For the hypothesis put what you think will happen
125 mile *1gallon/35 mi = 135/35 = (27/7) gallon gasoline
27/7 gallon * 1 L/0.264 gallon = 14.6 L gasoline
14.6 L gasoline * 2.5kg CO2/1L gasoline= 36.5 kg CO2
36.5 kg CO2 * 1lb/0.454 kg = 80.4 lb
Answer: 80.4 lb CO2
Use M1V1 = M2V2 to solve
3(V1) = 2.8 * 1.6
3(V1) = 4.48
V1 = 1.493 L of stock solution
Answer:
Approximately 10,5
Explanation:
The question is not really very specific, because it would need the percentages of those isotopes in the nature. As they are not shown, it should be the median of those two numbers.
atomic weight ≈
= 10,5
If you check a periodic table, you'll see it's actually 10,8, but that's because of the thing I told you at first (percentages missing).
Hope I could help.
Answer:
3,29L
Explanation:
3.29L = V2
Formula: V1/T1 = V2/T2
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Given:
V1 = 3.0 L V2 = ?
T1 = 310 K T2 = 340 K
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Plugin:
(X stands in place of V2 just to make it easier to look at)
[3.0L / 310K = X / 340K]
(3.0L / 310K = 0.01L/K)
0.01L/K = X / 340K
(multiply 340K on both sides, it cancels out on the right)
0.01L/K * 340K = X
(0.01L/K * 340K = 3.29L)
**3.29L = X**
[or]
**3.29L = V2**