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Ugo [173]
4 years ago
11

you are given an unknown gaseous binary compound. when 10.0g of the compound isburned in excess oxygen, 16.3g of water is produc

ed. the compound has 1.38 times that of oxygen gas at the same conditions of temperature and pressure. give a possible identity for the compound.
Chemistry
1 answer:
Flauer [41]4 years ago
5 0
Since the compound has 1.38 time that of oxygen gas at the same conditions of temperature and pressure, we have the relationship:
MW/MWoxygen = 1.38
MW = 44.16

Since there is water formed during the reaction, the formula of the compound must be:
XaHb
where a and b are the coefficients of each element.

If the compound reactions with oxygen forming water and an oxide of the element X, the combustion reaction must be:
XaHb + ((2a + (b/2))/2) O2 = a (XO2) + (b/2)(H2O)
Using dimensional analysis:
10 (1/44.16) (b/2 / 1) (18) = 16.3
Solving for b:
b = 8

The compound now is XaH8. Most probably, the compound is C3H8 since it has a molecular formula of 44 and it reacts with O2 to form water and CO2.
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Some chemical reactions releases energy other store energy what important chemical reaction storage energy?
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Explanation:

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Read the lab procedure for a controlled experiment that looks at the effect of heat on the circumference of bicycle tires.
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5 0
2 years ago
What is the percent composition of NaHCO3?
zhuklara [117]

Answer:

                Option-C (27.36% Na, 1.20% H, 14.30% C, and 57.14% O)

Explanation:

<em>Percent Composition</em> is defined as the <u><em>%age by mass of each element present in a compound</em></u>. Therefore, it is a relative amount of each element present in a compound.

Calculating Percent Composition of NaHCO₃:

1: Calculating Molar Masses of all elements present in NaHCO₃:

              a) Na  =  22.99 g/mol

             b) H  =  1.01 g/mol

              c) C  =  12.01 g/mol

              d) O₃  =  16.0 × 3 =  48 g/mol

2: Calculating Molecular Mass of NaHCO₃:

              Na  =  22.99 g/mol

             H    =  1.01 g/mol

              C    =  12.01 g/mol

              O₃  =  48 g/mol

                       ----------------------------------  

Total                  84.01 g/mol

3: Divide each element's molar mass by molar mass of NaHCO₃ and multiply it by 100:

For Na:

                 =  22.99 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

                 =  27.36 %

For H:

                 =  1.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

                 =  1.20 %

For C:

                 =  12.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

                 =  14.29 % ≈ 14.30 %

For O:

                 =  48.0 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

                 =  57.13 % ≈ 57.14 %

8 0
3 years ago
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