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levacccp [35]
4 years ago
13

How do we turn petroleum into energy we use?

Chemistry
1 answer:
arlik [135]4 years ago
7 0

You don't "turn" it into energy; petroleum HAS stored energy (chemical energy).However, you can turn it into ANOTHER TYPE OF ENERGY; usually this is done by burning the petroleum, and using it to drive machinery.

Since burning fuels is wasteful (the efficiency is limited, in theory, to the Carnot efficiency of a heat engine), other options are being explored, such as chemical reactions in a fuel cell. But such technology is not yet used on a large scale.

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Why is the wavelength of 633 nm used to analyze the standard solutions and drink samples?
DanielleElmas [232]

Answer:

The correct option is;

A. Corresponding with orange light, is the wavelength that blue solutions absorb the most

Explanation:

The wavelength of 633 nm is the wavelength of red light emitted by the helium-neon laser also known as the He-Ne laser

According to absorption spectroscopy and the color wheel, a solution or drink sample that is efficient at absorbing light having a wavelength of 633 nm will appear blue, because, light at 633 nm is red and very close to orange which is the complementary color blue light.

4 0
3 years ago
When a solid becomes a liquid, its particles are _____.
Alexeev081 [22]

Answer:

Further apart

Explanation:

7 0
3 years ago
The following data are obtained for the decomposition of N2O5 at a certain temperature: 2N2O5(g) ↔ 4NO2(g) + O2(g) Time(s) 0 200
cricket20 [7]

Answer:

K = 0.0003 s⁻¹

Explanation:

The reaction is

2N₂O₅ = 4NO₂ + O₂

The equilibrium is

K[N_{2} O_{5} ]=-\frac{1}{2} \frac{d[N_{2} O_{5} ]}{dt} \\\int\limits^y_x {d[N_{2} O_{5} ]}{} \, =-2k\int\limits^t_0 {dt} \,\\ln[N_{2} O_{5} ]=-2kt+ln[N_{2} O_{5} ]

a plot is attached

From the plot, the slope is -0.0006

-2 * K = -0.0006

K = 0.0003 s⁻¹

Download xlsx
6 0
3 years ago
. How much energy is lost by a 30.0g sample of water that decreases in temperature from 56.7C to 25.0C?
lbvjy [14]

Answer:

Q = -3980.9 j

Explanation:

Given data:

Mass of sample = 30 g

Initial temperature = 56.7 °C

Final temperature = 25 °C

Specific heat of water = 4.186 j/g.°C

Amount of heat released = ?

Formula:

Q = m.c.ΔT

Q = heat released

m = mass of sample

c = specific heat of given sample

ΔT = change in temperature

Solution:

ΔT = T2 -T1

ΔT = 25 °C - 56.7 °C = - 31.7°C

Q = m.c.ΔT

Q = 30 g × 4.186 j/g.°C ×  - 31.7°C

Q = -3980.9 j

8 0
3 years ago
What is the molecular formula of the hydrocarbon whose molar mass is 536 g/mol and contains 89.55 w/w% carbon?
leonid [27]

Given the percentage composition of HC as C → 81.82 % and H → 18.18 %

So the ratio of number if atoms of C and H in its molecule can will be:

C : H = 81.82 12 : 18.18 1 C : H = 6.82 : 18.18 = 6.82 6.82 : 18.18 6.82 = 1 : 2.66 ≈ 3 : 8

So the Empirical Formula of hydrocarbon is:

C 3 H 8

As the mass of one litre of hydrocarbon is same as that of C O 2 The molar mass of the HC will be same as that of C O 2 i.e 44 g mol

Now let Molecular formula of the HC be ( C 3 H 8 ) n

Using molar mass of C and H the molar mass of the HC from its molecular formula is:

( 3 × 12 + 8 × 1 ) n = 44 n So 44 n = 44 ⇒ n = 1

Hence the molecular formula of HC is C 3 H 8

Does that help?

7 0
3 years ago
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