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vazorg [7]
3 years ago
12

How much pressure would 0.8 moles of a gas at 370K exert if it occupied 17.3L of space

Chemistry
2 answers:
dezoksy [38]3 years ago
7 0

Answer:

1.40 atm is the pressure for the gas

Explanation:

An easy problem to solve with the Ideal Gases Law:

P . V = n . R .T

T° = 370K

V = 17.3L

n = 0.8 mol

Let's replace data → P . 17.3L = 0.8mol . 0.082L.atm/mol.K . 370K

P = (0.8mol . 0.082L.atm/mol.K . 370K) / 17.3L = 1.40 atm

DerKrebs [107]3 years ago
3 0

Answer:

The pressure is 1.40 atm

Explanation:

Step 1: Data given

Number of moles = 0.8 moles

Temperature = 370 K

volume = 17.3 L

Step 2: Calculate the pressure

p*V = n*R*T

⇒ with p = the pressure = TO BE DETERMINED

⇒with V = the volume = 17.3 L

⇒with n = the number of moles of gas = 0.8 moles

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 370 K

p = (n*R*T)/V

p = (0.8 * 0.08206 * 370) / 17.3 L

p = 1.40 atm

The pressure is 1.40 atm

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How many grams of Al2O3are present in 7.1 x 10^22 molecules of Al2O3?
Alexus [3.1K]

Answer:

13.4g

Explanation:

we know that:

1 mole = 6.02 × 10²³ atoms

make the unknown number of moles = x

x = 7.1 × 10²² atoms

putting them both together:

1 mole = 6.02 × 10²³ atoms

x = 7.1 × 10²² atoms

Cross multiply:

6.02 × 10²³ x = 7.1 × 10²²

divide both sides by 6.02 × 10²³

x =  \frac{7.1  \times  10²² }{  6.02 × 10²³}

x =  \frac{71}{602}

we now have the number of moles of Al₂CO₃

to calculate the grams (mass):

moles =  \frac{mass}{relative \: formula \: mass}

mass = moles \:  \times  \: relative \: formula \: mass

add up all of the atomic masses of Al₂CO₃ to calculate relative formula mass:

(27 × 2) + 12 + (16 × 3) = 114

the grams (mass) of Al₂CO₃:

\frac{71}{602}  \times 114 = 13.4g

5 0
3 years ago
Which of the following would increase the rate of dissolving?
Pavlova-9 [17]

Answer:

increasing the temperature

Explanation:

it's not the 2nd one because that makes it take longer

and the 3rd and 4th one are the same thing so it can't be either.

5 0
3 years ago
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Answer: As sea levels rise, coastal communities will lose land and possibly see damages in infrastructure if the levels rise high enough

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3 years ago
Which of the following bases are strong enough to deprotonate CH3CH2CH2C≡CH (pKa = 25), so that equilibrium favors the products?
seraphim [82]

Answer:

a, and f.

Explanation:

To be deprotonated, the conjugate acid of the base must be weaker than the acid that will react, because the reactions favor the formation of the weakest acid. The pKa value measures the strength of the acid. As higher is the pKa value, as weak is the acid. So, let's identify the conjugate acid and their pKas:

a. NaNH2 will dissociate, and NH2 will gain the proton and forms NH3 as conjugate acid. pKa = 38.0, so it happens.

b. NaOH will dissociate, and OH will gain the proton and forms H2O as conjugate acid. pKa = 14.0, so it doesn't happen.

c. NaC≡N will dissociate, and CN will gain a proton and forms HCN as conjugate acid. pKa = 9.40, so it doesn't happen.

d. NaCH2(CO)N(CH3)2 will dissociate and forms CH3(CO)N(CH3)2 as conjugate acid. pKa = -0.19, so it doesn't happen.

e. H2O must gain one proton and forms H3O+. pKa = -1.7, so it doesn't happen.

f. CH3CH2Li will dissociate, and the acid will be CH3CH3. pKa = 50, so it happens.

6 0
3 years ago
A balloon filled with helium has a volume of 5.24 L at 290 ºC. The volume of the balloon decreases to 2.58 L after it is taken o
Neporo4naja [7]
Okay thanks for the update I will give you a call when you get home thanks
6 0
3 years ago
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