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3 years ago

Chlorine gas was first prepared in 1774 by the oxidation of NaCl with MnO2:

Chemistry

1 answer:

irina1246 [14]3 years ago

5 0

Answer:

0.5

Explanation:

2NaCl(s) + 2H2SO4(l) + MnO2(s) → Na2SO4(s) + MnSO4(s) + 2H2O(g) + Cl2(g)

Using ideal gas equation,

PV = nRT

28.7torr

Converting torr to atm,

= 0.0378atm

V = 0.597L

T = 27 °C

= 300 K

a) PV = nRT

(0.0378atm) * (0.597L) = n(0.0821) * (300k)

= 0.000915 mol

moles of water and chlorine = 0.000915 mol

From the above equation, the ratio of water to chlorine = 1 : 2

Therefore, mole of chlorine = 0.000915/2

= 0.000458 mol

mole fraction = moles of specie/moles of all the species present

= 0.000458/0.000915

= 0.5

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For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 atm for all species. For the reaction 2 NO ( g

enyata [817]

Answer:

ΔG = -61.5 kJ/mol (Spontaneous process)

Explanation:

2 NO (g) + O₂ (g) ⇄ 2NO₂ (g)

Let's apply the thermodynamic formula to calculate the ΔG

ΔG = ΔG° + R .T . lnQ

We don't know if the gases are at equilibrium, that's why we apply Q (reaction quotient)

ΔG = - 69 kJ/mol + 8.31x10⁻³ kJ/K.mol . 298K . ln Q

How can we know Q? By the partial pressures (Qp)

P NO = 0.450atm

PO₂ = 0.1 atm

PNO₂ = 0.650 atm

Qp = [NO₂]² / [NO]² . [O₂]

Qp = 0.650² / 0.450² . 0.1 = 20.86

ΔG = - 69 kJ/mol + 8.31x10⁻³ kJ/K.mol . 298K . ln 20.86

ΔG = -61.5 kJ/mol (Spontaneous process)

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3 years ago

At the same temperature, steam burns are often more severe than water burns because of water's high A. polarity B. heat of vapor

aivan3 [116]

Because of eater's high is A

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A chemical reaction produces 56.2 grams of iron (II) sulfide (FeS). How many moles of iron (II) sulfide does the reaction produc

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Answer:

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Explanation:

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3 years ago

How are energy and work related? A. Energy is the force needed to do work B. Work times energy is force C. Energy is the capacit

Svetllana [295]

Answer:

I believe the answer is A

Explanation:

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3 years ago

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Under certain conditions Argon gas diffuses at a rate of 3.2 cm per second under the same conditions an unknown gas diffuses at

kozerog [31]

Answer:

20 g/mol

Explanation:

We can use Graham’s Law of diffusion:

The rate of diffusion (r) of a gas is inversely proportional to the square root of its molar mass (M).

$r = \frac{1 }{\sqrt{M}}$

If you have two gases, the ratio of their rates of diffusion is

$\frac{r_{2}}{r_{1}} = \sqrt{\frac{M_{1}}{M_{2}}}$

Squaring both sides, we get

$(\frac{r_{2}}{r_{1}})^{2} = \frac{M_{1}}{M_{2}}$

Solve for M₂:

$M_{2} = M_{1} \times (\frac{r_{1}}{r_{2}})^{2}$

$M_{2} = \text{39.95 g/mol} \times (\frac{\text{3.2 cm/s}}{\text{4.5 cm/s}})^{2}= \text{39.95 g/mol} \times (0.711 )^{2}$

$= \text{39.95 g/mol} \times 0.506 = \textbf{20 g/mol}$

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3 years ago