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Arisa [49]
3 years ago
9

C0 + 02--> CO2 Balance the equation carbon dioxide forming

Chemistry
1 answer:
eimsori [14]3 years ago
8 0

Answer:

C + O2 - CO (limited oxygen)

2CO + O2-2 CO2 (excess oxygen)

Explanation:

In excess oxygen reacts with carbon to form carbon iv oxide

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Chlorophyll captures the sun's energy and is used as energy to complete the photosynthesis process
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Raoult's law states that the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent at the same temperature scaled by the mole fraction of the solvent present: Psolution=χsolventPosolvent.

5 0
3 years ago
Combustion reactions are exothermic. The heat of reaction for the combustion of 2-methylheptane, C8H18, is 1.306×103 kcal/mol. W
WARRIOR [948]

Answer:

11.45kcal/g

2.612 × 10³ kcal

Explanation:

When a compound burns (combustion) it produces carbon dioxide and water. The combustion of 2-methylheptane can be represented by the following balanced equation:

2 C₈H₁₈ + 25 O₂ ⇄ 16 CO₂ + 18 H₂O

It releases  1.306 × 10³ kcal every 1 mol of C₈H₁₈ that is burned.

<em>What is the heat of combustion for 2-methylheptane in kcal/gram?</em>

We know that the molar mass of C₈H₁₈ is 114.0g/mol. Then, using proportions:

\frac{1.306 \times 10^{3}Kcal}{1mol} .\frac{1mol}{114.0g} =11.45kcal/g

<em>How much heat will be given off if molar quantities of 2-methylheptane react according to the following equation? 2 C₈H₁₈ + 25 O₂ ⇄ 16 CO₂ + 18 H₂O</em>

In this equation we have 2 moles of C₈H₁₈. So,

2mol \times\frac{1.306 \times 10^{3}kcal }{1mol} =2.612\times 10^{3}kcal

3 0
3 years ago
Calculate the number of hydrogen atoms present in 40g of urea, (NH2)2CO
tekilochka [14]

Answer: There are 16.14 \times 10^{23} atoms of hydrogen are present in 40g of urea, (NH_{2})_{2}CO.

Explanation:

Given: Mass of urea = 40 g

Number of moles is the mass of substance divided by its molar mass.

First, moles of urea (molar mass = 60 g/mol) are calculated as follows.

Moles = \frac{mass}{molar mass}\\= \frac{40 g}{60 g/mol}\\= 0.67 mol

According to the mole concept, 1 mole of every substance contains 6.022 \times 10^{23} atoms.

So, the number of atoms present in 0.67 moles are as follows.

0.67 mol \times 6.022 \times 10^{23} atoms/mol\\= 4.035 \times 10^{23} atoms

In a molecule of urea there are 4 hydrogen atoms. Hence, number of hydrogen atoms present in 40 g of urea is as follows.

4 \times 4.035 \times 10^{23} atoms\\= 16.14 \times 10^{23} atoms

Thus, we can conclude that there are 16.14 \times 10^{23} atoms of hydrogen are present in 40g of urea, (NH_{2})_{2}CO.

7 0
3 years ago
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