The constituents of petroleum that are used for the following purposes are as follows:
- To make candles ----- Paraffin wax
- A solvent for dry cleaning ----- Petrol
- For surfacing roads ----- Bitumen
- Jet engine fuel ----- Kerosene
- For lubrication ----- Lubricating oil.
<h3>What are the constituents of petroleum?</h3>
The constituents of petroleum are LPG, bitumen, paraffin wax, lubricating oil, kerosene, diesel, etc. These compounds are a mixture of hydrocarbons.
Therefore, each constituent of petroleum that is used for the following purposes is mentioned above with proper names.
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Answer:
4 meters
Explanation:
4 centimeters and millimeters are too small, while 4 kilometers is too large.
Answer:
The chemical formula for potassium sulfide is K2 S. Sulfur atoms need two electrons to fill their 3p electron orbitals.
Explanation:
Answer:
6.82 kg
Explanation:
Given that the amount of water is 15L and we know that the density of water is ≈ 1kg/L. The mass of water is given by mass = volume x density, i.e,
mass = 15 x 1 = 15 kg. Also the specific heat capacity of water is 4.186 KJ/kg.
The sublimation enthalpy of dry ice is 571 KJ/kg.
Now, the amount of heat lost by water is entirely used up for the sublimation (conversion from soild to gas) of dry ice. And the heat (Q) lost by water is given as : Q = mCΔT, where m is the mass of water, C the specific heat capacity of water and ΔT the change in temperature.
Here, Q = 15 x 4.186 x (90 - 28) = 3892.98 KJ.
This amount of heat is taken up by the dry ice for its sublimation. Also the energy taken by dry ice (Q') for its sublimation is given by: Q' = m'L', where m' is the mass of dry ice, L' is the latent heat of sublimation (i.e, the amount of heat required per kg of a substance to sublime) of dry ice amd L' = 571 KJ/kg.
Now, Q' =m'L' = heat lost by water = 3892.98KJ.
And, m'L' = m' x 571 KJ/kg = 3892.98 KJ. (Dividing with 571)
Therefore, m' = 6.82 kg.