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grigory [225]
3 years ago
8

What is the value of X?

Chemistry
2 answers:
mylen [45]3 years ago
8 0
Hi there! The answer is x = 6

Since the length of both sides is the same, we get the following equation:
5x + 10 = 40

Subtract 10 from both sides
5x = 30

Divide both sides by 5.
x = 30 \div 5 = 6

pychu [463]3 years ago
3 0
The two sides are congruent (because of the straight line through the side)

set them equal to one another

5x + 10 = 40

Isolate the x, do the opposite of PEMDAS

5x + 10 = 40

subtract 10 from both sides

5x + 10 (-10) = 40 (-10)

5x = 30

divide 5 from both sides

5x/5 = 30/5

x = 30/5

x = 6

6 is your answer for x

hope this helps
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3 years ago
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What would the percentage of thorium-232 (parent compound) be in a rock that was dated 1.4 billion years?
trasher [3.6K]

Answer:

93.33%

Explanation:

Thorium is a radioactive element with a molecular weight of 232g/mol. Thorium is very stable and can have 14.05 billion years of half-life period. Every half-life passed, the parent compound mass will be decayed by half. If the age of the rock is 1.4 billion years, then the amount of the parent compound will be:r = ½^(time elapsed/half-life)

r = ½^(1.4bil years/14.05 bil years)

r = ½^(0.0996)= 0.9333

r= 93.33%

The rock will have 93.33% of the parent compound and 6.67% of the decayed compound.

5 0
4 years ago
Which colorless and odorless gas, produced by radioactive decay of Uranium-238, is considered to be a cancer-causing agent?
DiKsa [7]
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7 0
3 years ago
Calculate the mass in amu of 75 atoms of aluminum
lyudmila [28]
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4 0
3 years ago
At a certain temperature, 0.3411 0.3411 mol of N 2 N2 and 1.661 1.661 mol of H 2 H2 are placed in a 2.50 2.50 L container. N 2 (
Sveta_85 [38]

<u>Answer:</u> The equilibrium constant for the above reaction is 1.31

<u>Explanation:</u>

We are given:

Initial moles of nitrogen gas = 0.3411 moles

Initial moles of hydrogen gas = 1.661 moles

Equilibrium moles of nitrogen gas = 0.2001 moles

For the given chemical reaction:

                   N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)

<u>Initial:</u>         0.3411      1.661

<u>At eqllm:</u>     0.3411-x  1.661-3x      2x

Evaluating for 'x', we get:

\Rightarrow (0.3411-x)=0.2001\\\\\Rightarrow x=0.3411-0.2001=0.141

Volume of the container = 2.50 L

The expression of K_c for the above equation follows:

K_c=\frac{[NH_3]^2}{[N_2]\times [H_2]^3}

We are given:

[NH_3]=\frac{2\times 0.141}{2.50}=0.1128M

[N_2]=\frac{0.2001}{2.5}=0.08004M

[H_2]=\frac{1.661-(3\times 0.141)}{2.5}=0.4952M

Putting values in above expression, we get:

K_c=\frac{(0.1128)^2}{0.08004\times (0.4952)^3}\\\\K_c=1.31

Hence, the equilibrium constant for the above reaction is 1.31

5 0
3 years ago
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