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4 years ago

Octane C8H18 is an ingredient in gasoline. How many carbon atoms are in 20 kg of octane? help please

Chemistry

2 answers:

abruzzese [7]4 years ago

4 0

Answer:

Explanation:

Octane is C8H18, carbon is C. Therefore, for every mol of C8H18 there will be 8 mols of carbon.

Convert 20 kg of octane to mols octane using molar mass of octane.

Molar Mass octane = (8*12.011)+(18*1.008)=114.23 g/mol, where 8 and 18 are the atoms in the molecule and 12.011 and 1.008 are the respective molar masses of carbon and hydrogen.

20kg C8H18 *1000g/kg * mol C8H18/114.23g C8H18 = 175.1 mols octane.

Now convert that number of mols to mols carbon:

175.1 mol C8H18 * (8 mol C/mol C8H18)=1400.8 mol C.

Now remember that a mol of any given atom=6.022*10^23 atoms.

So 1400.8 mol C=1400.8*(6.022*10^23) atoms C, which is equal to your final answer, 8.4*10^26 atoms. (round to 2 significant digits since the initial number given, 20, has 2 significant digits.)

The main thing you want to look for here is your conversion factors. Once you have those down it is simple algebra from there, so be sure to practice those!

eduard4 years ago

3 0

Answer:

Octane is a hydrocarbon and an alkane with the chemical formula C 8 H 18, and the condensed structural formula CH 3 (CH 2) 6 CH 3.Octane has many structural isomers that differ by the amount and location of branching in the carbon chain. One of these isomers, 2,2,4-trimethylpentane (commonly called iso-octane) is used as one of the standard values in the octane rating scale.

Chemical formula: C₈H₁₈

Molar mass: 114.232 g·mol−1

Melting point: −57.1 to −56.6 °C; −70.9 to −69.8 °F; 216.0 to 216.6 K

Solubility in water: 0.007 mg dm−3 (at 20 °C)

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tigry1 [53]

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4 years ago

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Fynjy0 [20]

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3 years ago

A 0.9440 g sample of a mixture of NaCl and KCl is dissolved in water, and the solution is then treated with an excess of AgNO3 t

Gnom [1K]

Answer : The percent by mass of NaCl and KCl are, 18.11 % and 81.88 % respectively.

Explanation :

As we know that when a mixture of NaCl and KCl react with excess $AgNO_3$ then the silver ion react with the chloride ion in both NaCl and KCl to form silver chloride.

Let the mass of NaCl be, 'x' grams and the mass of KCl will be, (0.9440 - x) grams.

The molar mass of NaCl and KCl are, 58.5 and 74.5 g/mole respectively.

First we have to calculate the moles of NaCl and KCl.

$\text{Moles of }NaCl=\frac{\text{Mass of }NaCl}{\text{Molar mass of }NaCl}=\frac{xg}{58.5g/mole}=\frac{x}{58.5}moles$

$\text{Moles of }KCl=\frac{\text{Mass of }KCl}{\text{Molar mass of }KCl}=\frac{(0.9440-x)g}{74.5g/mole}=\frac{(0.9440-x)}{74.5}moles$

As, each mole of NaCl and KCl gives one mole of chloride ions.

So, moles of chloride ions in NaCl = $\frac{x}{58.5}moles$

Moles of chloride ions in KCl = $\frac{(0.9440-x)}{74.5}moles$

The total moles of chloride ions = $\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles$

Now we have to calculate the moles of AgCl.

As we know that, this amount of chloride ion is same as the amount chloride ion present in the AgCl precipitate. That means,

Moles of AgCl = Moles of chloride ion = $\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles$

Now we have to calculate the moles of AgCl.

The molar mass of AgCl = 143.32 g/mole

$\text{Moles of }AgCl=\frac{\text{Mass of }AgCl}{\text{Molar mass of }AgCl}=\frac{1.903g}{143.32g/mole}=0.0133moles$

Now we have to determine the value of 'x'.

Moles of AgCl = $\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles$

0.0133 mole = $\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles$

By solving the term, we get the value of 'x'.

$x=0.171g$

The mass of NaCl = x = 0.171 g

The mass of KCl = (0.9440 - x) = 0.9440 - 0.171 = 0.773 g

Now we have to calculate the mass percent of NaCl and KCl.

$\text{Mass percent of }NaCl=\frac{\text{Mass of }NaCl}{\text{Total mass of mixture}}\times 100=\frac{0.171g}{0.9440g}\times 100=18.11\%$

$\text{Mass percent of }KCl=\frac{\text{Mass of }KCl}{\text{Total mass of mixture}}\times 100=\frac{0.773g}{0.9440g}\times 100=81.88\%$

Therefore, the percent by mass of NaCl and KCl are, 18.11 % and 81.88 % respectively.

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4 years ago

Your body uses about 24 grams of glucose to run cellular respiration. How many molecules does your body need for cellular respir

stealth61 [152]

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3 years ago

A solution is prepared by adding .8 g of iron sulfate to 450mL of water

Anastasy [175]

Molar mass FeSO₄ =151.908 g/mol

Volume = 450 mL / 1000 = 0.45 L

1 mol ----------- 151.908 g
n mol ----------- 0.8 g

n = 0.8 * 1 / 151.908

n = 0.00526 moles

M = n / V

M = 0.00526 / 0.45

M = 0.01168 mol/L⁻¹

hope this helps!

7 0

4 years ago