Answer:
Mass of carbon dioxide produced = 52.8 g
Explanation:
Given data:
Mass of carbon react = 14.4 g
Mass of oxygen = 56.5 g
Mass of oxygen left = 18.1 g
Mass of carbon dioxide produced = ?
Solution:
C + O₂ → CO₂
Number of moles of C:
Number of moles = mass/ molar mass
Number of moles = 14.4 g/ 12 g/mol
Number of moles = 1.2 mol
18.1 g of oxygen left it means carbon is limiting reactant.
Now we will compare the moles of C with CO₂.
C : CO₂
1 : 1
1.2 : 1.2
Mass of CO₂:
Mass = number of moles × molar mass
Mass = 1.2 mol × 44 g/mol
Mass = 52.8 g
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Answer:</h2>
We will need to know Avogadro's number and the molar mass of sucrose for this problem to do dimensional analysis.
- Avogadro's number: 6.022 × 10²³ molecules
- Molar mass of sucrose: 342.2965 g/mol
250g × 
× 
= 4.398 molecules
There are <em>4.398 sucrose molecules </em>in 250 grams of sucrose.
Answer:
13. Index Fossils
14. Index Fossils
15. C
16. Geologists use index fossils to define geological periods.
17. C
18. A
The big advantage to using continuous compounding to express growth rates is it avoids the problem of asymmetry in growth rates:
For example, if we use the normal definition and $100 grows to $105 in one time period, that's a growth rate of $105/$100 - 1 = 5% But if $105 decreases to $100, that's a growth rate of $100/$105 - 1 = -4.76%
The problem of asymmetry is those two growth rates, 5% and -4.75% are not equal up to a sign.
But if you use continuous compounding the growth rate in the first case is ln(105/100) = 0.04879.
And the growth rate in the second is ln (100/105) = -0.04879.
Those two growth rates are definitely the negative of each other.<span>
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