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3 years ago

Can you imagine a situation in which the reserve of a fuel might be increased?

2 answers:

6 0

The reserve of a fossil fuel might be increased if mining or extraction techniques improve so that more of fuel can be extracted for a profit

hammer [34]3 years ago

6 0

The reserve of a fossil fuel could be increased if mining or the extraction techniques of removing fossil fuels improve so that more of fuel can be extracted for more profit.

You might be interested in

If there's an oil spill and there's a fire in the ocean why can't the ocean water just put out the fire?

Lesechka [4]

Answer:

cause the oil is on fire nit the water oil is less dense than water so it floats on top of it

Explanation:

I hope I helped you

6 0

2 years ago

When 6.13 g of a certain molecular compound X are dissolved in 90. g of formamide (NH2COH), the freezing point of the solution i

belka [17]

Answer:

321.6 g/Mol

Explanation:

mass of solvent in kilograms = 90g/1000 = 0.09 Kg

Given that;

ΔTf = Kf . m . i

Where;

Kf = freezing point constant = 4.25 °C/Kg mol

m = molality of the solution

i = Van't Hoff factor = 1 (since the substance is molecular)

ΔTf = freezing point of pure solvent - freezing point of solution

freezing point of pure solvent = 3 °C

ΔTf = 3 °C - 2.1 °C

ΔTf = 0.9 °C

0.9= 4.25 * 6.13/M/0.09 * 1

0.9= 26.0525/M * 1/0.09

0.9 = 26.0525/0.09 M

0.9 * 0.09M = 26.0525

M = 26.0525/0.9 * 0.09

M= 321.6 g/Mol

8 0

3 years ago

How much heat, in kJ, will be absorbed by a 25.0 g piece of aluminum (specific heat = 0.930 J/g･°C) as it changes temperature fr

faust18 [17]

Answer:

quantity of heat=mc*theta

=25*0.930(76-25)

=25*0.930*51

=1185.75J

=11.9kJ

4 0

2 years ago

In the laboratory a student combines 47.8 mL of a 0.321 M aluminum nitrate solution with 21.8 mL of a 0.366 M aluminum iodide so

alex41 [277]

Answer: The final concentration of aluminum cation is 0.335 M.

Explanation:

Given: $V_{1}$ = 47.8 mL (1 mL = 0.001 L) = 0.0478 L

$M_{1}$ = 0.321 M, $V_{2}$ = 21.8 mL = 0.0218 L, $M_{2}$ = 0.366 M

As concentration of a substance is the moles of solute divided by volume of solution.

Hence, concentration of aluminum cation is calculated as follows.

$[Al^{3+}] = \frac{M_{1}V_{1} + M_{2}V_{2}}{V_{1} + V_{2}}$

Substitute the values into above formula as follows.

$[Al^{3+}] = \frac{M_{1}V_{1} + M_{2}V_{2}}{V_{1} + V_{2}}\\= \frac{0.321 M \times 0.0478 L + 0.366 M \times 0.0218 L}{0.0478 L + 0.0218 L}\\= \frac{0.0153438 + 0.0079788}{0.0696}\\= 0.335 M$