Answer:
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At STP 32 g of O₂ would occupy by the same volume as 4 g of He
<h3>Further explanation</h3>
Complete question
At STP 32 g of O₂ would occupy by the same volume as:
- 4.0 g of He
- 8.0 g of CH₄
- 64 g of H₂
- 32 g of SO₂
Standard Conditions
Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.
So the gas will have the same volume if the number of moles is the same
mol of 32 grams of O₂ :





<em>So mol of 4 g He = mol of 32 g O₂</em>
There are four F atoms on the products side.
Since two more F atoms are required on the reactant side, you multiply the number of F2 molecules by two.
So 2 should be placed in front of F2
Let the ratio of grams of hydrogen per gram of carbon in methane be M, we know that:
M = 0.3357 g / 1 g
Next, lets represent the grams of hydrogen per gram of carbon in ethane be E. The final piece of information we have is:
M / E = 4/3
If we cross multiply,
3M = 4E
Now, substituting the value of M from earlier and solving for E,
E = (3 * 0.3357) / 4
E = 0.2518
There are 0.2518 grams of hydrogen per gram of carbon in ethane.
Answer:
is b
Explanation:
es la b porque estpy estudiando lo mismo