Answer is: mass of unused sulfur is 5.87 grams.
Balanced chemical reaction: C + 2S → CS₂.
m(C) = 12.0 g; mass of carbon.
m(S) = 70.0 g; mass of sulfur.
n(C) = m(C) ÷ M(C).
n(C) = 12 g ÷ 12 g/mol.
n(C) = 1 mol; amount of substance.
n(S) = m(S) ÷ M(S).
n(S) = 70 g ÷ 32.065 g/mol.
n(S) = 2.183 mol.
From chemical reaction: n(C) : n₁(S) = 1 : 2.
n₁(S) = 1 mol · 2 = 2 mol.
Δn(S) = n(S) - n₁(S).
Δn(S) = 2.183 mol - 2 mol.
Δn(S) = 0.183 mol; amount of unused sulfur.
Δm(S) = 0.183 mol · 32.065 g/mol.
Δm(S) = 5.87 g.
The positively charged protons would repel one another and fly apart, destroying the nucleus.
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do you mean fractional distillation of crude oil ?
if you do then this is what happens:
basically, crude oil is made up of a lot of hydrocarbons so it's put into a fracitonating column to separate these compounds into fractions. the top of the fractionating column is colder whereas the bottom of the column is hotter.
gases, petrol, diesel and kerosene come out near the top of the top of the column because they are shorter chains. substances near the top of the column also have a lower boiling point (since it's cooler and there's less intermolecular forces).
the substances at the top of the column are typically useful fuels because they have a higher ease of ignition (higher up=easier to ignite, lower down = harder to ignite). Also, substances at the top have a low viscosity so they can flow easily.
substances that come out near the bottom of the fracitonating column include fuel oil and bitumen.
these are longer chains of hydrocarbons and bitumen is used for road surfacing.
these substances have a high viscosity (harder to flow) and have a high boiling point since they have more intermolecular forces which require a lot of energy to break. Also the bottom of the fractionating column is warmer and this is where these fractions are released.
Answer:
4.2 x 10⁴ mL
Explanation:
Data Given:
Density (d) of air = 1.19 x 10⁻³g/mL
Mass of the air (m) = 50 g
Volume of the air (V) = ?
Solution:
Formula will be used
d = m/V
As we have to find volume so rearrange the above equation
V = m/d . . . . . . . . . . . (1)
Put values in above equation 1
V = 50 g / 1.19 x 10⁻³g/mL
V = 4.2 x 10⁴ mL
So,
volume of dry air = 4.2 x 10⁴ mL
