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2 years ago

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A 55-g sample of a gaseous fuel mixture contains 0.43 mole fraction propane; the remainder of the mixture is butane. What are th

e masses of propane and butane in the sample?

1 answer:

Andru [333]2 years ago

7 0

I know if you swallow gasoline you will probably die.

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Glade air freshener gel “disappearing” is an example of

netineya [11]

Answer:

Vaporization

Explanation:

Vaporization is the change of a specie to the gaseous state. To 'disappear' in this case simply means to change to the gaseous state.

Substances with high vapour pressure tend to be easily converted to vapour phase. Hence if Glade air freshener gel 'disappears' easily, then it has a high vapour pressure and is easily converted to vapour (gas).

8 0

1 year ago

A 0.1153-gram sample of a pure hydrocarbon was burned in a C-H combustion train to produce 0.3986 gram of CO2and 0.0578 gram of

Mars2501 [29]

<u>Answer:</u> The mass of carbon and hydrogen in the sample is 0.1087 g and 0.0066 g respectively and the percentage composition of carbon and hydrogen in the sample is 94.27 % and 5.72 % respectively.

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon and hydrogen follows:

$C_xH_y+O_2\rightarrow CO_2+H_2O$

where, 'x' and 'y' are the subscripts of carbon and hydrogen respectively.

We are given:

Mass of $CO_2=0.3986g$

Mass of $H_2O=0.0578g$

Mass of sample = 0.1153 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.3986 g of carbon dioxide, $\frac{12}{44}\times 0.3986=0.1087g$ of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

So, in 0.0578 g of water, $\frac{2}{18}\times 0.0578=0.0066g$ of hydrogen will be contained.

To calculate the percentage composition of a substance in sample, we use the equation:

$\%\text{ composition of substance}=\frac{\text{Mass of substance}}{\text{Mass of sample}}\times 100$ ......(1)

<u>For Carbon:</u>

Mass of sample = 0.1153 g

Mass of carbon = 0.1087 g

Putting values in equation 1, we get:

$\%\text{ composition of carbon}=\frac{0.1087g}{0.1153g}\times 100=94.27\%$

<u>For Hydrogen:</u>

Mass of sample = 0.1153 g

Mass of hydrogen = 0.0066 g

Putting values in equation 1, we get:

$\%\text{ composition of hydrogen}=\frac{0.0066g}{0.1153g}\times 100=5.72\%$

Hence, the mass of carbon and hydrogen in the sample is 0.1087 g and 0.0066 g respectively and the percentage composition of carbon and hydrogen in the sample is 94.27 % and 5.72 % respectively.

8 0

1 year ago

2.A solid block with a length of 6.0 cm, a width of 3.0 cm, and a height of 3.0 cm has a mass of 146 g. What is the block’s dens

posledela

2: <span>Volume V = a*b*c = 6.0*3.0*3.0 = 54.0 cm^3 density ρ = mass/volume = 146/54 = 2.70 g/cm^3
3: Volume = (27.8 -21.2) cm^3
mass = 22.4 g
density = 22.4/(27.8-21.2) g/cm^3

</span>

3 0

2 years ago

Read 2 more answers

Titanium dioxide, TiO₂, reacts with carbon and chlorine to give gaseous TiCl₄: TiO₂+2C+2CI₂−TiCI₄+2CO The reaction of 7.39 kg ti

Dima020 [189]

Answer:

17.57kg of $TiCl_{4}$ and its percentage yield is 81.0%

Explanation:

Through the reaction you can get the theoretical amount of $TiCl_{4}$ that must be produced.

$7.39kgTiO_{2}x\frac{1kmolTiO_{2} }{79.867kgTiO_{2}}x \frac{1kmolTiCl_{4}}{1kmolTiO_{2}}x\frac{189.867kgTiCl_{4} }{1kmolTiCl_{4}}=17.57kgTiCl_{4}$

If the amount obtained is less than the theoretical amount, it means that the initial sample was not 100% pure. Now the actual amount obtained is compared with the theoretical amount using a percentage

$yield=\frac{actual amount}{theoretical amount}x100= \frac{14.24kg}{17.57kg}x100$ =81.0%

3 0

2 years ago

When uranium-235 atoms undergo fission, ________ is/are produced?

elena-s [515]

Smaller atoms ; free neutrons and energy

8 0

1 year ago