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2 years ago

A compound is made of 49.31% C, 43.79% 0 and the rest hydrogen. The molar

Chemistry

1 answer:

Alexeev081 [22]2 years ago

8 0

Explanation:

49.31 + 43.79 = 93.10

finding the percentage of hydrogen in the compound hence 100-93.10 =6.9

therefore 6.9 percent of hydrogen is present in the compound

assuming that the compound is in 100 g so the elements present would be 49.31g of C , 43.79g of O and 6.9g of H

no. of moles of C = given mass÷molar mass

= 49.31 ÷ 12

=4.11 mol

no of moles of O = givem mass ÷molar mass

=43.79÷ 16

=2.73 mol

no. of moles of H =6.9÷1

=6.9 mol

dividing the no. of moles of each element by the least count of the elemnt to get the ratio of the simple whole numbr

therefore

so by dividing we get 1:1:2 of C:O:H

the empirical formula is CH2O

empirical formula mass of CH2O IS

= (12)+(2×1)+(16)

= 12+2+16

=30 G/MOL

n= molar mass of the compound ÷ empirical formula mass of the compound

so, n= 146.1 ÷30

n=5

molecular formula = n×empirical formula

molecular formula=C5H10O5 (ribose is the compound)

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Answer:

The catalyzed reaction will take time of $5.11\times 10^{-8} years$ .

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According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

The expression used with catalyst and without catalyst is,

$\frac{K_2}{K_1}=\frac{A\times e^{\frac{-Ea_2}{RT}}}{A\times e^{\frac{-Ea_1}{RT}}}$

$\frac{K_2}{K_1}=e^{\frac{Ea_1-Ea_2}{RT}}$

where,

$K_2$ = rate of reaction with catalyst

$K_1$ = rate of reaction without catalyst

$Ea_2$ = activation energy with catalyst = 59.0 kJ/mol = 59000 J/mol

$Ea_1$ = activation energy without catalyst = 184 kJ/mol = 184000 J/mol

R = gas constant

T = temperature = $600K$

Now put all the given values in this formula, we get:

$\frac{K_2}{K_1}=e^{\frac{184,000 kJ-59000 kJ}{R\times 300}}=7.632\times 10^{10}$

The reaction enhances by $7.632\times 10^{10}$ when catalyst is present.

Time taken by reaction without catalyzed = 3900 years

Time taken by reaction with catalyzed = x

$x=\frac{3900 year}{7.632\times 10^{10}}=5.11\times 10^{-8} years$

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How many electrons would an uncharged atom of nitrogen have

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Answer:

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