Consider the reaction below. 2al2o3 → 4al 3o2 what is the mole ratio of oxygen to aluminum? o2:al = :4

The mechanical advantages of the machine are 3.

The amount that a force is boosted by employing a machine is known as the mechanical advantage. It is determined by dividing the applied force by the machine's force ( force exerted by the machine).

That is the mechanical advantage of any machine is equal to the ratio between the output force and the input force.

Mechanical advantages M.A. = Input force / Output force

We have,

Input force: 52 N

Output force: 156 N

Therefore,

Mechanical advantage = Input force / Output force.

52/156 = 3

Hence, the mechanical advantages of the machine are 3.

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The Protective Equipment is a Face Shield.

Hydrochloric acid is a dangerous chemical, it can be handled with proper care. As we know Hydrochloric acid is corrosive. If acid touches the skin, eyes, and internal organs then the damage can not be treated, or even in some cases it is fatal.

If hydrochloric acid is used in the workplace, then there is a need for access to an eye-flush station.

If acid is used at home, then need to follow some safety instructions. If acid is diluted with water, still there is a need for protection. For example: wear protective clothes and gloves.

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Answer:

it would do wonders for the health of those using them and, by extracting energy from the fuel more efficiently, would be more environmentally sustainable too.

Answer:

R = 8.314 pKa*L/mol*K

The value for the ideal gas constant (R) is approximately 8.314 kPa·L/(mol·K).

To determine the value for the ideal gas constant (R) when pressure (P) is in kilopascals (kPa), temperature (T) is in kelvins (K), volume (V) is in liters (L), and amount of gas (n) is in moles, we need to use the appropriate units for R based on these measurements.

The ideal gas constant, R, can be expressed in various units. The most common units for R are:

R = 0.0821 L·atm/(mol·K) (atmospheres, liters, moles, and kelvins)

However, since you provided the measurements in kilopascals, liters, moles, and kelvins, we need to use a different value for R that is consistent with these units:

R = 8.314 kPa·L/(mol·K)

Therefore, when pressure is in kilopascals, volume is in liters, amount of gas is in moles, and temperature is in kelvins, the value for the ideal gas constant (R) is approximately 8.314 kPa·L/(mol·K).

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The Ksp of the solution is 3.2 * 10^-8.

The Ksp is called the solubility product . We know it to be the equilibrium constant that shows the extent to which a substance is dissolved in solution.

Now given that;

CaF2(s) ⇔ Ca^2+(aq)  +  2F^-(aq)

Ksp = [ Ca^2+] [F^-]^2

Ksp = [0.002] [0.004]^2

Ksp = 3.2 * 10^-8

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The total atmospheric pressure of the final product in the flask is 18.36atm

Atmospheric pressure can be defined as the force exerted on a surface by the air above it as gravity pulls it to earth.

Furthermore, the standard atmosphere is a unit of pressure which is also defined as:

101,325 Pa =760mmHg

So therefore, the total pressure of the final product in the flask is 18.36atm

Complete question:

Gaseous iodine pentafluoride, IF₅, can be prepared by the reaction of solid iodine and gaseous fluorine:

I₂(s) + 5F₂(g) → 2IF₅(g)

A 10.00 L flask is charged with 19.0 grams of I₂ and 2.00 atm of F₂ at 25°C. The flask is heated to 100°C until one of the reagents is completely consumed. What will be the total pressure (in atm) of the final products in the flask at 100°C? (Assume ideal gas behavior)

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Answer:

Any base with an ability to dissolve in water are called alkalis. All alkalis are bases, however, not all bases are alkalis.

Answer:

b) metal ion indicator

Explanation:

The metal ion indicator is a type of volumetric analysis wherein the colored complex is used to determine the endpoint of the titration. Therefore, it can be used to detect the end-point of an EDTA titration.

It is a combination reaction.

            P + O₂ → P₂O₅

Combination reactions are those in which two or more reactants combine to create a single product.

