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4 years ago

How many days are in 2.5 years?

Chemistry

2 answers:

Talja [164]4 years ago

6 0

The answer to this question is 192.5

Hunter-Best [27]4 years ago

4 0

I believe the answer is, 912.5 days.
I hope this helped!

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What is one way that electric charges are different from magnetic poles?

sashaice [31]

The answer is A. As the first statement is a true statement. Hope this help you

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3 years ago

What does the exosphere do??

vovikov84 [41]

In the case of bodies with substantial atmospheres, such as Earth's atmosphere, the exosphere is the uppermost layer, where the atmosphere thins out and merges with interplanetary space. It is located directly above the thermosphere.

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4 years ago

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Use the drop-down menus to identify the electromagnetic waves.

mezya [45]

Answer:

Label A - Radio waves

Label C - Infrared

Label D - Visible light

Label G- Gamma rays

Explanation:

Electromagnetic spectrum is an arrangement of electromagnetic waves in order of their varying frequencies and wavelengths,
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3 years ago

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What is the diffrence between kinetic and potential energy

Aneli [31]

Potential Energy is the stored energy in an object or system because of its position or configuration. Kinetic energy of an object is relative to other moving and stationary objects in its immediate environment.

4 0

3 years ago

A chemist wants to find Kc for the following reaction at 751 K: 2NH3(g) + 3 I2 (g) LaTeX: \Longleftrightarrow ⟺ 6HI(g) + N2(g) K

charle [14.2K]

Answer: The equilibrium constant for the total reaction is $4.09\times 10^{-6}$

Explanation:

We are given:

$K_{c_1}=0.282\\\\K_{c_2}=41$

We are given two intermediate equations:

Equation 1: $N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g);K_{c_1}=0.282$

The expression of $K_{c_1}$ for the above equation is:

$K_{c_1}=\frac{[NH_3]^2}{[N_2][H_2]^3}$

$0.282=\frac{[NH_3]^2}{[N_2][H_2]^3}$ .......(1)

Equation 2: $H_2(g)+I_2(g)\rightleftharpoons 2HI(g);K_{c_2}=41$

The expression of $K_{c_2}$ for the above equation is:

$K_{c_2}=\frac{[HI]^2}{[H_2][I_2]}$

$41=\frac{[HI]^2}{[H_2][I_2]}$ ......(2)

Cubing both the sides of equation 2, because we need 3 moles of HI in the main expression if equilibrium constant.

$(41)^3=\frac{[HI]^6}{[H_2]^3[I_2]^3}$

Now, dividing expression 1 by expression 2, we get:

$\frac{K_{c_1}}{K_{c_2}}=\left(\frac{\frac{[NH_3]^2}{[N_2][H_2]^3}}{\frac{[HI]^6}{[H_2]^3[l_2]^3}}\right)\\\\\\\frac{0.282}{68921}=\frac{[NH_3]^2[I_2]^3}{[N_2][HI]^6}$

$\frac{[NH_3]^2[I_2]^3}{[N_2][HI]^6}=4.09\times 10^{-6}$

The above expression is the expression for equilibrium constant of the total equation, which is:

$2NH_3(g)+3I_2(g)\rightleftharpoons 6HI(g)+N_2;K_c$

Hence, the equilibrium constant for the total reaction is $4.09\times 10^{-6}$

8 0

3 years ago