ingot by this conversion factor:
Weight in ounces = Weight / 28.35 g
= 35,380.1375 g / 28.35 g
= 1247.0461 ounces
Calculate the price of the ingot in USD:
The current price of gold is $1629.8 per ounce, so we can multiply the weight of the ingot in ounces by this price:
Price in USD = Weight in ounces × Price per ounce
= 1247.0461 ounces × $1629.8/ounce
= $2,032,881.72278
Convert the price from USD to SAR:
The exchange rate is 9.6 R/$, so we can multiply the price in USD by this exchange rate:
Price in ZAR = Price in USD × Exchange rate = $2,032,881.72278 × 9.6 R/$ = 19,533,171.507 R
Therefore, the price of the ingot in South African Rand (SAR) is approximately 19,533,171.507 R.
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the relative formula masses (Mr) are: CaCo3 = 100; CaO =56 ; Co2=44
describe how this experiment could be used to provide evidence for the law of conservation of mass.
[6 marks]
include your answer:
-method
-which measurements should eb taken
-how the student could show evidence for the conservation for mass
The law of conservation of mass states that in a chemical reaction, the total mass of the reactants is equal to the total mass of the products. To provide evidence for this law, we can perform an experiment in which calcium carbonate ([tex]CaCO_3[/tex]) is decomposed to produce calcium oxide (CaO) and carbon dioxide ([tex]CO_2[/tex] ), and then measure the masses of the reactants and products.
Method:
Weigh a sample of [tex]CaCO_3[/tex] using a balance.
Heat the [tex]CaCO_3[/tex] in a crucible until it decomposes to CaO and [tex]CO_2[/tex]. The [tex]CO_2[/tex] gas will escape, leaving only CaO in the crucible.
Allow the crucible to cool and then weigh it again to determine the mass of the CaO produced.
Collect the [tex]CO_2[/tex] gas that is released during the reaction in a gas syringe or other collection device. Measure the volume of [tex]CO_2[/tex] gas produced, and calculate its mass using its molecular weight.
Which measurements should be taken:
The following measurements should be taken:
The mass of the [tex]CaCO_3[/tex] used as a reactant.
The mass of the CaO produced as a product.
The volume of [tex]CO_2[/tex] gas produced during the reaction.
The temperature and pressure of the [tex]CO_2[/tex] gas to allow for the calculation of its mass.
How the student could show evidence for the conservation of mass:
To show evidence for the law of conservation of mass, the student can compare the mass of the [tex]CaCO_3[/tex] used as a reactant to the total mass of the products, which includes the mass of CaO produced and the mass of [tex]CO_2[/tex] gas released.
The sum of the masses of CaO and [tex]CO_2[/tex] should be equal to the mass of the [tex]CaCO_3[/tex] used as a reactant, within experimental error. This will provide evidence that the mass of the reactants is conserved and equals the mass of the products, as required by the law of conservation of mass.
Additionally, the student could calculate the theoretical yield of CaO and CO2 based on the balanced equation for the reaction, and compare this to the actual yield obtained from the experiment. Any difference between the theoretical and actual yields could be due to experimental error, but the comparison can still provide additional evidence for the conservation of mass.
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in the quantum model of the atom , what does the shape of an atomic orbital represent ?
Answer:
In a quantum model of the atom, the shape of an electron orbital represents the probability distribution of finding an electron in a particular region of space around the nucleus of an atom. In other words, an electron orbital is a three-dimensional space around the nucleus where there is a high probability of finding an electron with a given energy level.
Different orbitals can have different shapes, such as spherical, hourglass-shaped, or more complex shapes.
Electron Orientations:
s = 1 orbital orientation
p = 3 orientations ([tex]6e^-[/tex])
d = 5 orientations ([tex]10e^-[/tex])
f = 7 orientations ([tex]14e^-[/tex])
Will you answer this for me ?