A synthesis reaction is another name for a combination reaction. The overall form of the reaction is as follows:

                                        X + Y → XY

There can be a combination reaction between two elements, two compounds, or a compound and an element.

Examples:

H₂ + Cl₂ → 2HCl

P + O₂ → P₂O₅

C + O₂ → CO₂

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Since Qp>Kp , the reaction is not at equilibrium.

The equilibrium constant shows the extent to which reactants are converted into products.

Now we have to obtain the Qp as follows;

Qp =[CH3OH]/[CO] [H2]^2

Qp = 0.265/(0.265) (0.265)^2

Qp = 14.2

Now we know that Kp = 6.09×10−3, Since Qp>Kp , the reaction is not at equilibrium.

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Answer:

Iron ore, coke and lime stone are some raw materials used for extracting iron.

Hope it helps! :)

A precipitation reaction occurs when 50. 0 mL of 0. 50 M BaCl₂ is mixed with 75. 0 mL of 0. 75 M Na₂CO₃ (aq). The only precipitate is the BaCO₃ (s) formed. (a) The balanced chemical equation is BaCl₂ + Na₂CO₃ → BaCO₃ + 2NaCl. (b) The limiting reagent is Na₂CO₃.

The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.

Now we have to write the balanced equation

BaCl₂ + Na₂CO₃ → BaCO₃ + 2NaCl

To calculate number of moles when molarity and volume are given as:

Number of moles = Molarity × Volume

Now,

Number of moles of BaCl₂ = Molarity  × Volume

                                             = 0.50 M  × 0.05 L         [1 ml = 0.001 L]

                                             = 0.025 moles

Number of moles of Na₂CO₃ = Molarity  × Volume

                                                = 0.75 M × 0.075 L     [1 ml = 0.001 L]

                                                = 0.05625 moles

The limiting reagent is Na₂CO₃ here.

Thus from the above conclusion we can say that A precipitation reaction occurs when 50. 0 mL of 0. 50 M BaCl₂ is mixed with 75. 0 mL of 0. 75 M Na₂CO₃ (aq). The only precipitate is the BaCO₃ (s) formed. (a) The balanced chemical equation is BaCl₂ + Na₂CO₃ → BaCO₃ + 2NaCl. (b) The limiting reagent is Na₂CO₃.

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Question: A precipitation reaction occurs when 50. 0 mL of 0. 50 M BaCl₂ is mixed with 75. 0 mL of 0. 75 M Na₂CO₃ (aq). The only precipitate is the BaCO₃ (s) formed.

a) Write the balanced equation that describes this reaction.

b) Which chemical is the limiting reactant

Iron (iii) chloride is obtained by vapor condensation from the reaction between chlorine gas and iron fillings.

Iron (ii) chloride can be formed from iron fillings in the laboratory as follows:

Chlorine gas is introduced into a reaction vessel containing iron fillings and the iron (iii) chloride vapor formed is obtained by condensation.

In conclusion, iron (iii) chloride is formed by the the direct combination of iron fillings and chlorine gas.

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The order of the students from the most reliable to the least reliable is C. Student 4, Student 1, Student 2, Student 3.

the student with the most reliable data is the one that has the least variation in their data in the various trials.

this student is therefore student 4.

the next student would be student 1 and student 2 would be the third most reliable.

student 3 has large variations in their trials which makes their data the least reliable.

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If the concentration of fluoride anion and aluminum cation was increased to 5M, then there would be decrease in Ecell by 0.055 volts

The standard reduction potential of Fluoride anion and aluminum cation are -

2Al -----> 2Al3+ + 6e-           [tex]E^o[/tex]cell = +1.66

6e- + 3F2 -----> 6F-               [tex]E^o[/tex]cell = +2.87

The complete reaction is -

2Al + 3F2 ------> 2Al3+  +  6F-    [tex]E^o[/tex] = +4.53

Using Nernst Equation :-

E = [tex]E^o[/tex] – 0.0592/n*log[Al3+]^2[F-]^6

n = 6 (n = number of transferred electrons)

E = +4.53 - 0.0592/6*log(5)^2(5)^6

E = +4.53 - 0.00987*log(25)(1.56 x 10^4)

E = +4.53 – 0.00987*log(3.9 x 10^5)

E = + 4.53 -0.00987(5.59)

E = + 4.53 - 0.055

E = +4.47

The change in Ecell = 5 - 4.47 = 0.055V

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Answer:

See below.

