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pishuonlain [190]
2 years ago
14

HELP ASAPPPP""""A 0.0200 M NaCl solution was formed when 38.0 grams of NaCl was dissolved in enough water. What was the total vo

lume of the solution formed? (5 points)
32.5 liters
30.3 liters
26.7 liters
22.1 liters
Chemistry
2 answers:
Ymorist [56]2 years ago
8 0

Answer: The volume of the solution formed will be 32.5 L. Explanation: Molarity is defined as the amount of solute present in 1 L of solution. The equation used to determine molarity of the solution follows: We are given: Molarity of the solution = 0.02 mol/L Mass of NaCl = 38 grams Molar mass of NaCl = 58.5 g/mol Putting values in above equation, we get: Hence, the volume of the solution formed will be 32.5 L.

Mrrafil [7]2 years ago
5 0

Answer:

32.5

Explanation:

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The value of Ka for nitrous acid (HNO2) at 25 ∘C is 4.5×10−4 .a. Write the chemical equation for the equilibrium that correspond
kvv77 [185]

Answers and Explanation:

a)- The chemical equation for the corresponden equilibrium of Ka1 is:

2. HNO2(aq)⇌H+(aq)+NO−2

Because Ka1 correspond to a dissociation equilibrium. Nitrous acid (HNO₂) losses a proton (H⁺) and gives the monovalent anion NO₂⁻.

b)- The relation between Ka and the free energy change (ΔG) is given by the following equation:

ΔG= ΔGº + RT ln Q

Where T is the temperature (T= 25ºc= 298 K) and R is the gases constant (8.314 J/K.mol)

At the equilibrium: ΔG=0 and Q= Ka. So, we can calculate ΔGº by introducing the value of Ka:

⇒ 0 = ΔGº + RT ln Ka

   ΔGº= - RT ln Ka

   ΔGº= -8.314 J/K.mol x 298 K x ln (4.5 10⁻⁴)

  ΔGº= 19092.8 J/mol

c)- According to the previous demonstation, at equilibrium ΔG= 0.

d)- In a non-equilibrium condition, we have Q which is calculated with the concentrations of products and reactions in a non equilibrium state:

ΔG= ΔGº + RT ln Q

Q= ((H⁺) (NO₂⁻))/(HNO₂)

Q= ( (5.9 10⁻² M) x (6.7 10⁻⁴ M) ) / (0.21 M)

Q= 1.88 10⁻⁴

We know that   ΔGº= 19092.8 J/mol, so:

ΔG= ΔGº + RT ln Q

ΔG= 19092.8 J/mol + (8.314 J/K.mol x 298 K x ln (1.88 10⁻⁴)

ΔG= -2162.4 J/mol

Notice that ΔG<0, so the process is spontaneous in that direction.

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Its mass is about the same as that of a proton

Explanation:

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If a system has a reaction quotient of 2.13 ✕ 10−15 at 100°C, what will happen to the concentrations of COBr2, CO, and Br2 as th
qaws [65]

This is an incomplete question, here is a complete question.

Consider the following equilibrium at 100°C.

COBr_2(g)\rightleftharpoons CO(g)+Br_2(g)

K_c=4.74\times 10^4

Concentration at equilibrium:

[COBr_2]=1.58\times 10^{-6}M

[Co]=2.78\times 10^{-3}M

[Br_2]=2.51\times 10^{-5}M

If a system has a reaction quotient of 2.13 × 10⁻¹⁵ at 100°c, what will happen to the concentrations of COBr₂, Co and Br₂ as the reaction proceeds to equilibrium?

Answer : The concentrations of Co and Br₂ decreases and the concentrations of COBr₂ increases.

Explanation :

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given balanced chemical reaction is,

COBr_2(g)\rightleftharpoons CO(g)+Br_2(g)

The expression for reaction quotient will be :

Q=\frac{[CO][Br_2]}{[COBr_2]}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

Q=\frac{(2.78\times 10^{-3})\times (2.51\times 10^{-5})}{(1.58\times 10^{-6})}=4.42\times 10^{-2}

The given equilibrium constant value is, K_c=4.74\times 10^4

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

There are 3 conditions:

When Q>K_c that means product > reactant. So, the reaction is reactant favored.

When Q that means reactant > product. So, the reaction is product favored.

When Q=K_c that means product = reactant. So, the reaction is in equilibrium.

From the above we conclude that, the Q that means product < reactant. So, the reaction is product favored that means reaction must shift to the product (right) to be in equilibrium.

Hence, the concentrations of Co and Br₂ decreases and the concentrations of COBr₂ increases.

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