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4 years ago

16 Which change results in the formation of different substances?(1) burning of propane

Chemistry

2 answers:

Fofino [41]4 years ago

8 0

Answer:The correct answer is option (1).

Explanation:

Burning of propane in the presence of oxygen will form carbon-dioxide, water and heat.The chemical reaction can represented as in chemical equation:

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O+heat$

This is an example of chemical change where new product is formed.

Where as in melting of sold sodium chloride, deposition of carbon-dioxide gas and solidification of water ;phase of the substance is changing that is a physical change is taking place.

Hence, the correct option is (1).

Trava [24]4 years ago

6 0

1. The burning of propane creates different substances: propane and oxygen combine to create water and carbon dioxide:

C3H8 + 5 O2 becomes 3 CO2 + 4 H2O + heat

2. When NaCl melts, it is changing state, but it is still the same NaCl.

3. When gaseous CO2 deposits it is changing state, but it is still the same CO2.

4. When (presumably liquid) water solidifies, it is changing state, but it is still the same water.

So the answer is 1.

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Ammonia, NH3NH3 , can react with oxygen to form nitrogen gas and water. 4NH3(aq)+3O2(g)⟶2N2(g)+6H2O(l) 4NH3(aq)+3O2(g)⟶2N2(g)+6H

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36.37% is the percent yield of the reaction.

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1)0.650 L nitrogen gas , at 295 K and 1.01 bar.

Let the moles of nitrogen gas be n.

Pressure of the gas ,P= 1.01 bar = 0.9967 atm (1 bar = 0.9869 atm)

Temperature of the gas = T = 295 K

Volume of the gas = V = 0.650 L

Using an ideal gas equation:

$PV=nRT$

$n=\frac{PV}{RT}=\frac{0.9967 atm\times 0.650 L}{0.0821 atm L/mol K\times 295 K}=0.0267 mol$

2) Moles of ammonia gas= $\frac{2.53 g}{17 g/mol}=0.1488 mol$

Moles of oxygen gas = $\frac{3.53 g}{32 g/mol}=0.1101 mol$

According to reaction ,3 mol of oxygen reacts with 4 mol of ammonia.

Then,0.1101 mol of oxygen will react with:

$\frac{4}{3}\times 0.1101 mol=0.1468 mol$ of ammonia.

Hence, oxygen gas is in limiting amount and act as limiting reagent.

3) Theoretical yield of nitrogen gas :

According to reaction, 3 mol of oxygen gas gives 2 moles of nitrogen gas.

Then 0.1101 mol of oxygen will give:

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Theoretical yield of nitrogen gas = 0.0734 mol

Experimental yield of nitrogen as calculated in part (1) = 0.0267 mol

Percentage yield:

$\frac{\text{Experiential yield}}{\text{Theoretical yield}}\times 100$

Percentage yield of the reaction:

$\frac{ 0.0267 mol}{0.0734 mol}\times 100=36.37\%$

36.37% is the percent yield of the reaction.

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