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GREYUIT [131]
3 years ago
6

Is/are determined by a genetic trait passed on from parents to their children

Chemistry
1 answer:
Darya [45]3 years ago
7 0

Answer:genes

Explanation:

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Suppose that 0.1000 mole each of H2and I2are placed in a 1.000-L flask, stoppered, and the mixture is heated to 425oC. At equili
Katen [24]

<u>Answer:</u> The value of equilibrium constant for the given reaction is 56.61

<u>Explanation:</u>

We are given:

Initial moles of iodine gas = 0.100 moles

Initial moles of hydrogen gas = 0.100 moles

Volume of container = 1.00 L

Molarity of the solution is calculated by the equation:

\text{Molarity of solution}=\frac{\text{Number of moles}}{\text{Volume}}

\text{Molarity of iodine gas}=\frac{0.1mol}{1L}=0.1M

\text{Molarity of hydrogen gas}=\frac{0.1mol}{1L}=0.1M

Equilibrium concentration of iodine gas = 0.0210 M

The chemical equation for the reaction of iodine gas and hydrogen gas follows:

                         H_2+I_2\rightleftharpoons 2HI

<u>Initial:</u>                0.1    0.1

<u>At eqllm:</u>          0.1-x   0.1-x   2x

Evaluating the value of 'x'

\Rightarrow (0.1-x)=0.0210\\\\\Rightarrow x=0.079M

The expression of K_c for above equation follows:

K_c=\frac{[HI]^2}{[H_2][I_2]}

[HI]_{eq}=2x=(2\times 0.079)=0.158M

[H_2]_{eq}=(0.1-x)=(0.1-0.079)=0.0210M

[I_2]_{eq}=0.0210M

Putting values in above expression, we get:

K_c=\frac{(0.158)^2}{0.0210\times 0.0210}\\\\K_c=56.61

Hence, the value of equilibrium constant for the given reaction is 56.61

6 0
3 years ago
Which reaction should be used to convert propene into an alkyl halide?
skelet666 [1.2K]
Substitution that is it 
4 0
3 years ago
Read 2 more answers
Periodic table <br> fill in the box :)
Natalija [7]

Answer:

I took a screen shot of your image, and wrote on top of it!

Explanation:

Hope this helps! :)

6 0
2 years ago
Mrs. Stark weighs 70kg. How much Potential Energy does she have if she stands on the roof of an 80m tall building?
8_murik_8 [283]
E = mgh

E = 70*9.81*80

E = 54936 J
5 0
3 years ago
Explain the difference in the boiling point of 2-methylpropane and 2-iodo-2-methylpropane in terms of both molecular polarity an
Len [333]
The Boiling Point of 2-methylpropane is approximately -11.7 °C, while, Boiling Point of <span>2-iodo-2-methylpropane is approximately 100 </span>°C.

As both compounds are Non-polar in nature, So there will be no dipole-dipole interactions between the molecules of said compounds.

The Interactions found in these compounds are London Dispersion Forces.

And among several factors at which London Dispersion Forces depends, one is the size of molecule.

Size of Molecule:
                          There is direct relation between size of molecule and London Dispersion forces. So, 2-iodo-2-methylpropane containing large atom (i.e. Iodine) experience greater interactions. So, due to greater interactions 2-iodo-2-methylpropane need more energy to separate from its partner molecules, Hence, high temperature is required to boil them.
6 0
3 years ago
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