According to the equation you have 1 mole of C2H4 and 3 moles of O2.
1 • (22.4L / 270L) = 3 • (22.4L / x)
1/270L = 3/x
x = 3(270) / 1
x = 810 L
810 Liters of oxygen will react with 270 liters of ethene (C2H4) at STP
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Answer:
There are 23076 peanut M&M's in 53.768 kg of M&M's.
Explanation:
First we <u>convert 53.768 kg into g</u>:
- 53.768 kg * 1000 = 53768 g
Then we <u>divide the total mass of M&M's by the mass of one peanut M&M,</u> in order to calculate the answer:
So there are 23076 peanut M&M's in 53.768 kg of M&M's.
Answer:
94.2 g/mol
Explanation:
Ideal Gases Law can useful to solve this
P . V = n . R . T
We need to make some conversions
740 Torr . 1 atm/ 760 Torr = 0.974 atm
100°C + 273 = 373K
Let's replace the values
0.974 atm . 1 L = n . 0.082 L.atm/ mol.K . 373K
n will determine the number of moles
(0.974 atm . 1 L) / (0.082 L.atm/ mol.K . 373K)
n = 0.032 moles
This amount is the weigh for 3 g of gas. How many grams does 1 mol weighs?
Molecular weight → g/mol → 3 g/0.032 moles = 94.2 g/mol