Answer:
Explanation:
balance
2 C2H6 + 7 O2 --> 4 CO2 + 6 H2O
given 360 g H20 (g)
required =586.67 g CO2
360 g H20 x (1mole/18 g H20) X (4 mole CO2/6 moles H20) X (44g CO2/1mole) =586.67 g CO2
Determine the [H+] , [OH−], and pOH of a solution with a pH of 7.41
at 25 °C. [H+]=
M
[OH−]=
M
pOH=
Answer:
Explanation:
H+ = 1 X 10^-7.41 = 3.89 X 10^ -8
POH = 14-7.41 = 6.59
OH- = 1 x 10 ^-6.59 = 2.57 X 10^ -7
The [H+] and [OH−] concentrations of the solution are approximately 2.38 × 10^(-7) M, and the pOH is 6.59.
The pH of a solution is a measure of the concentration of hydrogen ions ([H+]) in the solution. The pH scale ranges from 0 to 14, with a pH of 7 considered neutral. A pH of 7.41 indicates that the solution is slightly basic. To calculate the [H+], [OH−], and pOH of the solution, we can use the relationship:
pH + pOH = 14
Given that the pH is 7.41, we can subtract it from 14 to find the pOH:
pOH = 14 - 7.41 = 6.59
Since pH + pOH = 14, we can also determine the [OH−] by taking the antilogarithm of the pOH value:
[OH−] = 10^(-pOH)
[OH−] = 10^(-6.59)
[OH−] ≈ 2.38 × 10^(-7) M
Since the solution is neutral, the concentration of [H+] will be equal to the concentration of [OH−]:
[H+] = [OH−] ≈ 2.38 × 10^(-7) M
Therefore, the [H+] and [OH−] concentrations of the solution are approximately 2.38 × 10^(-7) M, and the pOH is 6.59.
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Will you answer this ?
The mass (in grams) of carbon dioxide, CO₂ produced, given that 360 g of water vapor was produced from the reaction is 586.67 g
How do i determine the mass of carbon dioxide, CO₂ produced?First, we shall obtain the mass of C₂H₆ that reacted to produce 360 g of water vapor, H₂O. This is obtain as follow:
2C₂H₆ + 7O₂ -> 4CO₂ + 6H₂O
Molar mass of H₂O = 18 g/molMass of H₂O from the balanced equation = 6 × 18 = 108 g Molar mass of C₂H₆ = 30 g/molMass of C₂H₆ from the balanced equation = 2 × 30 = 60 gFrom the balanced equation above,
108 g of H₂O were obtained from 60 g of C₂H₆
Therefore,
360 g of H₂O will be obtain from = (360 × 60) / 108 = 200 g of C₂H₆
Thus, we can mass of C₂H₆ that reacted is 200 g
Finally, we shall obtain the mass of carbon dioxide, CO₂ produced. This is shown below:
2C₂H₆ + 7O₂ -> 4CO₂ + 6H₂O
Molar mass of C₂H₆ = 30 g/molMass of C₂H₆ from the balanced equation = 2 × 30 = 60 gMolar mass of CO₂ = 44 g/molMass of CO₂ from the balanced equation = 4 × 44 = 176 gFrom the balanced equation above,
60 g of C₂H₆ reacted to produce 176 g of CO₂
Therefore,
200 g of C₂H₆ will react to produce = (200 × 176) / 60 = 586.67 g of CO₂
Thus, the mass of carbon dioxide, CO₂ produced is 586.67 g
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1. Draw up schemes for the formation
of bonds between the atoms of the following
elements:
C and P; So; Mg u Si
2.
What kind of bond and type
of crystal
lattice
do you
follow me:
CaO, C, SiO2, Fe, K3N
Assume their physical
properties.
