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harina [27]
3 years ago
8

An object is described as being 350 K. What does this indicate about the object?

Chemistry
1 answer:
LuckyWell [14K]3 years ago
3 0
A would be the answer to this question
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I need help on this plsss
Firlakuza [10]

Answer:

An engine that creates ignites fuel with highly compressed air.

6 0
3 years ago
Read 2 more answers
What is the pressure of a sample of gas at a volume of .335 L if it occupies 1700 mL at 850 mm Hg?
dangina [55]

Answer:

Pressure = 4313.43mmHg

Explanation:

P1 = ?

V1 = 0.335L

V2 = 1700mL =1700*10^-3L = 1.7L

P2 = 850mmhg

From Boyle's law, the volume of a fixed mass of gas is inversely proportional to its pressure provided that temperature remains constant.

P = k / v

K = pv. P1V1 = P2V2 = P3V3 =........=PnVn

P1V1 = P2V2

Solve for P1,

P1 = (P2*V2) / V1

P1 = (850 * 1.7) / 0.335

P1 = 4313.43mmHg

The pressure of the gas was 4313.43mmHg

7 0
3 years ago
Consider the reaction: P(s) + 5/2 Cl2(g)PCl5(g) Write the equilibrium constant for this reaction in terms of the equilibrium con
Pani-rosa [81]

Answer: The equilibrium constant for the overall reaction is K_a\times K_b

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios.

a) P(s)+\frac{3}{2}Cl_2(g)\rightarrow PCl_3(g)

K_a=\frac{[PCl_3]}{[Cl_2]^{\frac{3}{2}}}

b) PCl_3(g)+Cl_2(g)\rightarrow PCl_5(g)

K_b=\frac{[PCl_5]}{[Cl_2]\times [PCl_3]}

For overall reaction on adding a and b we get c

c) P(s)+\frac{5}{2}Cl_2(g)\rightarrow PCl_5(g)

K_c=\frac{[PCl_5]}{[Cl_2]^\frac{5}{2}}

K_c=K_a\times K_b=\frac{[PCl_3]}{[Cl_2]^{\frac{3}{2}}}\times \frac{[PCl_5]}{[Cl_2]\times [PCl_3]}

The equilibrium constant for the overall reaction is K_a\times K_b

4 0
3 years ago
How many atoms does 32 grams of sulfur contain?
Lelechka [254]
32 grams of sulfur will contain 6.022 X 1023 sulfur atoms.
4 0
3 years ago
The pre-exponential constant and activation energy for the diffusion of iron in cobalt are 1.7 x 10-5 m2/s and 273,300 J/mol, re
Kitty [74]

<u>Answer:</u> The temperature of the system will be 1622 K

<u>Explanation:</u>

The equation relating the pre-exponential factor and activation energy follows:

\log D=\log D_o-\frac{E_a}{2.303RT}

where,

D = diffusion coefficient = 2.7\times 10^{-14}m^2/s

D_o = pre-exponential constant = 1.7\times 10^{-5}m^2/s

E_a = activation energy of iron in cobalt = 273,300 J/mol

R = Gas constant = 8.314 J/mol.K

T = temperature = ?

Putting values in above equation, we get:

\log (2.7\times 10^{-14})=\log (1.7\times 10^{-5})-\frac{273,300}{2.303\times 8.314\times T}\\\\T=1622K

Hence, the temperature of the system will be 1622 K

8 0
3 years ago
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