How do wheels reduce the force of friction

Which statement describes a chemical property of sodium

Alenkasestr [34]

Answer:

Option b is show the chemical property of sodium....

4 0

3 years ago

What is a by-product of a nuclear power plant?

Mariana [72]

Answer:

There are four categories of byproduct material: Radioactive material that results from the fissioning, or splitting apart, of enriched uranium or plutonium in nuclear reactors. Examples include cobalt-60, cesium-137 and iridium-192. Tailings or waste produced by processing uranium or thorium from ore.

6 0

3 years ago

I need number 5,6 please thanks

bulgar [2K]

5. b is the right answer

6. c is the right answer

6 0

3 years ago

Consider the following reaction: AB(s)⇌A(g)+B(g) . At equilibrium, a 10.9 L container at 650 K contains A(g) at a pressure of 0.

irina1246 [14]

Answer:

pA = 0.095 atm

pB = 0.303 atm

Explanation:

Step 1: the reaction

AB(s) ⇔ A(g) + B(g)

Kp = pA * pB

⇒ with Kp = equilibrium constant

Kp = 0.126 * 0.23 ⇒ Kp = 0.02898

Since the container will be compressed to half of its original volume, means that he pressure will be doubled.

⇒pA = 0.252

⇒pB =0.46

To establish this equilibrium, each pressure has to be lowered by x

⇒pA = 0.252 - x

⇒pB = 0.46 - x

Kp = 0.02898 = (0.252 - x)(0.46-x)

0.02898 = 0.11592 - 0.252x -0.46x + x²

-x² + 0.712x - 0.08694 = 0

D= b² - 4ac

⇒ D = 0.712² -4*(-1) *(-0.08694) = 0.506944‬ -0.34776‬ =0.159184

x = (-b ± √D)/2a

x = (-0.712 ± √0.159184)/(2*-1) = (-0.712 ± 0.398978696)/-2

x = 0.156510652 or x= 0.555489348

x = 0.555489348 is impossble or the pressure would be negative

x=0.156510652

pA =0.252 - 0.156510652 = 0.095489348 atm

pB = 0.46 - 0.156510652 = 0.303489348 atm

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