A) 7.110 x 10^2
B) 2.39 x 10^-1
C) 9.0743 x 10^4
D) 1.342 x 10^2
E) 5.499 x 10^-2
F) 1.00000 x 10^4
G) 7.38592 x 10^-7
percent by mass of carbon = 27,27 %
percent by mass of oxygen = 72.72 %
Explanation:
To calculate the percent composition of carbon dioxide (CO₂) we use the following algorithm.
Molecular mass of CO₂ = atomic weight of carbon × number of carbon atoms + atomic weight of oxygen + number of oxigen atoms
Molecular mass of CO₂ = 12 × 1 + 16 × 2 = 44 g / mole
Considering 1 mole of CO₂ we devise the following reasoning:
If in 44 g of CO₂ we have 12 g of carbon and 32 g of oxygen
Then in 100 g of CO₂ we have X g of carbon and Y g of oxygen
X = (100 × 12) / 44 = 27,27 % carbon
Y = (100 × 32) / 44 = 72.72 % oxygen
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The answer is to this question is A. California.
Answer:
V₁ = 208.3 mL
Explanation:
Given data:
Initial molarity of HCl = 6.0 M
Final volume = 500 mL
Final molarity = 2.5 M
Volume of initial solution required = ?
Solution:
Formula:
M₁V₁ = M₂V₂
Now we will put the values in formula.
6.0 M × V₁ = 2.5 M ×500 mL
6.0 M × V₁ = 1250 M.mL
V₁ = 1250 M.mL / 6.0 M
V₁ = 208.3 mL
I think the answer would be A because O is oxygen and it has 7. Although it’s in parentheses and has a 2 on the outside of those parentheses, so you would multiply and 7 x 2 = 14. 14 is larger than the other ones.
Hopefully I’m right and hopefully that helps.