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ICE Princess25 [194]
2 years ago
7

Question 2

Chemistry
1 answer:
san4es73 [151]2 years ago
4 0

Explanation:

its hard to explain unless we know what the question fully asks..

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Burning a compound of calcium, carbon, and nitrogen in oxygen in a combustion train generates calcium oxide , carbon dioxide , n
mylen [45]

The question is incomplete, here is the complete question:

Burning a compound of calcium, carbon, and nitrogen in oxygen in a combustion train generates calcium oxide (CaO), carbon dioxide (CO_2), nitrogen dioxide (NO_2), and no other substances. A small sample gives 2.389 g CaO, 1.876 g CO_2, and 3.921 g NO_2 Determine the empirical formula of the compound.

<u>Answer:</u> The empirical formula for the given compound is CaCN_2

<u>Explanation:</u>

The chemical equation for the combustion of compound having calcium, carbon and nitrogen follows:

Ca_xC_yN_z+O_2\rightarrow CaO+CO_2+NO_2

where, 'x', 'y' and 'z' are the subscripts of calcium, carbon and nitrogen respectively.

We are given:

Mass of CaO = 2.389 g

Mass of CO_2=1.876g

Mass of NO_2=3.921g

We know that:

Molar mass of calcium oxide = 56 g/mol

Molar mass of carbon dioxide = 44 g/mol

Molar mass of nitrogen dioxide = 46 g/mol

<u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 1.876 g of carbon dioxide, \frac{12}{44}\times 1.876=0.5116g of carbon will be contained.

<u>For calculating the mass of nitrogen:</u>

In 46 g of nitrogen dioxide, 14 g of nitrogen is contained.

So, in 3.921 g of nitrogen dioxide, \frac{14}{46}\times 3.921=1.193g of nitrogen will be contained.

<u>For calculating the mass of calcium:</u>

In 56 g of calcium oxide, 40 g of calcium is contained.

So, in 2.389 g of calcium oxide, \frac{40}{56}\times 2.389=1.706g of calcium will be contained.

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Calcium =\frac{\text{Given mass of Calcium}}{\text{Molar mass of Calcium}}=\frac{1.706g}{40g/mole}=0.0426moles

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.5116g}{12g/mole}=0.0426moles

Moles of Nitrogen = \frac{\text{Given mass of Nitrogen}}{\text{Molar mass of Nitrogen}}=\frac{1.193g}{14g/mole}=0.0852moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0426 moles.

For Calcium = \frac{0.0426}{0.0426}=1

For Carbon = \frac{0.0426}{0.0426}=1

For Nitrogen = \frac{0.0852}{0.0426}=2

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of Ca : C : N = 1 : 1 : 2

Hence, the empirical formula for the given compound is CaCN_2

3 0
3 years ago
What is the difference between speed and acceleration?<br> Please help :)
ioda

Answer:

Speed is how fast you are going at the moment but acceleration is how fast you are building up speed to get to a set point

Explanation:

4 0
3 years ago
Which of the following methods can be used to remove water from a mixture of sand and water? A. Centrifuge B. Filtration C. Floa
wlad13 [49]
The correct answer is b. filtration
3 0
3 years ago
Read 2 more answers
Predict the products for nacl+cabr2​
Ray Of Light [21]

nabr + cacl2 are the products

7 0
3 years ago
A radioactive element reduces to 5.00% of its initial mass in
Stolb23 [73]

The half-life in months of a radioactive element that reduce to 5.00% of its initial mass in 500.0 years is approximately 1389 months

To solve this question, we'll begin by calculating the number of half-lives that has elapsed. This can be obtained as follow:

Amount remaining (N) = 5%

Original amount (N₀) = 100%

<h3>Number of half-lives (n) =?</h3>

N₀ × 2ⁿ = N  

5 × 2ⁿ = 100

2ⁿ = 100/5

2ⁿ = 20

Take the log of both side

Log 2ⁿ = log 20

nlog 2 = log 20

Divide both side by log 2

n = log 20 / log 2

<h3>n = 4.32</h3>

Thus, 4.32 half-lives gas elapsed.

Finally, we shall determine the half-life of the element. This can be obtained as follow.

Number of half-lives (n) = 4.32

Time (t) = 500 years

<h3>Half-life (t½) =? </h3>

t½ = t / n

t½ = 500 / 4.32

t½ = 115.74 years

Multiply by 12 to express in months

t½ = 115.74 × 12

<h3>t½ ≈ 1389 months </h3>

Therefore, the half-life of the radioactive element in months is approximately 1389 months

Learn more: brainly.com/question/24868345

8 0
2 years ago
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