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Helen [10]
3 years ago
12

Introduction: Reaction rates are also influenced by surface area and concentration. The surface area of a solid is a measure of

how much of the solid is exposed to other substances. The concentration of a substance is a measure of how many molecules of that substance are present in a given volume. Question: How do surface area and concentration affect reaction rates
Chemistry
1 answer:
Yuki888 [10]3 years ago
7 0

Answer:

See explanation

Explanation:

Surface area has to do with the number of solid particles that are exposed at a given time and is capable of colliding with other reactant particles. When more surface area is exposed for reaction, then it means that more particles are likely to collide with each other leading to faster chemical reaction rates. When few particles are exposed for reaction (low surface area) then less collisions occur and the rate of reaction is decreased.

Similarly, concentration refers to the amount of substance present. The greater the amount of substance present, the greater the likelihood of collision between particles and the greater the rate of reaction and vice versa.

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TEA [102]

<u>Given:</u>

Change in internal energy = ΔU = -5084.1 kJ

Change in enthalpy = ΔH = -5074.3 kJ

<u>To determine:</u>

The work done, W

<u>Explanation:</u>

Based on the first law of thermodynamics,

ΔH = ΔU + PΔV

the work done by a gas is given as:

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Therefore:

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W = ΔU-ΔH = -5084.1 -(-5074.3) = -9.8 kJ

Ans: Work done is -9.8 kJ


6 0
3 years ago
A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 23.0 mL of
Georgia [21]

NaOH reacts with CH3COOH in 1:1 molar ratio to produce CH3COONa 

NaOH + CH3COOH → CH3COONa + H2O 

Mol CH3COOH in 52.0mL of 0.35M solution = 52.0/1000*0.35 = 0.0182 mol CH3COOH 

Mol NaOH in 19.0mL of 0.40M solution = 19.0/1000*0.40 = 0.0076 mol NaOH 

These will react to produce 0.0076 mol CH3COONa and there will be 0.0182 - 0.0076 = 0.0106 mol CH3COOH remaining in solution unreacted . Total volume of solution = 52.0+19.0 = 71mL or 0.071L 

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pKa acetic acid = - log Ka = -log 1.8*10^-5 = 4.74. 

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pH = 4.74 + (-0.14) 

pH = 4.60.

7 0
3 years ago
How many grams of sodium sulfide can be produced when 45.3 g Na react with 105 g S?
RoseWind [281]
The chemical reaction would be as follows:

<span>2Na + S → Na2S

We are given the amount of the reactants to be used in the reaction. We use these to calculate the amount of product. We do as follows:

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The limiting reactant would be Na. We calculate as follows:

1.97 mol Na ( 1 mol Na2S / 2 mol Na ) (78.04 g / mol ) = 76.87 g Na2S produced</span>
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