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DanielleElmas [232]
3 years ago
6

An ordered list of chemical substances is shown. Chemical Substances 1 CuO 2 O2 3 CO2 4 NO2 5 Fe 6 H2O Which substances in the l

ist can be some of the reactants and products in the same combustion reaction?
Chemistry
1 answer:
a_sh-v [17]3 years ago
7 0

Answer:

reactants: 2 O2

products: 3 CO2, 4 NO2, 6 H2O

Explanation:

In a combustion, a combustible material, which generally is composed of C, H, O, N, and S, is combusted, that is, react with oxygen after a spark was produced; obtaining fire, heat and subproducts, including ashes and gases.

Oxygen is always one of the reactants of a combustion.

If Nitrogen was present in the combustible, NO2 (or other nitrogen oxides) will be produced.

If Carbon was present in the combustible, CO2 will be produced (also CO can be produced).

If Hydrogen was present in the combustible, H2O will be produced.

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When the equation for the combustion of ethane, C2H6, is correctly balanced, the coefficient on oxygen is:
sergejj [24]
2C2H6 + 7O2 —> 4CO2 +6H2O

Oxygen’s coefficient is 7
5 0
3 years ago
Gas particles are pumped into a rigid steel container at a constant temperature. Which statement describes the change in pressur
Masteriza [31]

Answer:

The pressure inside the container would increase with each additional pump.

Explanation:

  • From the general gas law of ideal gases:

<em>PV = nRT,</em>

where, P is the pressure of the gas.

V is the volume of the gas.

n is the no. of moles of the gas.

R is the general gas constant.

T is the temperature of the gas.

  • As clear from the gas law; the pressure of the gas is directly proportional to the no. of moles of the gas.

<em>P α n.</em>

  • As gas particles are pumped into a rigid steel container, the no. of moles of the gas will increase.

So, the pressure of the gas will increase.

<em>Thus, the right choice is: The pressure inside the container would increase with each additional pump.</em>

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2 years ago
Complete the following chart to fill in the blanks
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7 0
3 years ago
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Mrrafil [7]

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Read 2 more answers
One way of obtaining pure sodium carbonate is through the decomposition of the mineral trona, Na3(CO3)(HCO3)·2H2O. 2Na3(CO3)(HCO
zhenek [66]
Percentage yield = (actual yield / theoretical yield) x 100%

The balanced equation for the decomposition is,
 2Na₃(CO₃)(HCO₃)·2H₂O(s) → 3Na₂CO₃(s) + CO₂(g) + 5H₂<span>O(g)

The stoichiometric ratio between </span>Na₃(CO₃)(HCO₃)·2H₂O(s)  and Na₂CO₃(s) is 2 : 3

The decomposed mass of Na₃(CO₃)(HCO₃)·2H₂O(s) = 1000 kg
                                                                                     = 1000 x 10³ g

Molar mass of Na₃(CO₃)(HCO₃)·2H₂O(s) = 226 g mol⁻¹
moles of Na₃(CO₃)(HCO₃)·2H₂O(s) = mass / molar mass
                                                         = 1000 x 10³ g / 226 g mol⁻¹
                                                         = 4424.78 mol

Hence, moles of Na₂CO₃ formed = 4424.78 mol x \frac{3}{2}
                                                     = 6637.17 mol

Molar mass of Na₂CO₃ = 106 g mol⁻¹

Hence, mass of Na₂CO₃ = 6637.17 mol x 106 g mol⁻¹
                                        = 703540.02 g
                                        = 703.540 kg

Hence, the theoretical yield of Na₂CO₃ =  703.540 kg
Actual yield of Na₂CO₃ = 650 kg

Percentage yield = (650 kg / 703.540 kg) x 100%
                            = 92.34%
7 0
3 years ago
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