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qwelly [4]
3 years ago
13

7. Calculate the amount of sodium hydroxide you will need to make 250 mL of a 0.1 M solution of sodium hydroxide.

Chemistry
1 answer:
tresset_1 [31]3 years ago
7 0

Answer:

1 g

Explanation:

Given data:

Volume of solution = 250 mL (250 mL × 1 L/1000 mL = 0.25 L)

Molarity of solution = 0.1 M

Amount of sodium hydroxide = ?

Solution:

Molarity = number of moles / volume in L

0.1 M = number of moles /0.25 L

Number of moles = 0.1 M × 0.25 L

            M = mol/ L

Number of moles = 0.025 mol

Mass of sodium hydroxide:

Mass = number of moles × molar mass

Mass = 0.025 mol ×40 g/mol

Mass = 1 g

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(a) the temperature on a warm summer day is 87 °f. what is the temperature in °c? (b) many scientific data are reported at 25
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              (a)  30.55 °C

              (b) 298 K and 77°F

              (c)  204.44 °C and 477.44 K

              (d)  -320.8 °F and -196 °C

Explanation:

Converting °C into °F;

                                   °F  =  °C × 1.8 + 32

Converting °F into °C;

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8 0
3 years ago
What is the Ka of a weak acid (HA) if the initial concentration of weak acid is 4.5 x 10-4 M and the pH is 6.87? (pick one)
rewona [7]

Answer:

Ka = 4.04 \times 10^{-11}

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Initial concentration of weak acid = 4.5 \times 10^{-4}\ M

pH = 6.87

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[H^+]=10^{-pH}

[H^+]=10^{-6.87}=1.35 \times 10^{-7}\ M

HA dissociated as:

HA \leftrightharpoons H^+ + A^{-}

(0.00045 - x)    x     x

[HA] at equilibrium = (0.00045 - x) M

x = 1.35 \times 10^{-7}\ M

Ka = \frac{[H^+][A^{-}]}{[HA]}

Ka = \frac{(1.35 \times 10^{-7})^2}{0.00045 - 0.000000135}

0.000000135 <<< 0.00045

Therefore, Ka = \frac{(1.35 \times 10^{-7})^2}{0.00045 } = 4.04 \times 10^{-11}

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