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LiRa [457]
3 years ago
14

Why we used petrol for vehicles not water?

Chemistry
2 answers:
SashulF [63]3 years ago
5 0

because  the water  in itself does not produce energy

motikmotik3 years ago
4 0

Answer:

Water isn't combustible. There is nothing you can add to it other than gasoline that will make it even remotely combustible. Now, through electrolysis, it can be broken down into hydrogen and oxygen, which could be burned in an internal combustion engine.

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Elements that have the following qualities: Lustrous, Solid at Room Temperatures, They react and form Basic Compounds, Great con
Tatiana [17]

Answer:

They are classified as METALS.

Explanation:

Elements are simple substances that cannot be chemically broken down into smaller substances. Based on different characteristics, they are classified into 3 namely:

- metals

- non metals and

- metalloid( mainly act as semi- conductor).

METALS are the type of elements that loses electrons to form positive ion, that is, they are electropositive elements. They are distinguished by the following characteristics:

- LUSTROUS: they have the ability to reflect light from its surface.

- At room temperature: Metals are solid are room temperature with the exception of Mercury which is liquid at room temperature.

- They react and form Basic Compounds

- Great conductor: most metals are great conductors of heat and electricity because they possess free electrons.

- Melting Point: they have high melting points.

8 0
3 years ago
Sinkholes can form when _____?
Oxana [17]
Water flows under the ground
8 0
3 years ago
Read 2 more answers
At sea level the average heigh of mercury in barometer is
liubo4ka [24]
Although 1013.25 mb (760 mm Hg) is considered to be the standard atmospheric pressure at sea level, it does not mean that the pressure at this level has this value, actually this being 1011 mb.
4 0
4 years ago
How many molecules are in 20,484 grams of H2O?​
alexandr1967 [171]

Answer:

3.4027x10^-22

Explanation:

4 0
3 years ago
(b) Use the first law of thermodynamic to calculate AU for the following situations: (i) A coiled spring unwinds producing 153 J
il63 [147K]

Answer:

(i) ΔU = 116 J

(ii) ΔU = 289 J

(iii) ΔU = 1 KJ

(iv) ΔU = 0 J

(v) ΔU = 3.25 KJ

Explanation:

first law:

  • ΔU = Q + W

(i) W = 153 J;  Q = - 37 J ( Q ( - ), losing friction )

⇒ ΔU = 153 - 37 = 116 J

(ii) W = 289 J; Q = 0 ( insulated)

⇒ ΔU = W = 289 J

(iii) Q = 1 KJ , W = 0 ( isovolumetric process)

⇒ ΔU = Q = 1 KJ

(iv) isothermal ( constant temperature )

  • ΔU = Cv * ΔT

∴ ΔT = 0° ( isothermal )

⇒ ΔU = 0 J

(v) isobaric ( constant pressure )

⇒ ΔU = Q + W

∴ Q = 15.6 KJ

∴ W = - ∫ P dV = - P ΔV;  W (-) the system performs a job and the volume increases

.

∴ P = 950 KPa * ( 1000 Pa / KPa ) = 950000 Pa = 950000 J/m³

∴ ΔV = 18 - 5 = 13 L * ( m³ / 1000 L ) = 0.013 m³

⇒ W = - ( 950000 J/m³) * ( 0.013 m³ ) = - 12350 J ( - 12.35 KJ )

⇒ ΔU = 15.6 KJ + ( - 12.35 KJ )

⇒ ΔU = 3.25 KJ

7 0
3 years ago
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