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Svetach [21]
3 years ago
11

Why is hydrogen not used as fuel

Chemistry
2 answers:
meriva3 years ago
7 0

Answer:

so just like type anything right?

Explanation:

PIT_PIT [208]3 years ago
5 0

Answer: First, hydrogen is not as energy-dense as other fuels, meaning that you need a whole lot of it to do a little bit of work. ... Fuel cells are far more efficient than internal combustion engines, and a hydrogen fuel cell has cleaner emissions than an internal-combustion hydrogen engine.

Explanation:

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What volume of 12 M HCl solution is needed to make 2.5 L of 1.0 M HCI?
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0.21 L of 12 M HCL

Explanation:

CV=CV

(12)(x)=(1.0)(2.5)

x=0.21

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The process in which the nucleus of an atom splits into two lighter atoms, releasing a large amount of energy, is nuclear .
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It's nuclear fission

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According to this balanced equation, how many grams of water (H.0) form in
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your simpal answer is 177.32

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Las Vegas Raiders' running back Josh Jacobs is 72 inches tall and has mass of 100 kg. Los
gladu [14]

Answer:

Josh is running at a speed of 9.09 yards per second.

Josh's velocity is 9.09 East.

Josh's force is 900 N.

Aaron Donald's force is 845 N.

Yes Josh scores the touchdown because he is faster and has more mass than Donald.

Explanation:

Josh scores the touchdown as he is heavier and faster than Donald.

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Force= mass x acceleration

Speed= distance divided by time.

Velocity= distance divided by time.

4 0
3 years ago
Calculate the amount of heat in kJ that is required to heat 25.0 g of ice from -25 °C to 105 °C in a closed vessel and sketch a
kolezko [41]

Answer:

The total amount of heat required for the process is 76.86 KJ

Explanation:

We can divide the process in 5 parts, in which we can calcule each amount of heat required (see attached Heating curve):

(1) Ice is heated from -25ºC to 0ºC. We can calculate the heat of this part of the process as follows. Note that we must convert J in KJ (1 KJ= 1000 J).

Heat (1) = mass ice x Specific heat ice x (Final temperature - Initial Temperature)

Heat (1) =25 g x 2.11 J/g.ºC x \frac{1 KJ}{1000 J} x (0ºC-(-25º)

Heat (1) = 1.32 KJ

(2) Ice melts at ºC (it becomes liquid water). This is heating at constant temperature (ºC), so we use the melting enthalphy (ΔHmelt) and we must use the molecular weight of water (1 mol H₂O = 18 g):

Heat (2) = mass ice x ΔHmelt

Heat (2)= 25 g  x  \frac{6.01KJ} {1 mol H2O} x \frac{1 mol H2O}{18 g}

Heat (2)= 8.35 KJ

(3) Liquid water is heated from 0ºC to 100 ºC:

Heat (3)= mass liquid water x Specific heat water x (Final T - Initial T)

Heat (3)= 25 g x 4.18 J/gºC x 1 KJ/1000 J x (100ºC - 0ºC)

Heat (3)= 10.45 KJ

(4) Liquid water evaporates at 100ºC (it becomes water vapor). This is a process at constant temperature (100ºC), and we use boiling enthalpy:

Heat (4)= mass water x ΔH boiling

Heat (4)= 25 g x \frac{40.67 KJ}{mol H20} x \frac{1 mol H20}{18 g}

Heat (4)= 56.49 KJ

(5) Water vapor is heated from 100ºC to 105ºC. We use the specific capacity of water vapor:

Heat (5)= mass water vapor x Specific capacity vapor x (Final T - Initial T)

Heat (5)= 25 g x 2.00 J/g ºC x 1 KJ/1000 J x (105ºC - 100ºC)

Heat (5)= 0.25 KJ

Finally, we calculate the total heat involved in the overall process:

Total heat= Heat(1) + (Heat(2) + Heat(3) + Heat(4) + Heat(5)

Total heat= 1.32 KJ + 8.35 KJ + 10.45 KJ + 56.49 KJ + 0.25 KJ

Total heat= 76.86 KJ

3 0
3 years ago
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