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tensa zangetsu [6.8K]
3 years ago
12

PLZ RESPONDE QUICKLY MY EDUCATION IS ON THE LINE

Chemistry
1 answer:
lyudmila [28]3 years ago
6 0

Answer:

N= PV/RT (temperature in kelvin); ( r is .0821 when pressure is in atm & r is 8.314 when pressure is in Kpa)

2.00 = P x 10.0/ .0821 x 300

after solving i got 4.926 atm

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Yall i think if someone can solve ths problem would save my life. its really URGENT
andre [41]
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6 0
3 years ago
How do I calculate the molecular weight of Mg(OH)2?
vodomira [7]

The molecular weight of Mg(OH)2 : 58 g/mol

<h3>Further explanation</h3>

Given

Mg(OH)2 compound

Required

The molecular weight

Solution

Relative atomic mass (Ar) of element : the average atomic mass of its isotopes  

Relative molecular weight (M)  : The sum of the relative atomic mass of Ar  

M AxBy = (x.Ar A + y. Ar B)  

So for Mg(OH)2 :

= Ar Mg + 2 x Ar O + 2 x Ar H

= 24 g/mol + 2 x 16 g/mol + 2 x 1 g/mol

= 24 + 32 + 2

= 58 g/mol

5 0
3 years ago
Which statement best describes insoluble substances?
Maksim231197 [3]

Answer:

Water does not remain when an insoluble substance is added

7 0
3 years ago
Read 2 more answers
Problem PageQuestionSteam reforming of methane ( ) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas,
Serhud [2]

The question is incomplete. Her eis the complete question.

Steam reforming methane  (CH4) produces "synthesis gas", a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a 125L tank with 20 mol of methane gas and 10 mol of water vapor at 38°C. He then raises the temperature, and when the mixture has come to equilibrium measures the amount of gas hydrogen to be 18 mol. Calculate the concentration equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to significant digits.

Answer: K_{c} = 2.10^{-2}

Explanation: The reaction for steam reforming methane is:

CH_{4} + H_{2}O ⇒ CO_{} + 3H_{2}

To calculate the concentration equilibrium constant, first calculate the molarity (\frac{mol}{L}) of each molecule of the reaction.

At 38°C: At the initial temperature, there no products yet

<u>Molarity of CH4</u>:

CH4 = \frac{20}{125} = 0.16M

<u>Molarity of H20</u>:

H2O = \frac{10}{125} = 0.08M

At final temperature:

<u>Molarity of H2</u>:

H2 = \frac{18}{125} = 0.144M

According to the chemical reaction, the combination of 1 mol of each reagents produces 1 mol of CO and 3 mols of H2, so, for the products, the ratio is 1:3.

<u>Molarity of CO</u>:

CO = \frac{0.144}{3} = 0.048M

For the reagents, the proportion is 1:1, but they had an initial concentration, so, when in equilibrium, the concentration will be:

<u>Molarity of CH4</u>:

CH4 = 0.16 - 0.048 = 0.112M

<u>Molarity of H2O</u>:

H20 = 0.08 - 0.048 = 0.032M

The equilibrium constant is given by:

K_{c} = \frac{[CO][H_{2}]^{3} }{[CH_{4}][H_{2}O ] }

K_{c} = \frac{0.048.0.144^{3} }{0.112.0.032}

K_{c} = 2.10^{-2}

The concentration equilibrium constant for the process is K_{c} = 2.10^{-2}.

4 0
4 years ago
PLEASE HELP WITH THIS QUESTION
gtnhenbr [62]

Answer:

C. Egg

Explanation:

has more higher concentration on water.

8 0
3 years ago
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