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sergiy2304 [10]
3 years ago
10

If all the matter in the universe was birthed seconds after the Big Bang, then what was likely triggered the bang?

Chemistry
1 answer:
ArbitrLikvidat [17]3 years ago
8 0

Answer:

Welp God made everything so that means the big bang never happened so that means techniqually the answer to your question is nothing ig.

hope this helps

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Which state of matter has a definite volume, but does not have a definite shape?
evablogger [386]

Answer:

Liquid, definite volume but no definite shape.

Explanation:

7 0
3 years ago
PLEASE HELP WORTH 50 POINTS NO LINKS I WILL MARK CORRECT ANSWER BRAINLIEST!!! complete this punnet square, enter your answer in
Dafna1 [17]

Answer:

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Explanation:

6 0
3 years ago
The element nitrogen has an atomic weight of 14.0 and consists of two stable isotopes nitrogen-14 and nitrogen-15. The isotope n
Alecsey [184]
The mass of N-15 is 15 amu.

Chemical elements are represented using their symbols, their mass number and their proton numbers. The mass number of this isotope of nitrogen is shown next to the symbol. Regardless of the mass of the isotopes, their proton number is the same, which is 7.
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# of protons = 48 <br><br> Cadmium <br> Titanium <br> Oxygen <br> Krypton
Dafna11 [192]

Answer:

Cadmium +p= 48

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Oxygen p=8

Krypton p=36

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Explanation:

7 0
2 years ago
2) Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800 mole of CO and 2.40 mole of H2
nadezda [96]

Answer:

Kc = 3.90

Explanation:

CO reacts with H_2 to form CH_4 and H_2O. balanced reaction is:

CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g)  +  H_2O(g)

No. of moles of CO = 0.800 mol

No. of moles of H_2 = 2.40 mol

Volume = 8.00 L

Concentration = \frac{Moles}{Volume\ in\ L}

Concentration of CO = \frac{0.800}{8.00} = 0.100\ mol/L

Concentration of H_2 = \frac{2.40}{8.00} = 0.300\ mol/L

                 CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g)  +  H_2O(g)

Initial            0.100      0.300             0   0

equi.            0.100 -x    0.300 - 3x     x    x

It is given that,

at equilibrium H_2O (x) = 0.309/8.00 = 0.0386 M

So, at equilibrium CO = 0.100 - 0.0386 = 0.0614 M

At equilibrium H_2 = 0.300 - 0.0386 × 3 = 0.184 M

At equilibrium CH_4 = 0.0386 M

Kc=\frac{[H_2O][CH_4]}{[CO][H_2]^3}

Kc=\frac{0.0386 \times 0.0386}{(0.184)^3 \times 0.0614} =3.90

8 0
3 years ago
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