Explanation:

Acceleration is the change in velocity over a period of time.  

Let vi be the initial velocity.

Let vf be the final velocity.

t is the time it took to go from vi to vf.

Acceleration is (vf-vi)/t

If the initial velocity was 2 cm/s and the final velocity was 10 cm/s, the object would have incresed it's velocity by (10 - 2) or 8 cm/s.  If iit took 10 seconds to reach vf, the acceleration would be:

(8 cm/s)/(10 s) = 0.8 cm/s^2

Magnesium and calcium have similar chemical properties because they have the same number of electrons in their valence shells.

The valence/outermost shell of magnesium and calcium both has two electrons. An element's chemical characteristics are dictated by how its electrons are arranged in their orbits around the nucleus. The same number of protons and electrons means that an element's chemical behavior is fundamentally determined by its atomic number, which explains why Mg and Ca share the same chemical properties.

On the periodic table, magnesium and calcium are both located in the second vertical column. Magnesium and calcium are alkaline earth metals, just like all Group 2 elements. Both of them are salt-forming soft, silver-colored metals that respond fast to halogen gases like fluorine, chlorine, and bromine.

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Orbits do the proton complete:

A proton exits the cyclotron 1. 0 ms after starting its spiral trajectory in the center of the cyclotron and 1.8 x [tex]10^{4}[/tex] orbits does the proton complete during this 1. 0 ms

Calculation:

(a) The kinetic energy of the proton is given by, K.E. = [tex]\frac{1}{2} m v^{2}[/tex]

v = [tex]\sqrt{\frac{2(K E)}{m}}[/tex]

v = [tex]\sqrt{\frac{2(6.5 \mathrm{MeV})\left(1.6 \times 10^{-19} \mathrm{~J} / \mathrm{eV}\right)}{1.672 \times 10^{-27} \mathrm{~kg}}}[/tex]= 3.5 [tex]\times 10^{7} \mathrm{~m} / \mathrm{s}[/tex]

The radius of the orbit is given by,

r = [tex]\frac{m v}{B q}=\frac{1.672 \times 10^{-27} \mathrm{~kg}\left(3.5 \times 10^{7} \mathrm{~m} / \mathrm{s}\right)}{(1.2 \mathrm{~T})\left(1.6 \times 10^{-19} \mathrm{C}\right)}[/tex] = 0.30 m

Hence, the diameter of the largest orbit just before the proton leaves the orbit is 0.60 m.

(b) The time required to complete one revolution is given by, T=[tex]\frac{2 \pi m}{B q}=\frac{2 \pi\left(1.672 \times 10^{-27} \mathrm{~kg}\right)}{(1.2 \mathrm{~T})\left(1.6 \times 10^{-19} \mathrm{C}\right)}[/tex] =5.48 x [tex]10^{-8}[/tex] s

The number of revolutions made by the proton is given by,

Tn = t

Therefore, n = t/T = [tex]\frac{1 \mathrm{~ms}}{5.48 \times 10^{-8} \mathrm{~s}}[/tex] = 1.8 x [tex]10^{4}[/tex]

Note: your question is incomplete, but most probably your full question was, A medical cyclotron used in the production of medical isotopes

accelerates protons to 6.5 MeV. The magnetic field in the

cyclotron is 1.2 T.

a. What is the diameter of the largest orbit, just before the protons

exit the cyclotron?

b. A proton exits the cyclotron 1.0 ms after starting its spiral

trajectory in the center of the cyclotron. How many orbits

does the proton complete during this 1.0 ms?