3. Specify which process
is depicted by the following diagram
(oxidation or reduction) and
make an electronic balance
corresponding to this scheme:
a) Cu0 -, Cu+2
b) S0
- S-2
B) Fe+3
Fe0
4. Make
up the redox reactions and
arrange the coefficients
by the electronic balance method:
A) H2O + CO2 - HCL +O2
b) Fe203 + H2 - Fe + H20
b) H2SO4 + S - SO2 + H2O
Schemes for the formation of bonds:
C and P; C + P → CPS; S + S → S₈Mg and Si: Mg + Si → Mg₂SiHow to setup schemes and bonds?The schemes for the formation of bonds between the atoms of the following elements are:
Carbon and phosphorus:
C + P → CP
This is an example of a covalent bond, which is a type of bond that is formed by the sharing of electrons. In this case, the carbon atom shares one electron with the phosphorus atom, forming a single covalent bond.
Sulfur:
S + S → S₈
This is an example of a molecular bond, which is a type of bond that is formed by the sharing of electrons between multiple atoms of the same element. In this case, the sulfur atoms share two electrons each, forming a double bond.
Magnesium and silicon:
Mg + Si → Mg₂Si
This is an example of an ionic bond, which is a type of bond that is formed by the transfer of electrons from one atom to another. In this case, the magnesium atom gives up two electrons to the silicon atom, forming a magnesium ion with a charge of +2 and a silicon ion with a charge of -4. These ions are then attracted to each other by the opposite charges.
2. The types of bonds and crystal lattices for the following elements:
CaO: ionic bond, ionic lattice
C: covalent bond, diamond lattice
SiO₂: covalent bond, tetrahedral lattice
Fe: metallic bond, body-centered cubic lattice
K₃N: ionic bond, cubic lattice
3. The processes depicted by the following diagrams, along with the corresponding electronic balances:
Cu0 → Cu⁺²: oxidation
Cu0 → Cu⁺² + 2e⁻
S0⁻ → S⁻²: reduction
S0⁻- + 2e⁻ → S⁻²
Fe⁺³ → Fe⁺²: reduction
Fe⁺³ + 1e⁻ → Fe⁺²
4. The redox reactions and the coefficients arranged by the electronic balance method:
H₂O + CO₂ → HCL + O₂
2H⁺ + ¹/₂O₂ → HCL
2e⁻ + 2H⁺ → H₂
¹/₂O₂ + 2e⁻ → O₂⁻
Fe₂O₃ + H₂ → Fe + H₂O
Fe₂O₃ + 3H₂ → 2Fe + 3H₂O
3Fe⁺³ + 6e⁻ + 6H⁺ → 2Fe + 6H₂O
2O₂⁻ + 6H⁺ → 4H₂O
H₂SO₄ + S → SO₂ + H₂O
2H⁺ + SO₄²⁻ → SO₂ + H₂O
2e⁻ + 2H+ → H₂
S → S²⁻ + 2e⁻
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Question 4 of 10
In what way does the shape of a molecule affect how the molecule is
involved with living systems?
OA. It determines what elements are in the molecule.
OB. It determines oxidation states present in the molecule.
OC. It determines how the molecule functions.
OD. It determines the weight of the molecule.
SUBMIT
Complete the following table; If the pressure of the gas is 250 mmHg, Volume is
34mL and the temperature is at 25°C.
Answer:
Explanation:
constant = 33.43 ml charles law; formula is v1/t1=v2/t2
129.77 mmhg= boyles law; = p1v1=p2v2
659.51 K ; gay-lussac law; P1/T1=P2/T2
20.79 mL; combined gas law; p1v1/t1=p2v2/t2
90.67 mmHg; combined gas law; P1V1/T1=P2V2/T2
How many How many molecules are there in 265 grams of FeF^3?
a. 1.1 x 1023
b. 1.3 x 1023
c. 1.4 x 1024
d. 2.8 x1024
2. How many molecules are there in 98 grams of FeF^3?
a. 1.4 x 1023
b. 5.2 x 1023
c. 1.4 x 1024
d. 5.2 x 1024
3. How many atoms are there in 6.2 grams of silver?
a. 1.2 x 1022
b. 3.5 x1022
c. 1.2 x1023
d. 3.5 x 1023
4. How many atoms are there in 54.2 grams of Manganese?
a. 5.9 x 1023
b. 1.3 x 1024
c. 5.9 x1024
d. 1.3 x 1025
. How many molecules are there in 250 grams of Cu(NO3)2?
