Answer:
True rod cell are best in the dark or dim light.
Answer:
moles of carbon dioxide produced are 410.9 mol.
Explanation:
Given data:
Mass of C₆H₁₄O₂ = 16.5 g
Moles of O₂ = 499 mol
Moles of CO₂ = ?
First of all we will write the balance chemical equation.
2C₆H₁₄O₂ + 17O₂ → 14CO₂ + 12H₂O
moles of C₆H₁₄O₂ = mass × molar mass
moles of C₆H₁₄O₂ = 16.5 g × 118 g/mol
moles of C₆H₁₄O₂ = 1947 mol
Now we compare the moles of CO₂ with moles of O₂ and C₆H₁₄O₂ from balance chemical equation.
O₂ : CO₂
17 : 14
499 : 14/17× 499 = 410.9 moles
C₆H₁₄O₂ : CO₂
2 : 14
1947 : 14/2× 1947 = 13629 moles
Oxygen will be limiting reactant so moles of carbon dioxide produced are 410.9 mol.
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Answer:
T2 = 2843.1 oK. This is a huge temperature. Check it for errors.
Explanation:
Remark
This is the same question as the other one I've answered. Only the numbers have been altered.
Givens
v1 = 56 mL
P1 = 1 atm
T1 = 273o K
v2 = 162
P2 = 3.6 atm
T2 = ?
Formula
Vi * P1 / T1 = V2 * P2/T2
Solution
Rearrange the formula so T2 is on the left
T2 = V2 P2 * T1 / (V1 * P1) Now just put the numbers in.
T2 = 162 * 3.6* 273 / (56 *1)
T2 = 159213.6/56
T2 = 2843.1