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3 years ago

What is the molar mass of a gas which occupies 48.7 L at 111oC under 698 torr of pressure and has a mass of 21.5 grams?

Chemistry

1 answer:

seropon [69]3 years ago

8 0

Answer:

Molar mass of the gas = 15.15 g/mol

Explanation:

PV = nRT

Where,

P = pressure

n = No. of moles

R = Gas constant

T = Temperature

P = 698 torr, 1 torr = 0.00131579 atm

$698\;torr = 698\times0.00131579 = 0.9184\;atm$

Temperature = 111 °C = 100 + 273.15 = 384.15 K

V = 48.7 L

R = 0.082057 L atm/mol K

Now, PV = nRT

$n = \frac{PV}{RT}$

$n = \frac{0.9184\times48.7}{0.082057\times384.15}$

=1.4189 mol

Molar mass = Mass/ No. of moles

= 21.5/1.4189

=15.15 g/mol

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Assuming that all the energy given off in the reaction goes to heating up only the air in the house, determine the mass of metha

nirvana33 [79]

Answer:

0.92 kg

Explanation:

The volume occupied by the air is:

$35.0m\times 35.0m \times 3.2m \times \frac{10^{3}L }{1m^{3} } =3.9 \times 10^{6} L$

The moles of air are:

$3.9 \times 10^{6} L \times \frac{1.00mol}{22.4L} =1.7 \times 10^{5}mol$

The heat required to heat the air by 10.0 °C (or 10.0 K) is:

$1.7 \times 10^{5}mol \times \frac{30J}{K.mol} \times 10.0 K = 5.1 \times 10^{7}J$

Methane's heat of combustion is 55.5 MJ/kg. The mass of methane required to heat the air is:

$5.1 \times 10^{7}J \times \frac{1kgCH_{4}}{55.5 \times 10^{6} J } =0.92kgCH_{4}$

3 0

3 years ago

How many grams of dry nh4cl need to be added to 2.50 l of a 0.800 m solution of ammonia, nh3, to prepare a buffer solution that

tia_tia [17]

Answer:

The correct answer is 574.59 grams.

Explanation:

Based on the given information, the number of moles of NH₃ will be,

= 2.50 L × 0.800 mol/L

= 2 mol

The given pH of a buffer is 8.53

pH + pOH = 14.00

pOH = 14.00 - pH

pOH = 14.00 - 8.53

pOH = 5.47

The Kb of ammonia given is 1.8 * 10^-5. Now pKb = -logKb,

= -log (1.8 ×10⁻⁵)

= 5.00 - log 1.8

= 5.00 - 0.26

= 4.74

Based on Henderson equation:

pOH = pKb + log ([salt]/[base])

pOH = pKb + [NH₄⁺]/[NH₃]

5.47 = 4.74 + log ([NH₄⁺]/[NH₃])

log([NH₄⁺]/[NH₃]) = 5.47-4.74 = 0.73

[NH₄⁺]/[NH₃] = 10^0.73= 5.37

[NH₄⁺ = 5.37 × 2 mol = 10.74 mol

Now the mass of dry ammonium chloride required is,

mass of NH₄Cl = 10.74 mol × 53.5 g/mol

= 574.59 grams.

8 0

3 years ago

Determine % yield if a student obtains 45 g of product in an experiment and the theoretical amount is determined to be 50 g. *

ycow [4]

Answer:

90%

Explanation:

Percentage yield = ?

Theoretical yield = 50g

Actual yield = 45g

To calculate the percentage yield of a compound, we'll have to use the formula of percentage yield which is the ratio between the actual yield to theoretical multiplied by 100

Percentage yield = (actual yield / theoretical yield) × 100

Percentage yield = (45 / 50) × 100

Percentage yield = 0.9 × 100

Percentage yield = 90%

The percentage yield of the substance is 90%

6 0

3 years ago

Which of the following could be a long-term health effect of chemical pesticide use?

Helen [10]

The correct Answer for this question is D

3 0

3 years ago

The first-order rate constant for the reaction of methyl chloride (CH3Cl) with water to produce methanol (CH3OH) and hydrochlori

Dvinal [7]

Answer:

K(48.5°C) = 1.017 E-8 s-1

Explanation:

CH3Cl + H2O → CH3OH + HCl

at T1 = 25°C (298 K) ⇒ K1 = 3.32 E-10 s-1

at T2 = 48.5°C (321.5 K) ⇒ K2 = ?

Arrhenius eq:

K(T) = A e∧(-Ea/RT)
Ln K = Ln(A) - [(Ea/R)(1/T)]

∴ A: frecuency factor

∴ R = 8.314 E-3 KJ/K.mol

⇒ Ln K1 = Ln(A) - [Ea/R)*(1/T1)]..........(1)

⇒ Ln K2 = Ln(A) - [(Ea/R)*(1/T2)].............(2)

(1)/(2):

⇒ Ln (K1/K2) = (Ea/R)* (1/T2-1/T1)

⇒ Ln (K1/K2) = (116 KJ/mol/8.3134 E-3 KJ/K.mol)*(1/321.5 K - 1/298 K)

⇒ Ln (K1/K2) = (13952.37 K)*(- 2.453 E-4 K-1)

⇒ Ln (K1/K2) = - 3.422

⇒ K1/K2 = e∧(-3.422)

⇒ (3.32 E-10 s-1)/K2 = 0.0326

⇒ K2 = (3.32 E-10 s-1)/0.0326

⇒ K2 = 1.017 E-8 s-1

7 0

3 years ago