a. 8.0 x 1022
b. 1.4 x 1023
c. 8.0 x 1023
d. 1.4 x 1024
How many grams are in 6.2 × 1024 molecules of water?
a. 16.5 grams
b. 18.5 grams
c. 165.3 grams
d. 185.3 grams
7. How many grams are in 2.3 x 1023 molecules of NO3?
a. 23.7 grams
b. 46. 2 grams
c. 237 grams d.
d. 462 grams
8. How many grams are in 9.7 x 1023 atoms of selenium?
a. 12.7 grams
b. 62.3 grams
c. 127.3 grams
d. 623 grams
9. How many grams are in 3.4 x 1024 atoms of osmium?
a. 52.2 grams
b. 107.4 grams
c. 203.7 grams
d. 410. 1 grams
10. How many grams are in 11 x 1022 molecules of oxygen?
a. 2.9 grams
b. 5.8 grams
c. 0.3 grams
d. 0.6 grams
The masses, moles, and number of molecules are as follows:
1. c. 1.4 x 10²⁴
2. b. 4.214 x 10²³ molecules
3. a. 1.2 x 10²²
4. a. 5.9 x 10²³
5. c. 8.0 x 10²³
6. d. 185.3 grams.
7. a. 23.7 grams
8. c. 127.3 grams
9. No answer in the option
10. a. 2.9 grams
What is the number of molecules?To determine the number of molecules in 265 grams of FeF₃:
First, calculate the number of moles of FeF₃:
moles = mass / molar mass
moles = 265 g / 139.839 g/mol
moles = 1.8939 mol
number of molecules = moles * Avogadro's number
number of molecules = 1.8939 mol * 6.022 x 10²³ molecules/mol
number of molecules = 1.138 x 10²⁴ molecules
2. To determine the number of molecules in 98 grams of FeF₃:
moles = 98 g / 139.839 g/mol = 0.7001 mol
number of molecules = 0.7001 mol * 6.022 x 10²³ molecules/mol
number of molecules = 4.214 x 10²³ molecules
3. To determine the number of atoms in 6.2 grams of silver:
moles = 6.2 g / 107.8682 g/mol = 0.0574 mol
number of atoms = 0.0574 mol * 6.022 x 10²³ atoms/mol
number of atoms = 3.457 x 10²² atoms
4. To determine the number of atoms in 54.2 grams of Manganese:
moles = 54.2 g / 54.938045 g/mol = 0.9876 mol
number of atoms = 0.9876 mol * 6.022 x 10²³ atoms/mol
number of atoms = 5.947 x 10²³ atoms
5. To determine the number of molecules in 250 grams of Cu(NO₃)₂:
moles = 250 g / (63.546 g/mol + 2 * 14.007 g/mol + 6 * 16.00 g/mol) = 250 g / 187.56 g/mol = 1.333 mol
number of molecules = 1.333 mol * 6.022 x 10^23 molecules/mol
number of molecules = 8.027 x 10^23 molecules
6. To determine the mass in grams of 6.2 x 10²⁴ molecules of water:
moles = (6.2 x 10²⁴ molecules) / (6.022 x 10²³ molecules/mol)
moles = 10.29 mol
mass = moles * molar mass
mass = 10.29 mol * 18.00 g/mol)
mass = 185.3 grams
7. To determine the mass in grams of 2.3 x 10²³ molecules of NO₃:
moles = (2.3 x 10²³ molecules) / (6.022 x 10²³ molecules/mol)
moles = 0.382 mol
mass = moles * molar mass
mass = 0.382 mol * (14.007 g/mol + 3 * 16.00 g/mol) = 23.7 grams
8. To determine the mass in grams of 9.7 x 10²³ atoms of selenium:
moles = (9.7 x 10²³ atoms) / (6.022 x 10²³ atoms/mol)
moles = 1.61 mol
mass = moles * molar mass
mass = 1.61 mol * 78.9718 g/mol = 127.3 grams
9. To determine the mass in grams of 3.4 x 10²⁴ atoms of osmium:
moles = (3.4 x 10²⁴ atoms) / (6.022 x 10²³ atoms/mol)
moles = 5.64 mol
mass = moles * molar mass
mass = 5.64 mol * 190.23 g/mol
mass = 1074.8 grams
10. To determine the mass in grams of 11 x 10²² molecules of oxygen:
moles = (11 x 10^22 molecules) / (6.022 x 10²³ molecules/mol)
moles = 0.182 mol
mass = moles * molar mass
mass = 0.182 mol * 16.00 g/mol
mass = 2.9 grams
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Suppose you were doing a titration where you start out with a basic solution of around 8.0 and you expect to keep adding an acid until the mixture has a pH of 3.0. Based on the indicator chart which pH indicator would be the best one to use. Describe the color change that would be observed
Based on the indicator chart, the pH indicator that would be the best one to use would be Thymol Blue.
