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pickupchik [31]
4 years ago
6

A student needs to prepare 50.0 mL of 0.80 M aqueous H2O2 solution. Calculate the volume of 4.6 M H2O2 stock solution that shoul

d be used.
Chemistry
1 answer:
Degger [83]4 years ago
7 0

Answer:

We need 8.7 mL of the stock solution

Explanation:

Step 1: Data given

Volume of the solution he wants to prepare = 50.0 mL = 0.050 L

Concentration of the solution he wants to prepare = 0.80 M

The concentration of the stock solution = 4.6 M

Step 2: Calculate the volume of the stock solution

C1*V1 = C2*V2

⇒with C1 = the concentration of the stock solution = 4.6 M

⇒with V1 = the volume of the stock solution = TO BE DETERMINED

⇒with C2 = the concentration of the prepared solution = 0.80 M

⇒with V2 = the volume of the prepared solution = 0.050 L

4.6 M * V2 = 0.80 M * 0.050 L

V2 = (0.80 M * 0.050 L) / 4.6M

V2 = 0.0087 L = 8.7 mL

We need 8.7 mL of the stock solution

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Answer:

Explanation:

The trick here is to realize that if you know the volume of a gas at STP, you can use the fact that

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Under STP conditions:

1 mole of an ideal gas = 22.7 L

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

In your case, you know that your sample of gas occupies

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under STP conditions, which are currently defined as a pressure of

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and a temperature of

0

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This means that your sample will contain

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⋅

molar volume of a gas at STP



1 mole gas

22.7

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=

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Now, the molar mass of the gas is the mass of exactly

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3.78 g

for every

0.10044

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1

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⋅

3.78 g

0.10044

moles

=

37.6 g

Since this is the mass of

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molar mass = 37.6 g mol

−

1

−−−−−−−−−−−−−−−−−−−−−−−

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4 years ago
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.

6 0
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