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Neko [114]
3 years ago
12

An industrial chemist introduces 8.1 atm h2 and 8.1 atm co2 into a 1.00-l container at 25.0°c and then raises the temperature to

700.0°c, at which keq = 0.534: h2(g) + co2(g) ⇔ h2o(g) + co(g) how many grams of h2 are present after equilibrium is established?
Chemistry
1 answer:
kirill115 [55]3 years ago
3 0
I reccomend searching it up
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Study the image and answer the question.
konstantin123 [22]
The image represents A COMPOUND because the molecules are BONDED CHEMICALLY.
A compound is a substance formed when two or more elements combine together chemically. In the process of chemical combination, the chemical bonds that were present in the participating elements will be broken and new chemical bonds will be formed in the product.
7 0
3 years ago
What combination of forces was the closest match? Why?
Virty [35]

Answer: COMBINED FORCES

When forces act in the same direction, they combine to make a bigger force. When they act in opposite directions, they can cancel one another out. If the forces acting on an object balance, the object does not move, but may change shape.

Explanation:

8 0
3 years ago
Matter that is composed of two or more different elements chemically combined in a fixed proportion is classified as...
Lilit [14]
It's called a compound because different elements are held together by a chemical bond.
4 0
3 years ago
2. if 0.20 m fe3 had been used instead of 0.020 m fe3 , how would the numerical value of the rate constant and the activation en
dezoksy [38]

the calculated value is Ea is 18.2 KJ and A is 12.27.

According to the exponential part in the Arrhenius equation, a reaction's rate constant rises exponentially as the activation energy falls. The rate also grows exponentially because the rate of a reaction is precisely proportional to its rate constant.

At 500K, K=0.02s−1

At 700K, k=0.07s −1

The Arrhenius equation can be used to calculate Ea and A.

RT=k=Ae Ea

lnk=lnA+(RT−Ea)

At 500 K,

ln0.02=lnA+500R−Ea

500R Ea (1) At 700K lnA=ln (0.02) + 500R

lnA = ln (0.07) + 700REa (2)

Adding (1) to (2)

700REa100R1[5Ea-7Ea] = 0.02) +500REa=0.07) +700REa.

=ln [0.02/0 .07]

Ea= 2/35×100×8.314×1.2528

Ea =18227.6J

Ea =18.2KJ

Changing the value of E an in (1),

lnA=0.02) + 500×8.314/18227.6

= (−3.9120) +4.3848

lnA=0.4728

logA=1.0889

A=antilog (1.0889)

A=12.27

Consequently, Ea is 18.2 KJ and A is 12.27.

Learn more about Arrhenius equation here-

brainly.com/question/12907018

#SPJ4

5 0
1 year ago
If 3.0 liters of oxygen gas react with excess carbon monoxide at STP, how many liters of carbon dioxide can be produced under th
mr_godi [17]

Answer:

The volume of CO2 produced is 6.0 L (option D)

Explanation:

Step 1: Data given

Volume of oxygen = 3.0 L

Carbon monoxide = CO = in excess

Step 2: The balanced equation

2 CO (g) + O2 (g) → 2 CO2 (g)

Step 3: Calculate moles of O2

1 mol of gas at STP = 22.4 L

3.0 L = 0.134 moles

Step 3: Calculate moles of CO2

For 2 moles CO we need 1 mol of O2 to produce 2 moles of CO2

For 0.134 moles O2 we'll have 2*0.134 = 0.268 moles CO2

Step 4: Calculate volume of CO2

1 mol = 22.4 L

0.268 mol = 22.4 * 0.268 = 6.0 L

The volume of CO2 produced is 6.0 L

8 0
3 years ago
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