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3 years ago

Hi:)) how to do 5? I’m not sure if it’s hydrogen carbonate or carbonate

Chemistry

1 answer:

zysi [14]3 years ago

8 0

Answer:

copper(ll) carbonate

Explanation:

Since the product is a salt which is copper(II) carbonate, water and carbon dioxide, this reaction is an acid + metal carbonate reaction.

Looking at the salt, Cu²⁺ has to be part of the reactants.

Hence, the missing compound there has to be copper(ll) carbonate, CuCO₃.

The balanced chemical equation would be:

CuCO₃ + 2HNO₃➙ Cu(NO₃)₂ +H₂O +CO₂

P.s. You left out CO₂ as a product in Q2 ;)

Just a recap of the main reactions you would've learnt:

1) Acid + base/ alkali ➙ salt + water

2) Acid + metal ➙ salt + hydrogen gas

3) Acid + metal carbonate ➙ salt + H₂O + CO₂

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3 years ago

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Ethyl butyrate, CH3CH2CH2CO2CH2CH3, is an artificial fruit flavor commonly used in the food industry for such flavors as orange

SIZIF [17.4K]

Answer:

A. 10.0 grams of ethyl butyrate would be synthesized.

B. 57.5% was the percent yield.

C. 7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

Explanation:

$CH_3CH_2CH_2CO_2H(l)+CH_2CH_3OH(l)+H^+\rightarrow CH_3CH_2CH_2CO_2CH_2CH_3(l)+H_2O(l)$

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According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

$\frac{1}{1}\times 0.08636 mol=0.08636 mol$ of ethyl butyrate

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Experimental yield = ?

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$100\%=\frac{\text{Experimental yield}}{10.0 g}\times 100$

Experimental yield = 10.0 g

10.0 grams of ethyl butyrate would be synthesized.

Theoretical yield of ethyl butyrate = 10.0 g

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Percentage yield of the reaction = ?

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$=\frac{5.75 g}{10.0 g}\times 100=57.5\%$

57.5% was the percent yield.

Moles of butanoic acid = $\frac{7.60 g}{88 g/mol}=0.08636 mol$

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

$\frac{1}{1}\times 0.08636 mol=0.08636 mol$ of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

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Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 78.0%

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$78.0\%=\frac{\text{Experimental yield}}{10.0 g}\times 100$

Experimental yield = 7.80 g

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2 years ago

What is the charge on an ion that has an atomic number of 16 and contains 14e-?

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