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Bond [772]
3 years ago
11

If the gas in the piston above has a volume of 20.0 L at a temperature of 25 C what is the volume of that gas when it is heated

to 100 C if the pressure remains constant

Chemistry
2 answers:
dexar [7]3 years ago
8 0

Answer:

15.98 L

Explanation:

First, you need to find T1, T2, V1 and V2.

T1 = 25 C = 298.15 K (25C + 273.15K)

T2 = 100 C = 373.15 K (100C + 273.15K)

V1 = 20. L

V2 = ? (we are trying to find)

Next, rearrange to fit the formula

V2 = V1 x T1 / T2

Next, fill in with our numbers

V2 = 20. L x 298.15 K / 373.15 K

Do the math and you should get...

15.98 L

- If you need more help or futher explanation please let me know. I would be glad to help!

Otrada [13]3 years ago
7 0

Answer: 25L

Explanation:

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Compute the percent ionic character of the interatomic bonds for the following compounds : a. TiO2 b. ZnTe c. CsCld. InSb e. MgC
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Answer:

a. 63.2%

b. 11.7%

c. 73.3%

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e. 55.5%

Explanation:

An ionic compound is a compound that is formed by ions, so one of the elements must donate electrons (which is the cation, the positive ion), and the other will receive these electrons (which is the anion, the negative ion).

The power of an element has to attract the electrons is called electronegativity, and so, as higher is the difference of electronegative of the elements, it is more probable that one of them will "still" the electrons and will form an ionic compound. The percent of this ionic character can be found by the Pauling's equation:

%IC = (1 - e^{-0.25*(x_A - x_B)^2}) *100%

Where x_A - x_B is the electronegativity difference of the elements. Thus, consulting an electronegativity table:

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%IC = (1 - e^{-0.25*(3.5 - 1.5)^2})*100%

%IC = 63.2%

b. x_{Zn} = 1.6

x_{Te} = 2.1

%IC = (1 - e^{-0.25*(2.1 - 1.6)^2})*100%

%IC = 11.7%

c. x_{Cs} = 0.7

x_{Cl} = 3.0

%IC = (1 - e^{-0.25*(3.0 - 0.7)^2})*100%

%IC = 73.3%

d. x_{In} = 1.7

x_{Sb} = 1.9

%IC = (1 - e^{-0.25*(1.9 - 1.7)^2})*100%

%IC = 0.995 %

e. x_{Mg} = 1.2

x_{Cl} = 3.0

%IC = (1 - e^{-0.25*(3.0 - 1.2)^2})*100%

%IC = 55.5%

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