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Readme [11.4K]
3 years ago
10

Chemical formula for calculating pressure​

Chemistry
1 answer:
Anastaziya [24]3 years ago
6 0

Answer:

First, let's review the ideal gas law, PV = nRT. In this equation, 'P' is the pressure in atmospheres, 'V' is the volumes in liters, 'N' is the number of particles in moles, 'T' is the temperature in Kelvin and 'R' is the ideal gas constant (0.0821 liter atmosphere per moles Kelvin).

Explanation:

I majored in Chemistry

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Select the correct answer. Which statement is true for a heating curve?
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Answer:

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Explanation:

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Sitting at a four-legged table there is: 1 grandma, 2 moms, 2 daughters and a granddaughter. How many legs are under the table?
lbvjy [14]

Answer:

The answer is 14 "legs"

Explanation:

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3 years ago
How much 10.0 M HNO must be added to 1.00 L of a buffer that is 0.0100 M acetic acid and 0.100 M sodium acetate to reduce the pH
Naya [18.7K]

Explanation:

It is given that molarity of acetic acid = 0.0100 M

Therefore, moles of acetic acid = molarity of acetic acid × volume of buffer

            Moles of acetic acid = 0.0100 M × 1.00 L

                                              = 0.0100 mol

Similarly, moles of acetate = molarity of sodium acetat × volume of buffer

                                           = 0.100 mol

When HNO_{3} is added, it will convert acetate to acetic acid.

Hence, new moles acetic acid = (initial moles acetic acid) + (moles HNO_{3})

                                                = 0.0100 mol + x

New moles of sodium acetate = (initial moles acetate) - (moles HNO_{3})

                                        = 0.100 mol - x

According to Henderson - Hasselbalch equation,

           pH = pK_{a} + log\frac{[conjugate base]}{[weak acid]}

             pH = pKa + log\frac{(new moles of sodium acetate)}{(new moles of acetic acid)}

           4.95 = 4.75 + log\frac{(0.100 mol - x)}{(0.0100 mol + x)}

       log\frac{(0.100 mol - x)}{(0.0100 mol + x)}  = 4.95 - 4.75

                                            = 0.20

\frac{(0.100 mol - x)}{(0.0100 mol + x)}  = antilog (0.20)

                                           = 1.6

Hence,    x = 0.032555 mol

Therefore, moles of HNO_{3} = 0.032555 mol

volume of HNO_{3} = \frac{moles HNO_{3}}{molarity of HNO_{3}}

                                = \frac{0.032555 mol}{10.0 M}

                                 = 0.0032555 L

or,                             = 3.25           (as 1 L = 1000 mL)

Thus, we can conclude that volume of HNO_{3} added is 3.26 mL.

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