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Sophie [7]
3 years ago
13

A compound with a molar mass of 60g/mol is 40.4% carbon, 6.7% hydrogen and 53.3% oxygen (by mass). determine the emperical and m

olecular formulas
Chemistry
1 answer:
Fittoniya [83]3 years ago
4 0

<span>A compound is found to be 40.0% carbon, 6.7% hydrogen and 53.5% oxygen. Its molecular mass is 60. g/mol. 
</span>Q1)
Empirical formula is the simplest ratio of whole numbers of components making up a compound.
the percentages have been given, therefore we can calculate for 100 g of the compound.

                                C                            H                        O
Mass in 100 g      40.0 g                       6.7 g                   53.5 g
Molar mass            12 g/mol                1 g/mol                 16 g/mol
Number of moles   40.0/12= 3.33         6.7/1 = 6.7          53.5/16 = 3.34
Divide by the least number of moles  
                             3.33/3.33 = 1           6.7/3.33 = 2.01   3.34/3.33 = 1.00
after rounding off
C - 1 
H - 2
O - 1

Empirical formula - CH₂O

Q2)
Molecular formula is the actual number of components making up the compound.
To find the number of empirical units we have to find the mass of one empirical unit.
Mass of one empirical unit = CH₂O - 12 + (1x2) + 16 = 30 g
Mass of one mole of compound = 60 g
Number of empirical units = 60 g / 30 g = 2
Therefore molecular formula - 2(CH₂O) 
 Molecular formula - C₂H₄O₂
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A .60 L sample of nitrogen is heated from 27℃ to 77℃ at a constant pressure. What is the final volume of the gas?
Anon25 [30]

Answer:

CHARLES' LAW

given:

= 600 mL = 0.6 L

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conversion:

= 600 mL (1 L / 1000 mL)

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= ( × ) ÷

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4 0
3 years ago
When 0.560 g of na(s) reacts with excess f2(g) to form naf(s), 13.8 kj of heat is evolved at standard-state conditions. what is
sergejj [24]
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3 years ago
Determine the rate law and calculate the rate constant for the chemical reaction.
Soloha48 [4]

Answer:

The rate law is [B]

Explanation:

In Trials 1 and 2, the concentration of B changes and A is the same so you can see how changes in B affect the rate. In this case, 0.300/0.150=2 and 7.11 x 10^-4 / 3.56 x 10^-4= 2. Since there 2^1=2, we can say that the reaction order of B is 1.

Similarly, if we look at trials 2 and 3, the concentration of B is constant, while A is changing. In this case, the rate has not changed at all with a change in concentration of A, so this means that A has 0 order.

Therefore, the rate law is just [B].

6 0
3 years ago
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