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Mila [183]
3 years ago
12

A mixture of gaseous sulfur dioxide and oxygen are added to a reaction vessel and heated to 1000 K where they react to form SO3

(g). If the vessel contained 0.669 atm SO2 (g), 0.395 atm O2 (g) and 0.0851 atom SO3 (g) after the system has reached equilibrium, what is the equilibrium constant, Kp, for the reaction
Chemistry
1 answer:
nasty-shy [4]3 years ago
5 0

Answer:

Kp = 0.0410

Explanation:

The reaction of gaseous sulfur dioxide and oxygen to form SO3 (g) is:

2SO₂(g) + O₂(g) ⇄ 2SO₃(g)

Kp is defined as the ratio of pressure of products and pressure of reactants:

Kp = \frac{P_{SO_3}^2}{P_{SO_2}^2P_{O_2}}

Kp = \frac{0.0851atm^2}{0.669atm^2*0.395atm}

Kp = 0.0410

I hope it helps!

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Which planet formed near the sub where the solar system l’s temperatures were very high
Harrizon [31]

Answer:

Mars

Explanation:

Terrestrial or inner planets like Mars and Venus were formed near the Sun where the solar system's temperatures were very high.

4 0
3 years ago
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Which two notations represent isotopes of the same element
Mumz [18]
The two notations that represent isotopes of the same element is the one that represented in option 1
The lower number is the number of protons while the upper number is the atomic weight

hope this helps
6 0
3 years ago
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How many moles of chlorine gas would occupy a volume of 35.5 L at a pressure of 100.0 kPa and a
Nikolay [14]

Answer:

this was the last of .the last 555543 years of 55my in

Explanation:

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4 0
3 years ago
Consider the following reaction 2 N2O(g) => 2 N2(g) + O2(g) rate = k[N2O]. For an initial concentration of N2O of 0.50 M, cal
den301095 [7]

Answer:

After 2.0 minutes the concentration of N2O is 0.3325 M

Explanation:

Step 1: Data given

rate = k[N2O]

initial concentration of N2O of 0.50 M

k = 3.4 * 10^-3/s

Step 2: The balanced equation

2N2O(g) → 2 N2(g) + O2(g)  

Step 3: Calculate the concentration of N2O after 2.0 minutes

We use the rate law to derive a time dependent equation.

-d[N2O]/dt = k[N2O]

ln[N2O] = -kt + ln[N2O]i

 ⇒ with k = 3.4 *10^-3 /s

⇒ with t = 2.0 minutes = 120s

⇒ with [N2O]i = initial conc of N2O = 0.50 M

ln[N2O] = -(3.4*10^-3/s)*(120s) + ln(0.5)

ln[N2O] = -1.101

e^(ln[N2O]) = e^(-1.1011)

[N2O} = 0.3325 M

After 2.0 minutes the concentration of N2O is 0.3325 M

3 0
3 years ago
Why isn't carbon-14 dating accurate for estimating the age of materials older than 50,000 years?
soldi70 [24.7K]
Carbon-14 is radioactive isotope of carbon.
Carbon is essential element of living cells. While the living cells are alive, the carbon contained in them are in equilibrium with the carbon in atmosphere. But, once the cell dies, the carbon-14 isotope undergoes radioactive decay. By measuring the carbon-14 in atmosphere to the carbon-14 in dead organism, we can calculate the time (or years) that organism have died.

However, carbon-14 dating technique is not accurate for estimating the age of materials older than 50,000 years old (above 40,000 years). This is because, 99% of carbon is carbon-12, 1% is carbon-13 and trace remaining is the carbon-14. This means, carbon-14 is found in very trace amount, in fact 1 part per trillion of carbon atoms present is carbon-14. The half of life of carbon-14 is 5,730 years. For dating the organism, we use the concept of half lives of the carbon-14 isotope in the dead organisms and calculate how many half life old the sample is. But as the years increases, the number of carbon-14 isotope becomes too low to detect and make accurate calculation.
This means, at some point the organism can simply run out of carbon-14.

Hence carbon-14 dating is not accurate for estimating age of materials older than 50,000 years old.

5 0
3 years ago
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