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Seconds (s)

Liters (L)

SI units relate to the International System of Units. These units are the base, metric units that are most commonly accepted for different measurements.

Common SI Units

The most common SI units are as follows:

*Note that gram (g) is not an SI unit

Each type of measurement will only have 1 SI unit. This is the unit that will be expected for most answers. Sometimes this means converting an answer into a different unit so it is more widely accepted.

Why SI Units are Important

SI units are important for the same reason that the metric system is used. It gives a standardized list of measurements that scientists across the world can use. Also, it makes it easier to compare scientific findings and studies when all of the measurements are given in the same units.

There are times when it is unrealistic to use SI units. For example, when discussing space, using meters will not be realistic due to the size of space. Also, when measuring the mass of small elements, kilograms are too large. But otherwise, SI units should be used.

The book value of a machine, as shown on the balance sheet, is not relevant in a decision concerning the replacement of that machine by another machine: TRUE

Therefore, the statement "the book value of a machine, as shown on the balance sheet, is not relevant in a decision concerning the replacement of that machine by another machine" is TRUE.

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Complete question:

The book value of a machine, as shown on the balance sheet, is not relevant in a decision concerning the replacement of that machine by another machine. (Ignore taxes.) TRUE or FALSE

Their fossilized shells formed these limestone rocks, which, like Gubbio's mountains, were once at the bottom of the ocean. Now exposed, they represent over a million years of a geological time period known as the Cretaceous.

Walter Alvarez found that forming a distinct boundary between the limestone of the two periods was a thin layer of red clay. Immediately below this clay boundary, the Cretaceous limestone was heavily populated with a wide mix of the fossils of tiny marine creatures called foraminifera, or “foram” for short.

On the Atlantic Coast, outside the town of Zumaia, Dutch geologist Jan Smit was studying the forams from a different ancient sea. Their fossilized shells formed these limestone rocks, which, like Gubbio's mountains, were once at the bottom of the ocean.

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The equilibrium constant K is 17.78 × 10²⁷ at 255 K for the following reaction under acidic conditions.

4H⁺ (aq) + MnO₂ (s) + 2Fe⁺² (aq) → Mn⁺² (aq) + 2Fe⁺³ (aq) + 2H₂O (l)

The relation between standard electrode potential and equilibrium constant is as follows:

G = -2.303 RT log K

-nFE° = -2.303 RT logK

[tex]\log K = \frac{nFE^{\circ}_{\text{cell}}}{2.303RT}[/tex]

Here,

n = 2

F = 96500

[tex]E^{\circ}_{\text{Cell}} = E^{\circ}_{\text{reduced}} - E^{\circ}_{\text{oxidized}}[/tex]

        = 0.77 V - 1.51 V

        = -0.74 V

R = 8.314 J/K mole

T = 255 K

Now put the values in above formula we get

[tex]\log K = \frac{nFE^{\circ}_{\text{cell}}}{2.303RT}[/tex]  

[tex]\log K = \frac{2 \times 96500 \times (-0.74)}{2.303 \times 8.314\ \text{J/K mole} \times 255\ K}[/tex]  

[tex]\log K = \frac{-142820}{4882.521}[/tex]

log K = 29.25

K = 17.78 × 10²⁷

The equilibrium constant K is 17.78 × 10²⁷ at 255 K for the following reaction under acidic conditions.

4H⁺ (aq) + MnO₂ (s) + 2Fe⁺² (aq) → Mn⁺² (aq) + 2Fe⁺³ (aq) + 2H₂O (l)

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Question: Use the appropriate standard reduction potentials below to determine the equilibrium constant at 255 K for the following reaction under acidic conditions.

4H⁺ (aq) + MnO₂ (s) + 2Fe⁺² (aq) → Mn⁺² (aq) + 2Fe⁺³ (aq) + 2H₂O (l)

The equation that relates equilibrium constants with cell potentials is

[tex]\log K = \frac{nFE^{\circ}_{\text{cell}}}{2.303RT}[/tex]

Use this formula, along with the information in the appendix of your book to solve for K.  

