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3 years ago

Give the mass percent of carbon in C14H19NO2?

Chemistry

2 answers:

Pachacha [2.7K]3 years ago

7 0

72.0726 is the percent for carbon

denpristay [2]3 years ago

4 0

Answer:

The mass percent of carbon in C₁₄H₁₉NO₂ is 72.1%.

Explanation:

To calculate the mass percent of carbon in C₁₄H₁₉NO₂ we have to follow two simple steps:

1st) Look for the atomic weight of each element in the Periodic Table and then multiply each one by its subscript in the molecule and finally sum all of them to find the molar weight of the molecule:

Carbon atomic weight . Carbon subscript = 12 . 14 = 168g
Hydrogen atomic weight . Hydrogen subscript = 1 . 19 = 19g
Nitrogen atomic weight . Nitrogen subscript = 14 . 1 = 14g
Oxygen atomic weight . Oxygen subscript = 16 . 2 = 32g

The molar weight of C₁₄H₁₉NO₂ is: 168g + 19g + 14g +32g = 233g.

2nd) Find the mass percent of carbon with a Rule of three:

If 233g represents the 100% of mass, the 168g of carbon in the molecule will represents the 72.1% of mass.

The Rule of three is:

233g ------------- 100%

168g ------------- = (168 . 100)/233 = 72.1% mass of carbon.

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2 years ago

A. Based on the activation energies and frequency factors, rank the following reactions from fastest to slowest reaction rate, a

deff fn [24]

Answer:

A) $E_{a} = 350KJ/mol$ , $E_{a} = 50KJ/mol$ , $E_{a} = 50KJ/mol$

A = 1.5× $10^{-7}s^{-1}$ , A = 1.9× $10^{-7} s^{-1}$ , A=1.5× $10^{-7} s^{-1}$

B) 4.469

Explanation:

From Arrhenius equation

$K=Ae^{\frac{E_{a} }{RT} }$

where; K = Rate of constant

A = Pre exponetial factor

$E_{a}$ = Activation Energy

R = Universal constant

T = Temperature in Kelvin

Given parameters:

$E_{a} =165KJ/mol$

$T_{1}=505K$

$T_{2}=525K$

$R=8.314JK^{-1}mol^{-1}$

taking logarithm on both sides of the equation we have;

$InK=InA-\frac{E_{a} }{RT}$

since we have the rate of two different temperature the equation can be derived as:

$In(\frac{K_{2} }{K_{1} } )=\frac{E_{a} }{R}(\frac{1}{T_{1} } -\frac{1}{T_{2} } )$

$In(\frac{K_{2} }{K_{1} } )=\frac{165000J/mol}{8.314JK^{-1}mol^{-1} }.(\frac{1}{505} -\frac{1}{525} )$

$In(\frac{K_{2} }{K_{1} } )$ = 19846.04×7.544× $10^{-5}$ = 1.497

$\frac{K_{2} }{K_{1} }$ = $e^{1.497}$ = 4.469

6 0

3 years ago

Convert 3.30 g of copper (II) hydroxide Cu(OH)2 to molecules.

ratelena [41]

Answer:

0.18× 10²³ molecules

Explanation:

Given data:

Mass of copper hydroxide = 3.30 g

Number of molecules = ?

Solution:

Number of moles = mass/molar mass

Number of moles = 3.30 g/97.56 g/mol

Number of moles = 0.03 mol

Avogadro number:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance. The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ molecules

0.03 mol × 6.022 × 10²³ molecules / 1mol

0.18× 10²³ molecules

6 0

3 years ago

Which equation describes a reduction?

ira [324]

The answer choice is going to be B.

7 0

4 years ago

2CO + O2 --> 2002

olga_2 [115]

8 moles I think I’m not sure

4 0

3 years ago