Answer:
Any of the six chemical elements that markup group1
of the periodic table.
Explanation:
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.
A mile.
For reference, it's about 1,607 or so meters, and 1km is 1,000.
Answer:
C₂ = 0.149 M
Explanation:
Given data:
Initial concentration = 0.407 M
Initial volume = 2.56 L
Final volume = 7.005 L
Final concentration = ?
Solution:
Formula:
C₁V₁ = C₂V₂
C₁ = Initial concentration
V₁ = Initial volume
C₂ = Final concentration
V₂ =Final volume
Now we will put the values.
0.407 M × 2.56 L = C₂ × 7.005 L
1.042 = C₂ × 7.005 L
C₂ = 1.042 M.L / 7.005 L
C₂ = 0.149 M
Answer:
I would say the answer is 2
