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3 years ago

Hydrogen may not be advantagous as a fuel because-------

Chemistry

1 answer:

sergeinik [125]3 years ago

7 0

Answer:

Hydrogen may not be advantageous as a fuel because...

- Its expensive

- Its difficult to store

- Its highly flammable

- Its dependent on fossil fuels

Explanation:

Its expensive - Not only is hydrogen gas expensive, but it also takes a lot of work to free from other elements. It is both expensive and time-consuming to produce.

Its difficult to store - Moving hydrogen is not an easy task. Moving anything more than small amounts of hydrogen was also very expensive, making it impractical.

Its highly flammable - When exposed to the atmosphere, hydrogen could potentially form explosive mixtures.

Its dependent on fossil fuels - Hydrogen energy itself is renewable. However, the process of separating it from oxygen uses non-renewable sources such as coal and oil.

~Hope this Helps!~

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Which process is similar to binary fission?

elixir [45]

Fission is similar to mitosis because they both involve splitting.

Answer: d. mitosis

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3 years ago

A 237g sample of molybdnum metal is heated to 100.1 0C and then dropped into an insulated cup containing 244 g of water at 10.0

miv72 [106K]

Answer:

The specific heat of molybdenum is 0.254 joules per gram-Celsius.

Explanation:

We consider the system formed by the molybdenum metal and water as our system, a control mass inside an insulated cup, that is, a container that avoids any energy and mass interactions between system and surroundings.

From statement we notice that metal is cooled down whereas water is heated. According to the First Law of Thermodynamics, we know that:

$Q_{metal} - Q_{water} = 0$

$Q_{metal} = Q_{water}$

Where:

$Q_{water}$ - Heat received by water, measured in joules.

$Q_{metal}$ - Heat released by metal, measured in joules.

Now we expand this identity by definition of sensible heat:

$m_{metal}\cdot c_{metal}\cdot (T_{m,o}-T) = m_{water}\cdot c_{water}\cdot (T-T_{w,o})$

The specific heat of the metal is cleared within equation above:

$c_{metal} = \frac{m_{water}\cdot c_{water}\cdot (T-T_{w,o})}{m_{metal}\cdot (T_{m,o}-T)}$

If we know that $m_{water} = 0.237\,kg$ , $m_{metal} = 0.244\,kg$ , $c_{water} = 4186\,\frac{J}{kg\cdot ^{\circ}C}$ , $T_{w,o} = 10\,^{\circ}C$ , $T_{m,o} = 100.10\,^{\circ}C$ and $T = 15.30\,^{\circ}C$ , the specific heat of molybdenum is:

$c_{metal} = \frac{(0.237\,kg)\cdot \left(4186\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (15.30\,^{\circ}C-10\,^{\circ}C)}{(0.244\,kg)\cdot (100.10\,^{\circ}C-15.30\,^{\circ}C)}$

$c_{metal} = 254.119\,\frac{J}{kg\cdot ^{\circ}C}$

The specific heat of molybdenum is 0.254 joules per gram-Celsius.

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3 years ago

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mars1129 [50]

Answer:

Explanation:

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2 years ago

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Which of these molecules is nonpolar? a.CH3CI b.CO c.O2 d.PF3

valentina_108 [34]

Answer:

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Explanation:

7 0

2 years ago

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Calculate the volume 3.00 moles of a gas will occupy at 24.0˚C and 1.003 atm. *

vitfil [10]

Answer: 72.93 litres

Explanation:

Given that:

Volume of gas (V) = ?

Temperature (T) = 24.0°C

Convert 24.0°C to Kelvin by adding 273

(24.0°C + 273 = 297K)

Pressure (P) = 1.003 atm

Number of moles (n) = 3 moles

Molar gas constant (R) is a constant with a value of 0.0821 atm L K-1 mol-1

Then, apply ideal gas equation

pV = nRT

1.003 atm x V = 3.00 moles x 0.0821 atm L K-1 mol-1 x 297K

1.003 atm•V = 73.15 atm•L

Divide both sides by 1.003 atm

1.003 atm•V/1.003 atm = 73.15 atm•L/1.003 atm

V = 72.93 L

Thus, the volume of the gas is 72.93 litres

5 0

3 years ago