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lara31 [8.8K]
3 years ago
8

How many is 10+24x+39x=3

Chemistry
1 answer:
Artemon [7]3 years ago
7 0

Answer:

-7

Explanation:

10+24X+39x=3

10+63x=3......................collect like terms

63x=3-10

63x=-7

x=-7.........................63divided by minus seven

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Which of these best describes the particle motion taking place as CO2 gas is exposed to freezing temperatures?
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The pOH of a solution is 6.0. Which statement is correct? Use p O H equals negative logarithm StartBracket upper O upper H super
castortr0y [4]

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The pH of the solution is 8.0.

Explanation:

taking the test rn

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If an ideal gas has a pressure of 4.03 atm,
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Answer:

PV=nRT where P=pressure in atm, V=volume is liters, n=numbber of moles, R=gas constant, 0.08206 L-atm/mole KL, and T=temperature in K (273 + C).  So (5.67atm)(99.39L)=n(0.08206 L-atm/mol.K)(328.94K), solve for n, the number of moles, n=20.9 moles.

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3 years ago
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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.

5 0
3 years ago
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