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finlep [7]
3 years ago
11

Determine the temperature of 1.42 mol of a gas contained in a 3.00-L container at a pressure of 123 kPa.

Chemistry
1 answer:
Schach [20]3 years ago
8 0

Answer:

31.2K

Explanation:

The following data were obtained from the question:

n = 1.42 mol

V = 3L

P = 123 kPa = 123000Pa

Recall: 101325Pa = 1atm

123000Pa = 123000/101325 = 1.21atm

R = 0.082atm.L/Kmol

T =?

Using the ideal gas equation PV = nRT, we can obtain the temperature as follows:

PV = nRT

T = PV/nR

T = (1.21 x 3)/(1.42 x 0.082)

T = 31.2K

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2 years ago
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yKpoI14uk [10]

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3 years ago
The Ksp of yttrium iodate, Y(IO3)3 , is 1.12×10−10 . Calculate the molar solubility, s , of this compound.
torisob [31]

Answer:

1.427x10^-3mol per L

Explanation:

Y(IO_{3} )_{3} ---- Y^{3+} +IO_{3} ^{3-}

I could use ⇌ in the math editor so I used ----

from the question each mole of Y(IO3)3 is dissolved  and this is giving us a mole of Y3+ and a mole of IO3^3-

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1.12x10^-10 = 27S^4

the value of s is 0.001427mol per L

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3 0
3 years ago
How are ironic bonds and covalent bonds different and how are they different?
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<span>2CuO(s) + C(s) </span>→ <span>2Cu(s) + CO</span>₂<span>(g)

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The mass of carbon is then:
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