Answer:
36.2 K
Explanation:
Step 1: Given data
- Initial pressure of the gas (P₁): 8.6 atm
- Initial temperature of the gas (T₁): 38°C
- Final pressure of the gas (P₂): 1.0 atm (standard pressure)
- Final temperature of the gas (T₂): ?
Step 2: Convert T₁ to Kelvin
We will use the following expression.
K = °C +273.15
K = 38 °C +273.15 = 311 K
Step 3: Calculate T₂
We will use Gay Lussac's law.
P₁/T₁ = P₂/T₂
T₂ = P₂ × T₁/P₁
T₂ = 1.0 atm × 311 K/8.6 atm = 36.2 K
Answer:
grams of solution = 551.98 g
Explanation:
Given data:
Percentage of solution = 32.9
Mass of solute = 181.6 g
Grams of solvent = ?
Solution:
Formula:
% = [grams of solute / grams of solution] × 100
Now we will put the values in formula.
32.9 = [ 181.6 g / grams of solution] × 100
grams of solution = 181.6 g × 100 / 32.9
grams of solution = 18160 g /32.9
grams of solution = 551.98 g
Answer:
449730.879 cal/g
Explanation:
Given data:
Mass of sample = 4.9 g
Change in temperature = 2.08 °C (275.23 k)
Heat capacity of calorimeter = 33.50 KJ . K⁻¹
Solution:
C(candy) = Q/m
Q = C (calorimeter) × ΔT
C(candy) = C (calorimeter) × ΔT / m
C(candy) = 33.50 KJ . K⁻¹ × 275.23 K / 4.90 g
C(candy) = 9220.205 KJ / 4.90 g
C(candy) = 1881.674 KJ / g
It is known that,
1 KJ /g = 239.006 cal/g
1881.674 × 239.006 = 449730.879 cal/g
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Therefore, (e) is the correct option here.