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Monica [59]
3 years ago
5

Two hundred grams of a substance requires 9.76kj of heat to raise its temperature from 25 C to 45 C. It has a specific heat of 2

.44J/g. Use the table to identify the substance.
A Ethanol
B Gasoline
C Ice
D Air
I think it's gasoline!

Chemistry
2 answers:
yulyashka [42]3 years ago
7 0
B. Gasoline
hope this helps.
elena55 [62]3 years ago
7 0
Its gasoline im sure ...
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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
Please explain to me how to answer no.2 questions
bulgar [2K]

Answer: -

IE 1 for X = 801

Here X is told to be in the third period.

So n = 3 for X.

For 1st ionization energy the expression is

IE1 = 13.6 x Z ^2 / n^2

Where Z =atomic number.

Thus Z =( n^2 x IE 1 / 13.6)^(1/2)

Z = ( 3^2 x 801 / 13.6 )^ (1/2)

= 23

Number of electrons = Z = 23

Nearest noble gas = Argon

Argon atomic number = 18

Number of extra electrons = 23 – 18 = 5

a) Electronic Configuration= [Ar] 3d34s2

We know that more the value of atomic radii, lower the force of attraction on the electrons by the nucleus and thus lower the first ionization energy.

So more the first ionization energy, less is the atomic radius.

X has more IE1 than Y.

b) So the atomic radius of X is lesser than that of Y.

c) After the first ionization, the atom is no longer electrically neutral. There is an extra proton in the atom.

Due to this the remaining electrons are more strongly pulled inside than before ionization. Hence after ionization, the radii of Y decreases.

4 0
3 years ago
Which of the following are likely to form a covakent bond
Dmitriy789 [7]

Answer:

For the most part, non-metals (excluding Nobel gases) are the most likely to form covalent bonds. Pure covalent bonds are formed between atoms with the same electronegativity, ie. they are trying to hold on to the electrons in the bond with the same strength.

5 0
3 years ago
Which of these is a chemical property of a substance?
dlinn [17]

Answer:

Reactivity

Explanation:

8 0
2 years ago
Will mark brainliest now
Aleks [24]
I think it’s water evaporation I’m not sure tho
4 0
3 years ago
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