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Svetach [21]
3 years ago
8

(30 Points) You need to produce a buffer solution that has a pH of 5.02. You already have a solution that contains 10. mmol (mil

limoles) of acetic acid. How many millimoles of acetate (the conjugate base of acetic acid) will you need to add to this solution? The pKa of acetic acid is 4.74. Express your answer numerically in millimoles.
Chemistry
2 answers:
chubhunter [2.5K]3 years ago
8 0
<span>By definition:

pH = pKa + log [acetate]/ [acetic acid]

so

5.02 = 4.74 + log [acetate] / 10 mmole

10mmole = 10/1000 = 0.01 mole

5.02 = 4.74 + log [acetate] / 0.01

5.02 - 4.74 = 0.28 = log [acetate] /0.01


10^0.28 = </span><span>1.90546</span> = [acetate] / 0.01 <span>
[acetate] = 0.019 mole 

= 19 millimoles

</span>
ElenaW [278]3 years ago
3 0
<span>pH = pKa + log [acetate]/ [acetic acid] ( Handerso-Hasselbalch equation)

5.35 = 4.74 + log acetate / 0.010 moles

5.35 - 4.74 =0.61 = log acetate / 0.010

10^0.61 =4.1 = acetate / 0.010

moles acetate = 0.041

millimoles acetate required = 41</span>
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solniwko [45]

Answer:

1. Volume as STP = 755 L

2. Outside temperature = 255 K

3. Percentage yield = 70.5%

Explanation:

1. At STP, pressure  = 101.3 kpa, temperature  = 0°C or 273.15 K

Using the general gas equation :

P1V1/T1 = P2V2/T2

P1 = 620 kpa

V1 = 140 L

T1 = 37°C or (273.15 + 37) K = 310.15 K

P2 = 101.3 kpa

V2 = ?

T2 = 273.15 K

V2 = P1V1T2/P2T1

V2 = 620 × 140 × 273.15 / 101.3 × 310.15

V2 = 755 L

2. Using Charles' gas law:

V1/T1 = V2/T2

V1 = 2.5 L

T1 = 290 K

V2 = 2.2 L

T2 = ?

T2 = V2T1/VI

T2 = 2.2 × 290 / 2.5

T2 = 255 K

3. Equation of reaction : 2 Al + 3 CuSO4 ---> Al2 (SO4)3 + 3 Cu

From equation of the reaction,  2 moles of Al produces 3 moles of Cu

Molar mass of Al = 27 g; Molar mass of Cu = 63.5 g

2 moles of Al = 2 × 27 g = 54 g; 3 moles of Cu = 3× 63.5 = 190.5 g

54 g of Al produces 190.5 g of Cu

1.87 g of Al will produce 190.5/54 × 1.87 g of Cu = 6.60 g of Cu

Percentage yield = actual yield /theoretical yield × 100%

Percentage yield = 4.65/6.60 × 100%

Percentage yield = 70.5%

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3 years ago
When the following reaction is completed and written in the form of a net ionic equation, which of the following elements will N
Alex Ar [27]

Answer:

d. N

Explanation:

Chemical equation:

Pb(NO₃)₂(aq)  + K₂SO₄(aq)  →  PbSO₄(s) + KNO₃(aq)

Balanced Chemical equation:

Pb(NO₃)₂(aq)  + K₂SO₄(aq)  → PbSO₄(s) + 2KNO₃(aq)

Ionic equation:

Pb²⁺(aq) + 2NO₃⁻(aq)  + 2K⁺(aq) + SO₄²⁻(aq)  → PbSO₄(s) + 2K⁺(aq) + 2NO₃⁻(aq)

Net ionic equation:

Pb²⁺(aq) + SO₄²⁻(aq)  → PbSO₄(s)

The NO₃⁻(aq) and K⁺(aq)are spectator ions that's why these are not written in net ionic equation. The PbSO₄ can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.  

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Answer:

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In a chemical reaction, the total mass of the product(s) would be less than the total weight of the reactant(s) because GASES, which constituted part of the mass of the reaction, WERE RELEASED INTO THE ATMOSPHERE. However, if the mass of the gas released can be accounted for, the amount of reactants and products must balance.

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Answer:

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