Answer:
41.16 moles of H2O
Explanation:
Ratio for the products-reactants is 1:6, so 1 mol of glucose is produced when plants use 6 moles of water.
Then, let's make a rule of three:
1 mol of glucose is produce by using 6 moles of water
6.86 moles of glucose are produced by the use of (6 . 6.86)/1 = 41.16 moles of H2O
Answer:
30 L H2
Explanation:
- 10 L N2 x <u>3 L H2</u> = 30 L H2
. 1 L N2
Try to verify my answer, Stoichiometry is not easy for me.
Answer:
80.07 g/mol
Explanation:
Sulfur's g/mol = 32.07
Oxygen's g/mol = 16.00
32.07 + 16.00(3) = 80.07 g/mol
There are 1000 mililiters in a liter, so 1000 ml for every liter, you have 5 liters, so:
5L*1000 = 5000 mL
The second volume : V₂= 0.922 L
<h3>
Further explanation
</h3><h3>Given
</h3>
7.03 Liters at 31 C and 111 Torr
Required
The second volume
Solution
T₁ = 31 + 273 = 304 K
P₁ = 111 torr = 0,146 atm
V₁ = 7.03 L
At STP :
P₂ = 1 atm
T₂ = 273 K
Use combine gas law :
P₁V₁/T₁ = P₂V₂/T₂
Input the value :
0.146 x 7.03 / 304 = 1 x V₂/273
V₂= 0.922 L
