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Maurinko [17]
3 years ago
10

In a closed system, how will a decrease in pressure affect the following reaction: N2O4(g) ⇌ 2NO2(g)?

Chemistry
1 answer:
timama [110]3 years ago
5 0

Answer:

Answer 'A'

Explanation:

In general, if the Σmolar volumes(g) reactants ≠ Σmolar volumes(g) products, a change in atmospheric pressure will shift the reaction equilibrium. If the pressure is increased, the rxn will shift toward the LOWER molar volume side of the rxn or if the pressure is decreased the rxn will shift toward the HIGHER molar volume side. For the reaction N₂O₄(g) ⇄ 2NO₂(g), Vm(N₂O₄(g)) < Vm(NO₂(g)) so, a decrease in atmospheric pressure would shift rxn toward the NO₂(g) side of the equation increasing the moles of NO₂(g). Also, note that if the Σmolar volumes(g) reactants = Σmolar volumes(g) products, no shift in equilibrium will occur regardless of changes in atmospheric pressure.

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classify each solution as either an acid or a base using the properties that are stated. what are two ways these solutions are s
USPshnik [31]
Probably too late to help but solution a is likely a base while solution b is an acid.

two similar things: both are aqueous solution and they are not neutral.
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3 years ago
Which type of community will have more competition than cooperation?
Vedmedyk [2.9K]

Answer:

a 10,00 year-old coral reef ecosystem

6 0
2 years ago
The production of NOx gases is an unwanted side reaction of the main engine combustion process that turns octane, C8H18, into CO
borishaifa [10]

Answer:

710,33 g NO2

Explanation:

2 C8H18 + 25 O2 → 16 CO2 + 18 H2O  

(800 g octane) / (114.2293 g C8H18/mol x (25/2)) = 87.54 mol O2 used to combust the octane

87.54 mol O2 x \frac{15}{85} = 15.44 mol O2 used to form NO2

O2 + 2NO → 2NO2

(15.44 mol O2) x (2/2) x (46.0056 g NO2/mol) = 710,33 g NO2

4 0
3 years ago
Read 2 more answers
At constant volume, the heat of combustion of a particular compound, compound A, is − 3039.0 kJ / mol. When 1.697 g of compound
melisa1 [442]

Answer:

13.85 kJ/°C

-14.89 kJ/g

Explanation:

<em>At constant volume, the heat of combustion of a particular compound, compound A, is − 3039.0 kJ/mol. When 1.697 g of compound A (molar mass = 101.67 g/mol) is burned in a bomb calorimeter, the temperature of the calorimeter (including its contents) rose by 3.661 °C. What is the heat capacity (calorimeter constant) of the calorimeter? </em>

<em />

The heat of combustion of A is − 3039.0 kJ/mol and its molar mass is 101.67 g/mol. The heat released by the combustion of 1.697g of A is:

1.697g.\frac{1mol}{101.67g} .\frac{(-3039.0kJ)}{mol} =-50.72kJ

According to the law of conservation of energy, the sum of the heat released by the combustion and the heat absorbed by the bomb calorimeter is zero.

Qcomb + Qcal = 0

Qcal = -Qcomb = -(-50.72 kJ) = 50.72 kJ

The heat capacity (C) of the calorimeter can be calculated using the following expression.

Qcal = C . ΔT

where,

ΔT is the change in the temperature

Qcal = C . ΔT

50.72 kJ = C . 3.661 °C

C = 13.85 kJ/°C

<em>Suppose a 3.767 g sample of a second compound, compound B, is combusted in the same calorimeter, and the temperature rises from 23.23°C to 27.28 ∘ C. What is the heat of combustion per gram of compound B?</em>

Qcomb = -Qcal = -C . ΔT = - (13.85 kJ/°C) . (27.28°C - 23.23°C) = -56.09 kJ

The heat of combustion per gram of B is:

\frac{-56.09 kJ}{3.767g} =-14.89 kJ/g

4 0
3 years ago
What is isotope <br>the percentage of the solid components of soil​
zubka84 [21]

Answer:

an isotope consists of two or more forms of the same elements that contains equal number of protons but different number of neutrons in their nuclei but differ in relative atomic mass but not in chemical properties

The percentage of soil components of the soil is: inorganic components

-soil particles

-mineral elements = 45%

-water or moisture= 25%

- air= 25%

organic components

-humus or decayed organic matter= 5%

-living organisms

4 0
3 years ago
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