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11111nata11111 [884]
3 years ago
11

Why is cracking a necessary part of the refinement of crude oil?

Chemistry
1 answer:
Yanka [14]3 years ago
3 0

Answer:

Option C.because the demand for the mid-range hydrocarbons (C7-C12) found in gasoline exceeds the amount produced by the distillation of crude oil.

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Which of the following is a homogeneous mixture?A) Molten ironB) Stainless steelC) Trail mixD) Water
ehidna [41]

Answer:

B) Stainless steel

Explanation:

Homogeneous mixture  -

It refers to the composition of various substance , which leads to a mixture of complete uniform composition , i.e. , the ratio of each and every component of the mixture is constant all over the mixture , is referred to as a homogeneous mixture.

For example ,

In the solution of sugar and water , where 1 tablespoon of sugar is dissolved in 100 mL of water ,

The ratio of sugar and water remains constant all over the mixture.

Hence, from the given options the homogeneous mixture is B) Stainless steel .

5 0
3 years ago
Which of these is not a challenge an organism encounters
bekas [8.4K]
Staying healthy
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6 0
4 years ago
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What are the properties of the reactants?​
Galina-37 [17]

Answer:

Reactants starts the chemical reaction and gives right to the product either by thermal decomposition or in the presence of a catalyst.

Explanation:

Reactants and products contain the same atoms, but they are rearranged during the reaction, so reactants and products are different substances.

6 0
3 years ago
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Empirical Formula of P3O4H2?
puteri [66]

Answer:

H2 P4 O1. Explanation: In order to calculate the Empirical formula , we will assume that we have started with 10 g of the compound.

Explanation:

3 0
3 years ago
A fuel-air mixture is placed in a cylinder fitted with a piston. The original volume is 0.310-L. When the mixture is ignited, ga
Soloha48 [4]

Answer:

V_2=11.35L

Explanation:

Knowing that the system is at constant pressure, the energy balance is:

\Delta H=Q-W

If all the energy (enthalpy of combustion) is transformed into work:

\Delta H=-W

Work:

W=-\int_{V1}^{V2}P*dV

W=-P*(V_2-V_1)

\Delta H=P*(V_2-V_1)

Calculating:

935J=84659 Pa*(V_2-3.1*10^{-4}m^3)

V_2=0.0113 m^3=11.35L

3 0
3 years ago
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