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KatRina
2 months ago
11

A 1.0 mol sample of X(g) and a 1.0 mol sample of Q(g) are introduced into an evacuated, rigid 10.0 L container and allowed to re

ach equilibrium at 50ºC according to the equation above. At equilibrium, which of the following is true about the concentrations of the gases?
X(g) + 2 Q(g) --> R(g) + Z(g)

Kc = 1.3 x 105 at 50ºC

[R] = ½ [Q]
[X] = [Q] = [R] = [Z]
[Q] = ½ [X]
[R] = [Z] > Q
Chemistry
2 answers:
castortr0y [3K]2 months ago
6 0
At equilibrium, [R] = [Z] > [Q]. Explanation: In analyzing the equilibrium conditions, the provided reaction of X(g) + 2 Q(g) ⇄ R(g) + Z(g) establishes that at a temperature of 50ºC, the equilibrium constant Kc = 1.3 x 105. This indicates a scenario where the concentrations of the products greatly exceed those of the reactants.
eduard [2.7K]2 months ago
4 0
At equilibrium, the valid statement is [R] = [Z] > [Q]. Explanation: Given the initial quantities of 1.0 moles each for X and Q within a 10.0 L chamber, the calculations yield molarities that show the concentration of products at equilibrium—indicating that [R] and [Z] surpass [Q].
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A mass of 5.4 grams of aluminum (al) reacts with an excess of copper (ii) chloride (cucl2) in solution, as shown below. 3cucl2 2
eduard [2782]

Answer:

The mass of copper generated is 19.07g

Explanation:

Let's write out the chemical equation;

3CuCl2 + 2Al → 2AlCl3 + 3Cu

3:   2:   2: 3

After confirming that the reaction is balanced, we can continue.

The problem asks for the mass of Cu produced from 5.4g of Al reacting.

Based on the equation, what is the relation between Al and Cu?

2 moles of Al would yield 3 moles of Cu

Expressing this in mass terms, we find;

mass = number of moles * molar mass

Mass of Aluminium = 2 * 26.98 = 53.98 g

Mass of Copper = 3 * 63.546 = 190.638g

This indicates that;

53.98 g of Al would react to create 190.638g of Cu

So, how much Cu would result from 5.4 g of Al?

This leads us to;

53.98 = 190.638

5.4 = x

By cross multiplication, we determine;

x = (190.638 * 5.4) / 53.98

x = 19.07 g

The mass of copper created is 19.07g

8 0
1 month ago
When the concentration of solute in a solid solution exceeds its solubility limit, a new solid solution or phase forms that has
eduard [2782]
The answer is true. A solid solution consists of a solid state solution formed by one or more solutes dissolved in a solvent, or a combination of two crystalline solids that coexist within a crystal lattice. Metal alloys, semiconductors, and moist solids are examples of such solid solutions.
8 0
2 months ago
What salt is produced from mixing csoh and h2co3
Anarel [2989]

Answer:

The resulting salt is Cesium Carbonate

Explanation:

2CsOH + H2CO3 → Cs2CO3 + 2H2O

3 0
2 months ago
First-order reaction that results in the destruction of a pollutant has a rate constant of 0.l/day. (a) how many days will it ta
Alekssandra [3086]

The rate equation for a first order reaction can be expressed as follows:

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}

In this context, k represents the reaction's rate constant, t denotes the time the reaction takes, A_{0} is the initial concentration, and A_{t} is the concentration at time t.

The rate constant of the reaction is 0.1 day^{-1}.

(a) If we start with an initial concentration of 100, when 90% of the substance is eliminated, the remaining quantity at time t will be 100-90=10. By substituting the values,

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{10}=23.03 days

The time required to destroy 90% of the substance amounts to 23.03 days.

(b) If the initial concentration is set at 100, when 99% is destroyed, the present amount at time t will be 100-99=1. By substituting the input values,

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{1}=46.06 days

This results in a duration of 46.06 days required to eradicate 99% of the chemical.

(c) Should the initial concentration be set at 100, with 99.9% of the chemical removed, the remaining quantity at time t will be 100-99.9=0.1. Substituting the values yields

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{0.1}=69.09 days

Thus, the time needed to eliminate 99.9% of the chemical is calculated as 69.09 days.

5 0
2 months ago
A 1.20 g sample of water is injected into an evacuated 5.00 l flask at 65°c. part of the water vaporizes and creates a pressure
alisha [2963]
Assuming the water vapor behaves as an ideal gas,


PV = nRT

For conversions, 760 mmHg = 101325 Pa and 1,000 L = 1 m³
(187.5 mmHg)(101325 Pa/760 mmHg)(5 L)(1 m³/1,000 L) = n(8.314 m³Pa/molK)(65+273 K)
Calculating for n,
n = 0.0445 mole of water

Considering the molar mass of water is 18 g/mol,
The mass of the vaporized water = 0.0445 * 18 = 0.8 g of water evaporated

Therefore,
The fraction of water that vaporized = 0.8/1.2 * 100 = 66.7%
3 0
2 months ago
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