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3 years ago

How many moles of chloroform, CHCl3, are required to fill a 253-mL flask at 100.0C and 940 torr?

Chemistry

2 answers:

aalyn [17]3 years ago

6 0

Answer:

relationship between four fundamental physical properties of gases: pressure PP, volume VV, number of moles nn, and the absolute temperature TT. The ideal gas constant is denoted by RR and its exact value depends on the units chosen for the other parameters. Note that the temperature is always in Kelvins (i.e., we use the absolute temperature). The relation is:

PV=nRT

muminat3 years ago

3 0

Answer:

0.0103 mole

Explanation:

Data obtained from the question include:

V (volume) = 253mL

Recall: 1000mL = 1L

Therefore, 253mL = 253/1000 = 0.253L

T (temperature) = 100°C = 100 + 273 = 373K

P (pressure) = 940 torr

Recall: 760torr = 1atm

Therefore, 940torr = 940/760 = 1.24atm

R (gas constant) = 0.082atm.L/Kmol

n (number of mole) =?

Using the ideal gas equation PV = nRT, the number of mole 'n' can be obtained as follow:

PV = nRT

1.24 x 0.253 = n x 0.082 x 373

Divide both side by 0.082 x 373

n = (1.24 x 0.253)/(0.082 x 373)

n = 0.0103 mole

Therefore, the number of mole of chloroform, CHCl3, required to fill the flask is 0.0103 mole

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3 years ago

What is the molar mass of Fe(NO3)3

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4 years ago

Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M.Express your answer to two significant f

N76 [4]

The question is incomplete, here is the complete question:

Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reaction

A + 2B ⇔ C

whose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:

The table is attached below as an image.

Answer: The initial rate for the formation of C at 25°C is $2.25\times 10^{-2}Ms^{-1}$

Explanation:

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$A+2B\rightleftharpoons C$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b$

where,

a = order with respect to A

b = order with respect to B

Expression for rate law for first trial:

$5.4\times 10^{-3}=k(0.30)^a(0.050)^b$ ....(1)

Expression for rate law for second trial:

$1.1\times 10^{-2}=k(0.30)^a(0.100)^b$ ....(2)

Expression for rate law for third trial:

$2.2\times 10^{-2}=k(0.50)^a(0.050)^b$ ....(3)

Dividing 2 by 1, we get:

$\frac{1.1\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.30)^a(1.00)^b}{(0.30)^a(0.050)^b}\\\\2=2^b\\b=1$

Dividing 3 by 1, we get:

$\frac{2.2\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.50)^a(0.050)^b}{(0.30)^a(0.050)^b}\\\\4.07=2^a\\a=2$

Thus, the rate law becomes:

$\text{Rate}=k[A]^2[B]^1$ ......(4)

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

$5.4\times 10^{-3}=k[0.30]^2[0.050]^1\\\\k=1.2M^{-2}s^{-1}$

Calculating the initial rate of formation of C by using equation 4, we get:

$k=1.2M^{-2}s^{-1}$

[A] = 0.50 M

[B] = 0.075 M

Putting values in equation 4, we get:

$\text{Rate}=1.2\times (0.50)^2\times (0.075)^1\\\\\text{Rate}=2.25\times 10^{-2}Ms^{-1}$

Hence, the initial rate for the formation of C at 25°C is $2.25\times 10^{-2}Ms^{-1}$

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3 years ago

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Nesterboy [21]

Answer:

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Explanation:

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