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Papessa [141]
3 years ago
11

(1.000) How many moles can be found in 83.80 grams of krypton?

Chemistry
1 answer:
marin [14]3 years ago
6 0

Answer: 1 mole Kr

Explanation: solution

83.80g Kr x 1mole Kr / 84 g Kr

= 0.997 or approximately 1 mole Kr

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Charles’ Law can be expressed mathematically as V=kT. If the constant k for 6.0 L of gas is 0.020 L/K (liters per kelvin), what
kozerog [31]

Answer: 300 K

Explanation:

Charles' Law: This law states that volume is directly proportional to the temperature of the gas at constant pressure and number of moles.

V\propto T    (At constant pressure and number of moles)

V=kT

Given : V= 6.0 L

k= 0.020 L/K

T=?

6.0=0.020LK^{-1}\times T

T=300 K

Thus temperature of the gas is 300 K.

3 0
2 years ago
Identify a cation. An atom that has gained a proton. An atom that has lost an electron. An atom that has gained an electron. An
iren2701 [21]

the answer would be B an atom that has lost an electron

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Some people say that only gay and bisexual men are likely to be infected with hub true false
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Answer:

They do have a larger chance, but others might be infected too.

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3 0
3 years ago
Read 2 more answers
Which scientist developed the first model of the atom that showed the structure of the inside of an atom
Eduardwww [97]

Answer:

Which scientist developed the first model of the atom that showed the structure of the inside of an atom

Ernest Rutherford

8 0
2 years ago
Using the van der Waals equation, determine the pressure of 500.0 g of SO2(g) in a 6.30-L vessel at 633 K. For SO2(g), a = 6.865
dem82 [27]

Answer:

The pressure is 58.75 atm.

Explanation:

From Vanderwaal's equation,

P = nRT/(V-nb) - n^2a/V^2

n is the number of moles of SO2 = mass/MW = 500/64 = 7.81 mol

R is gas constant = 0.0821 L.atm/mol.K

T is temperature of the vessel = 633 K

V is volume of the vessel = 6.3 L

a & b are Vanderwaal's constant = 6.865 L^2.atm/mol^2 and 0.0567 L/mol respectively.

P = (7.81×0.0821×633)/(6.3 - 7.81×0.05679) - (7.81^2 × 6.865)/6.3^2 = 69.30 - 10.55 = 58.75 atm

4 0
2 years ago
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