The answer is by "splitting water".
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Order.
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Answer:
19.6 J
Step-by-step explanation:
Before the ball is dropped, it has a <em>potential energy
</em>
PE = mgh
PE = 0.2 × 10 × 9.8
PE = 19.6 J
Just before the ball hits the ground, the potential energy has been converted into kinetic (<em>mechanical</em>) energy.
KE = 19.6 J
Given:
Q = 9.4 kJ/(kg-h), the heat production rate
c = 4.18 J/(g-K), the heat capacity
t = 2.5 h, amount of time
Note that
c = 4.18 J/(g-K) = 4180 J/(kg-K) = 4.18 kJ/kg-K)
Consider 1 kg of mass.
Then
Qt = cΔT
where ΔT is the increase in temperature (°K)
(1 kg)*(9.4 kJ/(kg-h))*(2.5 h) = (1 kg)*(4.18 kJ/(kg-K))*(ΔT K)
23.5 = 4.18 ΔT
ΔT = 23.5/4.18 = 5.622 K = 5.622 °C
Answer: 5.62 K (or 5.62 °C)
Answer:
a. 581.4 Pa
b. 3.33x10⁻⁴ mol/L
c. 3.49x10⁻⁴ mol/L
d. 0.015 g/L
Explanation:
a. By the Raoult's Law, the partial pressure of a component of a gas mixture is its composition multiplied by the total pressure, so:
pA = 0.9532*6.1
pA = 5.81452 mbar
pA = 5.814x10⁻³ bar
1 bar ----- 10000 Pa
5.814x10⁻³ bar--- pA
pA = 581.4 Pa
b. Considering the mixture as an ideal gas, let's assume the volume as 1,000 L, so by the ideal gas law, the total number of moles is:
PV = nRT
Where P is the pressure (610 Pa), V is the volume (1 m³), n is the number of moles, R is the gas constant (8.314 m³.Pa/mol.K), and T is the temperature.
n = PV/RT
n = (610*1)/(8.314*210)
n = 0.3494 mol
The number of moles of CO₂ is (V = 0.9532*1 = 0.9532 m³):
n = PV/RT
n = (581.4*0.9532)/(8.314*210)
n = 0.3174 mol
cA = n/V
cA = 0.3174/953.2
cA = 3.33x10⁻⁴ mol/L
c. c = ntotal/Vtotal
c = 0.3494/1000
c = 3.49x10⁻⁴ mol/L
d. The molar masses of the gases are:
CO₂: 44 g/mol
N₂: 28 g/mol
Ar: 40 g/mol
O₂: 32 g/mol
CO: 28 g/mol
The molar mass of the mixture is:
M = 0.9532*44 + 0.027*28 + 0.016*40 + 0.0008*28 = 43.36 g/mol
The mass concentration is the molar concentration multiplied by the molar mass:
3.49x10⁻⁴ mol/L * 43.36 g/mol
0.015 g/L