The color change would be from light blue to yellow to orange.
Why is this pH indicator best ?Thymol blue would be best because it would show you where your starting point is and then when you reach the desired pH value of 3. 0. Looking at the indicator chart, Thymol blue has a color of light blue between 8. 0 and 9. 0 so you will know you are at 8. 0 when the reaction starts.
As you add more acid, the color would move to yellow to let you know that it is getting more acidic. Once it reaches orange, the titration should stop.
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What is the energy associated with the formation of 2.55 g of 4He by the fusion of 3H and 1H?
Substance Mass (u)
4He 4.00260
3H 3.01605
1H 1.00783
The energy associated with the formation of 2.55 g of 4He by the fusion of 3H and 1H is -2.982 x 10⁻¹⁰ J.
The given masses of the isotopes can be converted to kilograms using the conversion factor: 1 u = 1.661 x 10⁻²⁷ kg.
Mass of 4He = 2.55 g = 2.55 x 10⁻³ kg
Mass of 3H = 3.01605 u = 3.01605 x 1.661 x 10⁻²⁷ kg/u
= 5.0099 x 10⁻²⁷ kg
Mass of 1H = 1.00783 u = 1.00783 x 1.661 x 10⁻²⁷ kg/u
= 1.6737 x 10⁻²⁷ kg
The balanced equation for the fusion reaction is;
3H + 1H → 4He
The molar mass of 4He is 4.0026 g/mol, which can be converted to kg/mol using the conversion factor: 1 g/mol = 1 x 10⁻³ kg/mol.
Molar mass of 4He = 4.0026 g/mol = 4.0026 x 10⁻³ kg/mol
The number of moles of 4He formed can be calculated from its mass;
n(4He) = m(4He) / M(4He)
= 2.55 x 10⁻³ kg / 4.0026 x 10⁻³ kg/mol
= 0.638 mol
From the balanced equation, 3 moles of H atoms react with 1 mole of He atoms to form 1 mole of He atoms. Therefore, the number of moles of H atoms required for the reaction is;
n(H) = 3/4 x n(4He)
= 3/4 x 0.638 mol
= 0.479 mol
The energy released in the reaction can be calculated using the mass-energy equivalence equation;
E = Δm c²
where Δm is change in mass, c is the speed of light.
The change in mass is;
Δm = [3H + 1H - 4He] = [5.0099 x 10⁻²⁷ kg + 1.6737 x 10⁻²⁷kg - 4.0026 x 10⁻³ kg]
= -3.315 x 10⁻²⁷ kg (negative because mass is lost in the reaction)
The energy released is;
E = (-3.315 x 10⁻²⁷ kg) c²
= (-3.315 x 10⁻²⁷ kg) (2.998 x 10⁸ m/s)²
= -2.982 x 10⁻¹⁰ J
The negative sign indicates that energy is released in the reaction (exothermic reaction).
Therefore, the energy associated is -2.982 x 10⁻¹⁰ J.
