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timurjin [86]
3 years ago
7

What are the products of the complete combustion of a hydrocarbon in excess oxygen

Chemistry
1 answer:
dangina [55]3 years ago
4 0
The products for the complete combustion of a hydrocarbon in excess air is carbon dioxide and water. Any hydrocarbon when reacted with oxygen will always yield the said products. Incomplete combustion, on the other hand, yields carbon monoxide and water.
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What is the mass of a sample of pure gold containing 3.00 x 1024 gold atoms?
miss Akunina [59]

Answer: The mass is 980.6g of Gold.

Explanation:

We begin by looking for the number of moles equivalent to 3.0 x 10^24 gold atoms.

Using the Avogadro's number,

6.02 x 10^23 atoms of gold make up 1 mole of gold.

3.0 x 10^24 atoms would make up: 1 / 6.02 x 10^23 x 3.0 x 10^24 = 4.98moles.

Now that we know the number of moles, we can then look for the mass using the formular:

Moles = mass/ molar mass

4.98 = mass / 196.9 (atomic mass of gold)

Making "mass" the subject of formula : mass = 4.98 x 196.9= 980.6g

8 0
3 years ago
A gas mixture contains 88 grams of oxygen gas, 88 grams of nitrogen gas, and 12 grams of nitrogen dioxide gas,
emmainna [20.7K]
BROO DEE YOU ALREADY KNOW WHAT IS STOP DA CAP, o.81
7 0
3 years ago
Read 2 more answers
How many miles of O2 can be produced by letting 12.00 miles of KC103 react?
Marat540 [252]
 2KClO3 --> 2KClO2 + O2 
    12                               6    (moles) 
The ratio of KClO3 and O2 is 2:1. This means 2 moles of KClO3 can create 1 mole of O2. So 12 moles of KClO3 will create 6 moles of O2. 
3 0
3 years ago
If 23.6 g of hydrogen gas reacts with 28.3 g of nitrogen gas, what is the maximum amount of product that can be produced?
Aleonysh [2.5K]

Answer:

34.3 g NH3

Explanation:

M(H2) = 2*1 = 2 g/mol

M(N2) = 2*14 = 28 g/mol

M(NH3) = 14 + 3*1 = 17 g/mol

23.6 g H2* 1 mol/2 g = 11.8 mol H2

28.3 g N2 * 1 mol/28 g = 1.01 mol N2

                                 3H2 + N2 ------> 2NH3

from reaction         3 mol    1 mol

given                   11.8 mol    1.01 mol

We can see that H2 is given in excess, N2 is limiting reactant.

                                 3H2 + N2 ------> 2NH3

from reaction                     1 mol         2 mol

given                                 1.01 mol      x

x = 2*1.01/1= 2.02 mol NH3

2.02 mol * 17g/1 mol ≈ 34.3 g NH3

8 0
3 years ago
How many moles are in 4.85 x 10^25 atoms of carbon?
Serjik [45]

Answer:

4.85 x 10^25

Explanation:

thats what i was told by a calculator

6 0
3 years ago
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