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zysi [14]
2 years ago
7

A student has a sample of 1.33 moles of fluorine gas that is contained in a 23.3 L container at 314 K. What is the pressure of t

he sample? The ideal gas constant is 0.0821 L*atm/mol*K. Round your answer to the nearest 0.01 and include units.
Chemistry
1 answer:
Gelneren [198K]2 years ago
4 0

1.4715 atm is the pressure of the sample 1.33 moles of fluorine gas that is contained in a 23.3 L container at 314 K.

What is an ideal equation?

The ideal gas equation, pV = nRT, is an equation used to calculate either the pressure, volume, temperature or number of moles of a gas. The terms are: p = pressure, in pascals (Pa).

Given data:

Volume (V) = 23.3 L

Number of mole (n) = 1.33 moles

Temperature (T) = 314 K

Gas constant (R) = 0.821 atm.L/Kmol

Pressure (P) =?

The pressure inside the container can be obtained by using the ideal gas equation as illustrated below:

PV = nRT

P × 23.3 L = 1.33 moles × 0.0821 ×314 K

P = 1.4715 atm

Therefore, the pressure of the sample is 1.4715 atm.

Learn more about the ideal gas equation:

brainly.com/question/23826793

SPJ1

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<h3>What is citric acid?</h3>

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8 0
1 year ago
Which of the following statements is true of combustion reactions?
enot [183]

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Combustion reactions release large amounts of heat. They have negative enthalpy. A negative enthalpy represents an exothermic reaction, releasing heat. This reaction is spontaneous and exothermic, since we can obtain energy from the reaction; the ΔG (free energy) is negative (So 1 is true).

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8 0
2 years ago
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¿Que es la energía térmica?
qaws [65]
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7 0
3 years ago
More efficient plumbing reduces water pollution by _______. a. using more water b. using less water c. using cleaner water d. us
Aliun [14]

Answer:

b. using less water

Explanation:

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In the United States of America, the agency which was established by US Congress and saddled with the responsibility of overseeing all aspects of pollution, environmental clean up, pesticide use, contamination, and hazardous waste spills is the Environmental Protection Agency (EPA). Also, EPA research solutions, policy development, and enforcement of regulations through the resource Conservation and Recovery Act.

In homes and offices, an effective and efficient design of pipes and other water-related facilities would go a long way to reduce water pollution because the water used are well managed and properly disposed into sinkholes.

Hence, more efficient plumbing reduces water pollution by using less water

6 0
2 years ago
50.0 mL of 0.200 M HNO2 is titrated to its equivalence point with 1.00 M NaOH. What is the pH at the equivalence point?
yKpoI14uk [10]

Answer:

8.279

Explanation:

The pH can be determined by hydrolysis of a conjugate base of weak acid at the equivalence point.

At the equivalence point, we have

$n_{NaOH}=n_{HNO_2}$

           = 25.00 x 0.200

           = 5.00 m-mol

           = 0.005 mol

Volume of the base that is added to reach the equivalence point is

$\frac{0.005}{1.00} \times 1000= 5.00 \ mL$

Number of moles of $NO^-_{2}=n_{HNO_2}$

                                           = 0.005 mol

Volume at the equivalence point is 25 + 5 = 30.00 mL

Therefore, concentration of $NO^-_{2}= \frac{5}{30}$

                                                        = 0.167 M

Now the ICE table :

            $NO^-_2 + H_2O \rightarrow HNO_3 + OH^-$

I (M)       0.167                   0            0

C (M)         -x                      +x          +x

E (M)      0.167-x                  x           x

Now, the value of the base dissociation constant is ,

$K_w=K_a \times K_b$            $(K_w \text{ is the ionic product of water })$

$K_b =\frac{K_w}{K_a}$

$K_b =\frac{1 \times 10^{-14}}{4.6 \times 10^{-4}}$

    = $2.174 \times 10^{-11}$

Base ionization constant, $K_b = \frac{\left[HNO_2\right] \left[OH^- \right]}{\left[NO^-_2 \right]}$

$2.174 \times 10^{-11}=\frac{x^2}{0.167 -x}$

$x= 1.9054 \times 10^{-6}$

So, $[OH^-]=1.9054 \times 10^{-6 } \ M$

pOH =- $\log[OH^-]$

       = $- \log(1.9054 \times 10^{-6} \ M)$

        =5.72

Now, since pH + pOH = 14

           pH = 14.00 - 5.72

                = 8.279

Therefore the ph is 8.279 at the end of the titration.

8 0
3 years ago
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