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Radda [10]
3 years ago
14

I need help with this question!!

Chemistry
1 answer:
s2008m [1.1K]3 years ago
3 0

Answer:

Fossil fuels..........

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Dvinal [7]

Answer:

Parallel Circuit

8 0
3 years ago
Read 2 more answers
When you decrease the volume of a gas, what happens to the temperature?
ale4655 [162]

the temperature is decreased

4 0
3 years ago
What mass of carbon monooxide must be burned to produce 175 kJ of heat under standard state condaitions?
Sphinxa [80]

Answer:

17.3124 grams

Explanation:

Given;

Amount of heat to be produced = 175 kJ

Molar mass of the carbon monoxide = 12 + 16 = 28 grams

Now,

The standard molar enthalpy of carbon monoxide = 283 kJ/mol

Thus,

To produce 175 kJ heat, number of moles of CO required will be

= Amount heat to be produced /  standard molar enthalpy of CO

or

= 175 / 283

= 0.6183

Also,

number of moles = Mass / Molar mass

therefore,

0.6183 = Mass / 28

or

Mass of the CO required = 0.6183 × 28 = 17.3124 grams

7 0
3 years ago
Graphic organizer: Use the terms in the word bank to complete the graphic organizer below
Valentin [98]

Answer:

1. Chemical reactions

2.Substances

3. Properties

4. Precipitate

5. Light

6. Temperature

7. Color

8. Gas

Explanation:

Chemical reactions are known to produce new substances which have new properties. During the chemical reactions, evidences that are seen are formation of precipitate or light gel, production of gas and change in temperature and color.

Chemical reactions occur when chemical substances come together to react because those substances are compatible with each other.

7 0
3 years ago
Calculate the value of the diffusion coefficient D (in m2/s) at 547°C for the diffusion of some species in a metal; assume that
svetlana [45]

Answer : The value of diffusion coefficient is, 2.97\times 10^{-16}m^2/s

Explanation :

Formula used :

D=D_o\times \exp \left(-\frac{Q_d}{RT}\right )

where,

D = diffusion coefficient = ?

D_o = 5.6\times 10^{-5}m^2/s

Q_d = 177kJ/mol

R = gas constant = 8.314 J/mol.K

T = temperature = 547^oC=273+547=820K

Now put all the given values in the above formula, we get:

D=5.6\times 10^{-5}m^2/s\times \exp \left(-\frac{177kJ/mol}{(8.314J/mol.K)\times (820K)}\right )

D=2.97\times 10^{-16}m^2/s

Thus, the value of diffusion coefficient is, 2.97\times 10^{-16}m^2/s

6 0
3 years ago
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