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Lyrx [107]
3 years ago
9

If a combustion reaction is reversed, what needs to happen to the H value?

Chemistry
1 answer:
zmey [24]3 years ago
3 0

Answer:

decomposition of the h value

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The difference between all of them in the make-up of the R group.



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What is the 7th element on the periodic table?
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Nitrogen

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How closely a measurement reflects the actual value is precision. True or False??
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7 0
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Urea, CH4N2O (s), is manufactured from NH3 (g) and CO2 (g). H2O (l) is another product of this reaction. An experiment is starte
Katarina [22]

Answer:

a. 4.41 g of Urea

b. 1.5 g of Urea

Explanation:

To start the problem, we define the reaction:

2NH₃ (g) +  CO₂ (g) → CH₄N₂O (s)  +  H₂O(l)

We only have mass of ammonia, so we assume the carbon dioxide is in excess and ammonia is the limiting reactant:

2.6 g . 1mol / 17g = 0.153 moles of ammonia

Ratio is 2:1. 2 moles of ammonia can produce 1 mol of urea

0.153 moles ammonia may produce, the half of moles

0153 /2 = 0.076 moles of urea

To state the theoretical yield we convert moles to mass:

0.076 mol . 58 g/mol = 4.41 g

That's the 100 % yield reaction

If the percent yield, was 34%:

4.41 g . 0.34 = 1.50 g of urea were produced.

Formula is (Yield produced / Theoretical yield) . 100 → Percent yield

3 0
3 years ago
If 15.6 grams of copper (ii) chloride react with 20.2 grams of sodium nitrate how many grams of sodium chloride can be formed? W
olasank [31]

Answer:

- 13.56 g of sodium chloride are theoretically yielded.

- Limiting reactant is copper (II) chloride and excess reactant is sodium nitrate.

- 0.50 g of sodium nitrate remain when the reaction stops.

- 92.9 % is the percent yield.

Explanation:

Hello!

In this case, according to the question, it is possible to set up the following chemical reaction:

CuCl_2+2NaNO_3\rightarrow 2NaCl+Cu(NO_3)_2

Thus, we can first identify the limiting reactant by computing the yielded mass of sodium chloride, NaCl, by each reactant via stoichiometry:

m_{NaCl}^{by\ CuCl_2}=15.6gCuCl_2*\frac{1molCuCl_2}{134.45gCuCl_2} *\frac{2molNaCl}{1molCuCl_2} *\frac{58.44gNaCl}{1molNaCl} =13.56gNaCl\\\\m_{NaCl}^{by\ NaNO_3}=20.2gNaNO_3*\frac{1molNaNO_3}{84.99gNaNO_3} *\frac{2molNaCl}{2molNaNO_3} *\frac{58.44gNaCl}{1molNaCl} =13.89gNaCl

Thus, we infer that copper (II) chloride is the limiting reactant as it yields the fewest grams of sodium chloride product. Moreover the formed grams of this product are 13.56 g. Then, we take 13.56 g of sodium chloride to compute the consumed mass sodium nitrate as it is in excess:

m_{NaNO_3}^{by\ NaCl}=13.56gNaCl*\frac{1molNaCl}{58.44gNaCl}*\frac{2molNaNO_3}{2molNaCl} *\frac{84.99gNaNO_3}{1molNaNO_3}=19.72gNaNO_3

Therefore, the leftover of sodium nitrate is:

m_{NaNO_3}^{leftover}=20.2g-19.7g=0.5gNaNO_3

Finally, the percent yield is computed via:

Y=\frac{12.6g}{13.56g} *100\%\\\\Y=92.9\%

Best regards!

6 0
3 years ago
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