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Mademuasel [1]
4 years ago
5

Hydrofluorocarbons (HFC) have replaced chlorofluorocarbon gases (CFC) in refrigerators.

Chemistry
1 answer:
g100num [7]4 years ago
7 0

Answer: 98.36g/mol

Explanation:Please see attachment for explanation

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What is the square root of 54554? What is the square root of 35654?
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√54554 = 233,56.
√35654 = 188,82.
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3 years ago
How many moles of water are contained in 0.250 mol of cuso4 5h2o?
Mamont248 [21]
1 mol CuSO₄·5H₂O ----------- 5 mol H₂O
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Many ionic compounds form crystal structures. How and why does this happen?
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6 0
3 years ago
Round to 4 significant figures.<br> 0.00238866
Nikitich [7]

Answer:

0.002389

Explanation:

look at the first non-zero digit if rounding to one significant figure

look at the digit after the first non-zero digit if rounding to two significant figures

draw a vertical line after the place value digit that is required

look at the next digit

if it's 5 or more, increase the previous digit by one

if it's 4 or less, keep the previous digit the same

fill any spaces to the right of the line with zeros, stopping at the decimal point if there is one

6 0
3 years ago
Read 2 more answers
Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2N
Montano1993 [528]

The question is incomplete, here is the complete question:

Urea (CH₄N₂O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH₃) with carbon dioxide as follows: 2NH₃(aq) + CO₂(aq) → CH₄N₂O(aq) + H₂O(l) In an industrial synthesis of urea, a chemist combines 135.9 kg of ammonia with 211.4 kg of carbon dioxide and obtains 178.0 kg of urea.

Determine the limiting reactant. (express your answer as a chemical formula)

<u>Answer:</u> The limiting reactant is ammonia (NH_3)

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For ammonia:</u>

Given mass of ammonia = 135.9 kg = 135900 g    (Conversion factor:  1 kg = 1000 g)

Molar mass of ammonia = 17 g/mol

Putting values in equation 1, we get:

\text{Moles of ammonia}=\frac{135900g}{17g/mol}=7994.12mol

  • <u>For carbon dioxide gas:</u>

Given mass of carbon dioxide gas = 211.4 kg = 211400 g

Molar mass of carbon dioxide gas = 44 g/mol

Putting values in equation 1, we get:

\text{Moles of carbon dioxide gas}=\frac{211400g}{44g/mol}=4804.54mol

The given chemical reaction follows:

2NH_3(aq.)+CO_2(aq,)\rightarrow CH_4N_2O(aq.)+H_2O(l)

By Stoichiometry of the reaction:

2 moles of ammonia reacts with 1 mole of carbon dioxide

So, 7994.12 moles of ammonia will react with = \frac{1}{2}\times 7994.12=3997.06mol of carbon dioxide

As, given amount of carbon dioxide is more than the required amount. So, it is considered as an excess reagent.

Thus, ammonia is considered as a limiting reagent because it limits the formation of product.

Hence, the limiting reactant is ammonia (NH_3)

5 0
3 years ago
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