Write the expression for the equilibrium constant Kp for the following reaction. Enclose pressures in parentheses and do NOT write the chemical formula as a subscript. For example, enter (PNH3 )2 as (P NH3)2. If either the numerator or denominator is 1, please enter 1 2 MoO3(s) ↔ 2 MoO2(s) + O2(g)

Answers

Answer 1

Answer:

The expression for the equilibrium constant Kp for the following reaction is $K_p = (po_2)$

Equilibrium constant:

It refers to the ratio of the concentration of products to the concentration of reactants where each raised to the power of their stoichiometric ratios. It is expressed as K.

Since the given expression is

MoO3(s) ↔ 2 MoO2(s) + O2(g)

So, the above expression should be considered for constant kp

Learn more about reaction here; brainly.com/question/24185208

Answer 2

Answer:

Answer: $K_p={(p_(O_2))}$

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. It is expressed as $K$

$K_p$ is the constant of a certain reaction at equilibrium for gaseous reactants and products.

For the given chemical reaction:

$MoO_3(s)\rightleftharpoons 2MoO_2(s)+O_2(g)$

The expression of for above equation follows:

$K_p={(p_(O_2))}$

As solids do not exert pressure, $MoO_3$ and $MoO_2$ are not involved.

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A cylinder is filled with 10.0L of gas and a piston is put into it. The initial pressure of the gas is measured to be 96.0kPa. The piston is now pulled up, expanding the gas, until the gas has a final volume of 45.0L. Calculate the final pressure of the gas. Be sure your answer has the correct number of significant digits.

Answers

Answer:

The final pressure of the gas is:- 21.3 kPa

Explanation:

Using Boyle's law

${P_1}* {V_1}={P_2}* {V_2}$

Given ,

V₁ = 10.0 L

V₂ = 45.0 L

P₁ = 96.0 kPa

P₂ = ?

Using above equation as:

${P_1}* {V_1}={P_2}* {V_2}$

${96.0\ kPa}* {10.0\ L}={P_2}* {45.0\ L}$

${P_2}=\frac {{96.0}* {10.0}}{45.0}\ kPa$

${P_2}=21.3\ kPa$

The final pressure of the gas is:- 21.3 kPa

The density for potassium is 0.856 g/cm3. What would be the mass of a 35 cm3 piece of potassium?

Answers

Answer:

30 grams

Explanation:

density = mass / volume => mass = density x volume

mass = 0.856 g/cm³ x 35 cm³ = 29.96 grams ≅ 30 grams 2 sig. figs.

Two samples of carbon tetrachloride were decomposed into their constituent elements. One sample produced 38.9 g of carbon and 463 g of chlorine, and the other sample produced 14.8 g of carbon and 144 g of chlorine. Part A Are these results consistent with the law of definite proportions?

Answers

Explanation:

Firstly, defining Law of definite proportion which states that in any chemical compound, the elements are the same and are in the same proportion by mass. It is also called Proust's law.

In sample A;

38.9 g of carbon

463 g of chlorine

In sample B;

14.8 g of carbon

144 g of chlorine.

Sample A, Ratio of mass of Carbon to mass of Chlorine

= 38.9 : 463

= 0.084

Sample B, Ratio of mass of Carbon to mass of Chlorine

= 14.8 : 144

= 0.103

These results above show that the ratio of the masses of Carbon to Chlorine in both samples A and B are not the same so therefore, the results are not consistent with the law of definite proportion.

You would like to make a 100 mL buffer solution at pH 8.00. Assuming you would like to accomplish this with a hypochlorous acid (HOCl) buffer (HOCl/NaOCl), Ka= 3.0 * 10-8. If the solution is 0.3 M in HOCl, what concentration of NaOCl would be necessary in the buffer solution to obtain a pH of 8.0?

Answers

Answer:

To obtain the pH of 8.0, the concentration of NaOCl needs to be 0.9 M in the 0.3 M HOCl solution

Explanation:

This problem can be solved by Henderson-Hasselbalch equation, which gives relation between the concentration of acid, its salt, pKa and the pH of the solution. This equation is given as,

$pH=-log(K_a)+log([NaOCl])/([HOCl])$

By placing the known variables in the above equation we get,

$8=-log(3*10^(-8))+log([NaOCl])/(0.3)$

$8-7.52=log([NaOCl])/([0.3])$

$10^(0.48)=([NaOCl])/(0.3)$

$[NaOCl]=10^(0.48)*{0.3}$

$[NaOCl]=0.9 M$

The above calculations show that the required concentration of NaOCl is 0.9 M.

What is the speed of a basketball that is thrown 18meters in 4 seconds?

Answers

Answer:

4.5m/s

Explanation:

speed = distance/time

speed = 18/4

speed = 4.5m/s

a length of #8 copper wire (radius = 1.63 mm) has a mass of 24.0 kg and a resistance of 2.061 ohm per km. what is the overall resistance of the wire

Answers

m = pVol

p : copper density

Vol: volume

Vol = Al

A: area

L: length

m = p * A * L

L = m/p*A

To find the total resistance

Rtotal = R per Km * L/1000

Actually I think the question is missing the copper density.

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