The pressure of the tyre after the temperature change is 0.89atm. Details about pressure can be found below.
<h3>How to calculate pressure?</h3>
The pressure of the tyre can be calculated using the following equation:
P1/T1 = P2/T2
Where;
- P1 = initial pressure
- P2 = final pressure
- T1 = initial temperature
- T2 = final temperature
According to this question, a tyre at 21°C (294K) has a pressure of 0.82 atm. Its temperature decreases to –3.5°C (269.5K).
0.82 × 294 = P2 × 269.5
241.08 = 269.5P2
P2 = 241.08 ÷ 269.5
P2 = 0.895atm
Therefore, the pressure of the tyre after the temperature change is 0.89atm.
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Answer:
0.39 mol
Explanation:
Considering the ideal gas equation as:
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
At same volume, for two situations, the above equation can be written as:-
Given ,
n₁ = 1.50 mol
n₂ = ?
P₁ = 3.75 atm
P₂ = 0.998 atm
T₁ = 21.7 ºC
T₂ = 28.1 ºC
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (21.7 + 273.15) K = 294.85 K
T₂ = (28.1 + 273.15) K = 301.25 K
Using above equation as:

Solving for n₂ , we get:
n₂ = 0.39 mol
PV = nRT
P = (nRT)/V
P = (0.3 mol × 0.08206 atm-l/(mol-K) × (273.15 + 30) K)/(0.5 l)
P = 14.9258934 atm
Answer:
a. 0.182
b. 1.009
c. 1.819
Explanation:
Henderson-Hasselbach equation is:
pH = pKa + log [salt / acid]
Let's replace the formula by the given values.
a. 3 = 3.74 + log [salt / acid]
3 - 3.74 = log [salt / acid]
-0.74 = log [salt / acid]
10⁻⁰'⁷⁴ = 0.182
b. 3.744 = 3.74 + log [salt / acid]
3.744 - 3.74 = log [salt / acid]
0.004 = log [salt / acid]
10⁰'⁰⁰⁴ = 1.009
c. 4 = 3.74 + log [salt / acid]
4 - 3.74 = log [salt / acid]
0.26 = log [salt / acid]
10⁰'²⁶ = 1.819
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