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How many moles of oxygen would be consumed to produce 68.1g water?
Answer:
To determine the number of moles of oxygen consumed to produce 68.1g of water, we need to know the amount of hydrogen in the water. Since water has two hydrogen atoms and one oxygen atom, the molar mass of water is 18.015 g/mol.
Using the molar mass of water, we can calculate the number of moles of water produced:
68.1 g ÷ 18.015 g/mol = 3.78 mol
Since there are two hydrogen atoms in each molecule of water, there must be twice as many moles of hydrogen as there are moles of water:
3.78 mol × 2 = 7.56 mol of hydrogen
Finally, we can use the balanced chemical equation for the reaction of hydrogen and oxygen to form water to determine the number of moles of oxygen consumed:
2H2 + O2 -> 2H2O
For every 1 mole of oxygen consumed, 2 moles of water are produced. Therefore, the number of moles of oxygen consumed is:
7.56 mol H2O ÷ 2 = 3.78 mol O2
So, 3.78 moles of oxygen would be consumed to produce 68.1g of water.
I need help solving this problem can anybody help thank you.
The system will react by changing the concentrations to restore equilibrium when the CO and CO2 concentrations in the chemical equation 2CO + O2 2CO2 are raised. The system will try to reduce the rise in CO and CO2 by moving the reaction towards the products side, in accordance with Le Chatelier's concept.
The reaction will go forward as the CO concentration rises, eating part of the extra CO and turning it into CO2. This change lowers the excess CO concentration and aids in reestablishing equilibrium.
However, since CO2 is already a product, its concentration does not have a direct impact on the reaction. However, to keep the stoichiometric balance, it can result in a somewhat greater concentration of CO.
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(b)
AU
(c)
MAT
w song will it take to deposit 10.47 grams of copper?
4) A voltaic cell consists of a copper electrode in a solution of copper(II) ions
and a palladium electrode in a solution of palladium(II) ions. The palladiu
is the cathode and its reduction potential is 0.951 V.
(a)
Write the half-reaction that occurs at the anode.
If E° is 0.609 V, what is the potential for the oxidation half-reaction
What is Keq for this reaction?
* LEA
a) The half-reaction that occurs at the anode is;
Cu(s) ----> Cu^2+(aq) + 2e
b) The Keq of the reaction is 3.75 * 10^20
What is the electrode potential?Electrode potential is a measure of the tendency of an electrode to gain or lose electrons when it is in contact with a solution containing ions.
For the oxidation half reaction;
E° = Ecathode - Eanode
Eanode = Ecathode - E°
Eanode = 0.951 - 0.609
= 0.342 V
Note that;
E° = 0.0592/nlogKeq
logKeq = E° * n/ 0.0592
= 0.609 * 2/0.0592
Keq =3.75 * 10^20
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A 25 L sample of oxygen gas (O2) has a mass of 48 grams and a pressure of 3.0 atm. What would be the temperature of the sample? Reminder: Use the equation PV=nRT, with the constant R = 0.0821 L atm/mol K.
A.
609 K
B.
305 K
C.
19.0 K
D.
1.60 x 10-2 K
The temperature of the oxygen gas sample is 609 K, which is approximately 336°C or 637°F. The answer is A.
We can use the ideal gas law equation, PV = nRT, to solve for the temperature of the oxygen gas sample.
First, we need to calculate the number of moles of oxygen gas present in the sample using its mass and molar mass:
n = m/M
where:
n = number of moles
m = mass (in grams)
M = molar mass (in g/mol)
The molar mass of oxygen gas (O2) is 32.00 g/mol.
n = 48 g / 32.00 g/mol = 1.50 mol
Next, we can rearrange the ideal gas law equation to solve for temperature (T):
T = (PV) / (nR)
where:
T = temperature (in Kelvin)
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = gas constant (0.0821 L atm/mol K)
Plugging in the given values, we get:
T = (3.0 atm x 25 L) / (1.50 mol x 0.0821 L atm/mol K)
T = 609 K
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