Keeping your arm straight in front of you, you rotate 90° to your left, and see the left side of the circle lit while the right side is dark. Half the ball is still lit up, but you can see only part of the lit area. As you continue to rotate, you see a different amount of the ball.
Answer:
Q14: 17,140 g = 17.14 kg.
Q16: 504 J.
Explanation:
<u><em>Q14:</em></u>
- To solve this problem, we can use the relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat absorbed by ice (Q = 3600 x 10³ J).
m is the mass of the ice (m = ??? g).
c is the specific heat of the ice (c of ice = 2.1 J/g.°C).
ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 100.0°C - 0.0°C = 100.0°C).
∵ Q = m.c.ΔT
∴ (3600 x 10³ J) = m.(2.1 J/g.°C).(100.0°C)
∴ m = (3600 x 10³ J)/(2.1 J/g.°C).(100.0°C) = 17,140 g = 17.14 kg.
<u><em>Q16:</em></u>
- To solve this problem, we can use the relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat absorbed by ice (Q = ??? J).
m is the mass of the ice (m = 12.0 g).
c is the specific heat of the ice (c of ice = 2.1 J/g.°C).
ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 0.0°C - (-20.0°C) = 20.0°C).
∴ Q = m.c.ΔT = (12.0 g)(2.1 J/g.°C)(20.0°C) = 504 J.
Heat produced = -13588.956 kJ
<h3>Further explanation</h3>
Given
The reaction of combustion of Methane
CH4(g)+2O2(g)→CO2(g)+2H2O(g) ΔH∘rxn=−802.3kJ
271 g of CH4
Required
Heat produced
Solution
mol of 271 g CH₄ (MW=16 g/mol0
mol = mass : MW
mol = 271 : 16
mol = 16.9375
So Heat produced :
= mol x ΔH°rxn
= 16.9375 mol x −802.3kJ/mol = -13588.956 kJ
Answer:
CH4
Explanation:
The number of moles of carbon and hydrogen has been given as follows:
C = 0.300 mol
H = 1.20 mol
Next, we divide each mole value by the smallest (0.300)
C = 0.300 ÷ 0.300 = 1
H = 1.20 ÷ 0.300 = 4
The empirical ratio of Carbon and Hydrogen is 1:4, hence, the empirical formula is CH4
