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3 years ago

Which chemical property deals with the ability of a substance being able to

Chemistry

2 answers:

Fantom [35]3 years ago

5 0

$\large\underline{\bf\red{Solution:-}}$

The chemical property deals with the ability of a substance being able to burn is called combustibility [ option b ] .

More to know :-

1) Combustion :-

Combustion is a chemical process in which a substance reacts rapidly with oxygen and gives out heat.

The fuel can be any one of the 3 states of matter that is - solid , liquid & gases .

Examples of combustible substances :-

Solid - Coal

Liquid - Petrol , Disel , etc.

Gas - Hydrogen , Oxygen , etc .

Mandarinka [93]3 years ago

3 0

Answer:

combust

Explanation:

did the test

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B) arteriosclerosis

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3 years ago

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The density of mercury, the only metal to exist as a liquid at room temperature, is 13.6 g/cm3. What is that density in pounds p

saw5 [17]

Answer:

0.492 lb/in^3

Explanation:

From Density of Mercury (13.6g/cm3) given,

Mass = 13.6g

Volume = 1cm3

Now we covert these variables as follows :

For Mass:

454 g = 1 lb

13.6g = 13.6/454 = 0.03lb

For volume:

2.54 cm = 1 in

1cm = 1/2.54 = 0.394in

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Density = Mass /volume

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3 years ago

How is Earth's surface most likely to change in a cold place that experiences rainfall?

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Answer:

I think the answer would have to be D

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3 years ago

compute the mass-specific enthalpy change associated with Nz that is undergoing a change in state from 400 k to 800 k

Kitty [74]

Answer:

The correct answer is "430 kJ/kg". A further explanation is given below.

Explanation:

The given values are:

T₁ = 400 k

T₂ = 800 k

The average temperature will be:

= $\frac{T_1+T_2}{2}$

= $\frac{400+800}{2}$

= $600 \ k$

From table,

At 600 k the $C p$ will be = 1.075

Now,

⇒ The specific enthalpy = $Cp(T_2-T_1)$

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8 0

3 years ago

What is deltaG at 298 K for the reaction C + O2 = CO

blondinia [14]

The answer for the following problem is mentioned below.

Therefore the Gibbs free energy (ΔG) for the reaction is -394.36 J

Explanation:

Given:

ΔH = -393.5 kJ/ mol

ΔS =0.0029 kJ/mol K

T = 298 K

To solve:

Gibbs free energy (ΔG)

We know;

ΔG = ΔH - TΔS

ΔG = -393.5 - (298 × 0.0029)

ΔG = -393.5 - 0.8642

ΔG = -394.36 J

Therefore the Gibbs free energy (ΔG) for the reaction is -394.36 J

8 0

3 years ago