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3 years ago

The reaction below will occur in a gaseous system at STP:

Chemistry

1 answer:

WARRIOR [948]3 years ago

4 0

Answer:

V = 12.5 L

Explanation:

Given data:

Volume of NO = 15.0 L

Temperature and pressure = standard

Volume of nitrogen gas produced = ?

Solution:

Chemical equation:

6NO + 4NH₃ → 5N₂ + 6 H₂O

Number of moles of NO:

PV = nRT

n = PV/RT

n = 1 atm × 15.0 L / 0.0821 atm.L /mol.K × 273.15 K

n = 15.0 atm.L / 22.43 atm.L /mol

n = 0.67 mol

now we will compare the moles of No and nitrogen gas.

NO : N₂

6 : 5

0.67 : 5/6×0.67 = 0.56

Volume of nitrogen gas:

PV = nRT

1 atm × V = 0.56 mol × 0.0821 atm.L /mol.K × 273.15 K

V = 12.5 atm.L / 1 atm

V = 12.5 L

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Answer:

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Explanation:

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What ion charge patterns are in the periodic table

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<span>the elements in groups1a, 2a, 3a have positive charges according to their group numbers</span>

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3 years ago

4. a compound called pyrene has the empirical formula c8h5. when 4.04 g of pyrene is dissolved in 10.00 g of benzene, the boilin

e-lub [12.9K]

The molecular mass of pyrene is 204.4 g/mol.

From;

ΔT = Kb m i

Where;

ΔT = boiling point elevation
Kb = boiling point constant
m = molality
i = Van't Hoff factor

Since the compound is molecular; i = 1

The number of moles of pyrene = 4.04 g/MM

Where; MM = molar mass of pyrene

molality = number of moles of pyrene/mass of solvent in Kg

The mass of solvent = 10 g or 0.01 Kg

molality = 4.04 g/MM/0.01

ΔT = Boiling point of solution - Boiling point of pure solvent

ΔT = 85.1°C - 80.1°C

ΔT = 5°C

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Learn more: brainly.com/question/2292439

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2 years ago

Which statement correctly describes protons?

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3 years ago

Read 2 more answers

Be sure to answer all parts. Hydrogen iodide decomposes according to the reaction 2 HI(g) ⇌ H2(g) + I2(g) A sealed 1.50−L contai

Zarrin [17]

Answer : The concentration of $HI$ and $I_2$ at equilibrium is, 0.0158 M and 0.00302 M respectively.

Explanation :

First we have to calculate the concentration of $H_2, I_2\text{ and }HI$

$\text{Concentration of }H_2=\frac{\text{Moles of }H_2}{\text{Volume of solution}}=\frac{0.00623mol}{1.50L}=0.00415M$

$\text{Concentration of }I_2=\frac{\text{Moles of }I_2}{\text{Volume of solution}}=\frac{0.00414mol}{1.50L}=0.00276M$

$\text{Concentration of }HI=\frac{\text{Moles of }HI}{\text{Volume of solution}}=\frac{0.0244mol}{1.50L}=0.0163M$

Now we have to calculate the value of equilibrium constant (K).

The given chemical reaction is:

$2HI(g)\rightleftharpoons H_2(g)+I_2(g)$

Initial conc. 0.0163 0.00415 0.00276

At eqm. (0.0163-2x) (0.00415+x) (0.00276+x)

As we are given:

Concentration of $H_2$ at equilibrium = 0.00467 M

That means,

(0.00415+x) = 0.00467

x = 0.00026 M

Concentration of $HI$ at equilibrium = (0.0163-2x) = (0.0163-2(0.00026)) = 0.0158 M

Concentration of $I_2$ at equilibrium = (0.00276+x) = (0.00276+0.00026) = 0.00302 M

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