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blsea [12.9K]
2 years ago
6

A 0.271g sample of an unknown vapor occupies 294ml at 140C and 874mmHg. The emperical formula of the compound is CH2. How many m

oles of CO2 would be formed if 3moles of the compound react with oxygen?
Chemistry
1 answer:
Phoenix [80]2 years ago
7 0
Using PV = nRT, we can calculate the moles of the sample.
874 mmHg = 116,524 Pa
n = PV/RT
n = 116,524 x 294 x 10⁻⁶ / 8.314 x (140 + 273)
n = 9.98 x 10⁻³ mol

moles = mass / Mr
Mr = 0.271/9.98 x 10⁻³
Mr = 27.2
Mass of empirical formula = 14
Repeat units = 27.2 / 14 ≈ 2

Formula of substance:
C₂H₄

Combustion equation:
C₂H₄ + 3O₂ → 2CO₂ + 2H₂O

1 mole produces 2 moles of CO₂, so 3 moles will produce 6 moles CO₂
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What is the most abundant gas in the atmosphere is _____. argon oxygen nitrogen hydrogen
Serga [27]
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7 0
3 years ago
Read 2 more answers
There are two different compounds of sulfur and fluorine.
weqwewe [10]
Atomic mass of F: 19.0 g/mol

Atomic mass of S: 32.1 g/mol

1.18 g F = [1.18 g / 19.0 g / mol] = 0.062  mol F

1 g S =  1 g/ 32.1 g/mol = 0.031  mol S

Divide by 0.031

0.062 mol F / 0.031 = 2  mol F

0.031 mol S / 0.031 = 1 mol S

SF2 Then X = 2

Verification:
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36 gF / 32.1 g S = 1.18 g F / g S



7 0
3 years ago
In the preparation of a certain alkyl halide, 10 g of sodium bromide (NaBr), 10 mL distilled water (H20), and 9 mL 3-methyl-1-bu
Novosadov [1.4K]

Percentage yield shows the amount of reactants converted into products. The percentage yield of the reaction is 51.7%.

The equation of the reaction is sown in the image attached. The reaction is 1:1 as we can see.

Number of moles of NaBr = 10 g/103 g/mol = 0.097 moles

We can obtain the mass of 3-methyl-1-butanol from its density.

Mass = density × volume

Density of 3-methyl-1-butanol =  0.810 g/mL

Volume of  3-methyl-1-butanol = 9 mL

Mass of 3-methyl-1-butanol = 0.810 g/mL × 9 mL

Mass of 3-methyl-1-butanol = 7.29 g

Number of moles of 3-methyl-1-butanol =  mass/molar mass =  7.29 g/88 g/mol = 0.083 moles

Since the reaction is 1:1 then the limiting reagent is 3-methyl-1-butanol

Mass of product 1-bromo-3-methylbutane = number of moles × molar mass

Molar mass of 1-bromo-3-methylbutane = 151 g/mol

Mass of product 1-bromo-3-methylbutane = 0.083 moles × 151 g/mol

= 12.53 g

Recall that % yield = actual yield/theoretical yield × 100

Actual yield of product = 6.48 g

Theoretical yield = 12.53 g

% yield = 6.48 g/12.53 g × 100

% yield = 51.7%

Learn more: brainly.com/question/5325004

7 0
2 years ago
SnO2 + 2 H2 ——&gt; Sn + 2 H2O
SpyIntel [72]

Answer:

0.15g

Explanation:

Given parameters:

Number of molecules of water = 1.2 x 10²¹ molecules

Unknown:

Mass of SnO₂  = ?

Solution:

To solve this problem, we have to work from the known to the unknown specie;

             SnO₂   +    2H₂    →   Sn  +   2H₂O

Ensure that the equation given is balanced;

       

Now,

          the known species is water;

                  6.02 x 10²³ molecules of water  = 1 mole

                   1.2 x 10²¹ molecules of water  = \frac{1.2 x 10^{21} }{6.02 x 10^{23} }    = 0.2 x 10⁻²moles

Number of moles of water  = 0.002moles

           From the balanced chemical equation:

         

             2 mole of water is produced from 1 mole of    SnO₂  

           0.002 moles of water will be produced from \frac{0.002}{2}  = 0.001moles

To find the mass;

           Mass  = number of moles x molar mass

Molar mass of  SnO₂ = 118.7 + 2(16) = 150.7g/mol

        Mass  =  0.001 x 150.7 = 0.15g

3 0
3 years ago
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