Answer:
3.51 g of oxygen per gram of gasoline is required.
Explanation:
Solution:
First of all we will write the balance chemical equation.
C8H18 + 12.5O2 → 8CO2 + 9H2O
This equation shows that,
1 mole of gasoline react with 12.5 mole of oxygen for complete burning.
mass of one mole of gasoline = 8×12 + 18×1 = 114 g
mass of 12.5 mole of oxygen = 12.5 (16×2) = 400 g
Formula:
mass of oxygen per gram of gasoline = (400 / 114) = 3.51
so, 3.51 g of oxygen require for per gram of gasoline.
Three of the intermolecular forces of attraction are roughly equal, the substances<span> will be </span>soluble in each other, <span>This means that </span>ionic<span> or </span>polar<span> solutes </span>dissolve<span> in </span><span>polar solvents</span>
Explanation:
20.0 moles= 80.1 or 80.05g
5.00 moles= 20.0g
1.20×1025moles= 4923.2g
1.00 moles= 4.00g
80.0 moles= 320.2g
Option B, C and D are correct.
<h3><u>Explanation:</u></h3>
Carbon is a non metallic element with the atomic number 6 and mass number 12. Whereas oxygen is also a non metallic element with the atomic number 8 and mass number of 16.
In carbon monoxide, the mass ratio of oxygen to carbon is 16:12 =1.33. This ratio is same for every sample of carbon monoxide, because carbon monoxide has the universal formula of CO.
Similarly In carbon dioxide , the mass ratio of oxygen to carbon is 32:12 =2.667. This ratio is same for every sample of carbon dioxide, because carbon dioxide has the universal formula of
.
Even we can see that the mass ratio of oxygen to carbon in carbon dioxide is just twice the mass ratio of oxygen to carbon in carbon monoxide, because of presence of twice as much as oxygen per molecule of carbon dioxide than carbon monoxide.