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Tpy6a [65]
3 years ago
13

A compound contains only carbon, hydrogen, and oxygen. Combustion of 139.1 g of the compound yields 208.6 g of CO2 and 56.93 g o

f H2O. The molar mass of the compound is 176.1 g/mol. *Each part of this problem should be submitted separately to avoid losing your work* 1. Calculate the grams of carbon (C) in 139.1 g of the compound: grams 2. Calculate the grams of hydrogen (H) in 139.1 g of the compound. grams 3. Calculate the grams of oxygen (O) in 139.1 g of the compound. grams
Chemistry
1 answer:
Fofino [41]3 years ago
3 0

Answer:

1. Mass of Carbon is 56.89g

2. Mass of Hydrogen is 6.33g

3. Mass of Oxygen is 75.88

Explanation:

The following were obtained from the question.

Mass of the compound = 139.1g

Mass of CO2 produced = 208.6g

Mass of H2O produced = 56.93

1. Determination of mass of Carbon (C). This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C = 12/44 x 208.6

Mass of C = 56.89g

2. Determination of the mass of Hydrogen (H). This is illustrated below:

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of H = 2/18 x 56.93

Mass of H = 6.33g

3. Determination of the mass of oxygen (O).

This is illustrated below:

Mass of the compound = 139.1g

Mass of C = 56.89g

Mass of H = 6.33g

Mass of O = Mass of compound - (mass of C + Mass of H)

Mass of O = 139.1 - (56.89 + 6.33)

Mass of O = 139.1 - 63.22

Mass of O = 75.88

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How many grams of water will be produced from 50 g hydrogen reacting with 50 g oxygen?
DanielleElmas [232]

We need to first come up with a balanced equation:


4H+O_{2} → 2H_{2}O


We know that the molar ratio of hydrogen to oxygen to water now is 4:1:2.


Converting the amount of grams given to moles is as follows:


Hydrogen: \frac{1mole}{1.008g}*50g=49.6mol


Oxygen: \frac{1mole}{15.999g}*50g=3.125mol


We know now that the limiting reactant is oxygen. We can then know that the number of moles of water are produced are double the number of moles of oxygen used due to the ratio that we established at the beginning - 4:1:2.


So we now can use 6.25 moles of water as the amount produced.


Then we convert moles of water to grams:


\frac{18.015g}{1mole} *6.25mol=112.59g


Now we know that there are 112.59g of water produced when we start with 50g of hydrogen and 50g of water.

4 0
2 years ago
the element carbon has two common isotopes: C-12 (12 U) AND C-13 (13.003355 U). IF THE AVERAGE ATOMIC MASS OF CARBON IS 12.0107
oksian1 [2.3K]

Answer:

The percent isotopic abundance of C- 12 is 98.93 %

The percent isotopic abundance of C- 13 is 1.07 %

Explanation:

we know there are two naturally occurring isotopes of carbon, C-12 (12u)  and C-13 (13.003355)

First of all we will set the fraction for both isotopes

X for the isotopes having mass 13.003355

1-x for isotopes having mass 12

The average atomic mass of carbon is 12.0107

we will use the following equation,

13.003355x + 12 (1-x) = 12.0107

13.003355x + 12 - 12x = 12.0107

13.003355x- 12x = 12.0107 -12

1.003355x = 0.0107

x= 0.0107/1.003355

x= 0.0107

0.0107 × 100 = 1.07 %

1.07 % is abundance of C-13 because we solve the fraction x.

now we will calculate the abundance of C-12.

(1-x)

1-0.0107 =0.9893

0.9893 × 100= 98.93 %

98.93 % for C-12.

7 0
3 years ago
Calculate in gramm the mass of 0.1 mole of hydrochloric acid(h=1 ,cl=35.5)​
babunello [35]

Answer:

mass of HCl = 3.65 g

Explanation:

Data Given:

Moles of hydrochloric acid HCl = 0.1 mole

Mass in grams of hydrochloric acid HCl = ?

Solution:

Mole Formula

                  no. of moles = Mass in grams / molar mass

To find Mass in grams rearrange the above Formula

                Mass in grams = no. of moles x molar mass . . . . . . . (1)

Molar mass of HCl = 1 + 35.5 = 36.5 g/mol

Put values in equation 1

              Mass in grams = 0.1 mole x 36.5 g/mol

              Mass in grams = 3.65 g

mass of HCl = 3.65 g

6 0
3 years ago
The specific heat of copper metal is 0. 385 J/(g °C). How much energy must be added to a 35. 0-gram sample of copper to change t
Rus_ich [418]

The amount of heat required for changing the temperature of copper has been 606 J. Thus, option B is correct.

Specific heat has been defined as the amount of heat required to raise the temperature of 1 gram of substance by 1 degree Celsius.

The heat required to raise the temperature has been expressed as:

\rm Heat=mass\;\times\;specific\;heat\;\times\;Change\;in\;temperature

<h3>Computation for the heat energy required</h3>

The given specific heat of copper has been \rm 0.385\;J/g^\circ C

The mass of copper has been, \rm 35\;g

The initial temperature of copper has been, \rm 20^\circ C

The final temperature of copper has been, \rm 65^\circ C

The change in temperature has been, \Delta T

\Delta T=\text{Final\;temperature-Initial\;temperature}\\\Delta T =65^\circ \text C-20^\circ \text C\\\Delta T=45^\circ \text C

Substituting the values for the heat required as:

\rm Heat=35\;g\;\times\;0.385\;J/g^\circ C\;\times\;45^\circ C\\Heat=606\;J

The amount of heat required for changing the temperature of copper has been 606 J. Thus, option B is correct.

Learn more about specific heat, here:

brainly.com/question/2094845

7 0
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True.

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