Answer:

-ide

Explanation:

Ionic compounds are made up of two parts: a cation and an anion. The cation is positively charged (less electrons) and the anion is negatively charged (more electrons). When naming these compounds, if the anions are nonmetals, they have their endings dropped in order to add the new ending: -ide.

Examples,

CaCl₂ ----> calcium chloride

BaF₂ -----> barium fluoride

MgO -----> magnesium oxide

The positive terminal of the battery should be connected to the anode .and Cu(s) → [tex]Cu^{+2} (aq)[/tex] + [tex]2e^{-}[/tex]  reaction will occur at the anode.

In electrolysis, the two wires or plates dipping into the electrolytic solution is called electrodes.

The battery is galvanic cell which produce electric current. An electrode connected to positive terminal of the battery is called anode. At anode, oxidation ( loss of electrons)  is takes place.

Cu(s) → [tex]Cu^{+2} (aq)[/tex] + [tex]2e^{-}[/tex]  

An electrode connected to negative terminal of the battery is called cathode. At cathode, reduction ( gain of electrons)  is takes place.

[tex]Zn^{+2} + 2e^{-}[/tex] → [tex]Zn(s)[/tex]

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The equilibrium of this reaction will shift to the left to favor the reverse reaction if the pressure is increased.

Balanced chemical equation

CH₄(g) + H₂O(g)  → CO(g) + 3H₂(g)

The principle suggests

The equilibrium will move to the side of the reaction where there are fewer moles of gas as pressure increases.

And the equilibrium will move toward the side of the reaction where there are more gas molecules when the pressure is reduced.

In the given chemical equation

The no. of moles of reactant = 2

The no of moles of the product = 4

Total moles of reactant is less than that of total moles of product

Therefore, If the pressure is increased the equilibrium of this reaction will shift to the left to favor the reverse reaction.

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Answer:

See attached image.

Explanation:

See attached image.

Remember to use the units in deciding steps to take.  Find the moles needed or delivered by using:

Molarity x Volume = moles  [(moles/L)*(L) = moles]

Molarity, M, is dedined as 1 mole/liter.  When working with a unit such as 6M, rewite it as 6 moles/liter, and then use unit conversions to guide how to find the solution, so to speak.

The freezing point of a solution is -1.16°C

Calculation,

The equation which give relation between freezing point depletion and molality is,

Δ[tex]T_{f} = K_{f} .m[/tex]      ... (i)

Moles of the solute = 22.0 g / 114.23 g/mole = 0.193 mole

Mass of the solvent in g = 148 g

Mass of the solvent in kg = 0.148 kg

Molality = 0.193 mole/ 0.148 kg = 1.3 m

Now, after putting the value of  [tex]K_{f}[/tex] and m in equation (i) we get

Δ[tex]T_{f} = K_{f} .m[/tex]  

Δ[tex]T_{f}[/tex]= 5.12°C/m×1.3 m = 6.66°C

Δ[tex]T_{f}[/tex]=  [tex]T_{f}[/tex] (solvent) - [tex]T_{f}[/tex] (solution)

[tex]T_{f}[/tex] (solution) = [tex]T_{f}[/tex] (solvent)  -Δ[tex]T_{f}[/tex]= 5.5 °C - 6.66°C = -1.16°C

The freezing point of a is -1.16°C

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The reactants are not good a good substitute for sodium azide in airbags because sodium bicarbonate and acetic acid produces far less volume of gas than sodium azide.

The chemical equation of the reactions is given below:

The molar volume of all gases is 22.4 L

Based on the mole ratio, sodium bicarbonate and acetic acid produces far less volume of gas than sodium azide. Thus, the reactants are not good a good substitute for sodium azide.

In conclusion, the molar volume of gases is constant.

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