Question:
<span>A sample of nitrogen gas had a volume of 500mL, a pressure in its closed container of 740 torr and a temperature of 25°c. what was the volume of gas when the temperature was changed to 50°c and the new pressure was 760 torr?
Answer:
Data Given:
V</span>₁ = 500 mL
P₁ = 740 torr
T₁ = 25 °C + 273 = 298 K
V₂ = ?
P₂ = 760 torr
T₂ = 50 °C + 273 = 323 K
Solution:
Let suppose the gas is acting Ideally, then According to Ideal Gas Equation,
P₁ V₁ / T₁ = P₂ V₂ / T₂
Solving for V₂,
V₂ = (P₁ V₁ T₂) ÷ (T₁ P₂)
Putting Values,
V₂ = (740 torr × 500 mL × 323 K) ÷ (298 K × 760 torr)
V₂ = 527.68 mL
Answer:
84.8%
Explanation:
Step 1: Given data
Bob measured out 1.60 g of Na. He forms NaCl according to the following equation.
Na + 1/2 Cl₂ ⇒ NaCl
According to this equation, he calculates that 1.60 g of sodium should produce 4.07 g of NaCl, which is the theoretical yield. However, he carries out the experiment and only makes 3.45 g of NaCl, which is the real yield.
Step 2: Calculate the percent yield.
We will use the following expression.
%yield = real yield / theoretical yield × 100%
%yield = 3.45 g / 4.07 g × 100% = 84.8%
1. L
Number one because the lines match up
42.34 g of water could be warmed from 21.4°C to 43.4°C by the pellet dropped inside it
Heat loss by the pellet is equal to the Heat gained by the water.
….(1)
where,
is the heat gained by water
is the heat loss by pellet
= mCΔT
where m = mass of water
C = specific heat capacity of water = 4.184 J/g-°C
ΔT = Increase in temperature
ΔT for water = 43.4 - 21.4 = 22°C
= m × 4.184 × 22 …. (2)
Now
=
×ΔT
where
= Heat capacity of pellet = 56J/°C
Δ T for pellet = 43.4 - 113 =- 69.6°C
= 56 × -69.6 = -3897.6 J
From equation (1) and (2)
-m× 4.184 × 22 =-3897.6
m= 42.34 g
Hence, 42.34 g of water could be warmed from 21.4 degrees Celsius to 43.4 degrees Celsius by the pellet dropped inside it.
Learn more about specific heat here brainly.com/question/16559442
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