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Harlamova29_29 [7]

3 years ago

7

hey i got a question for science its " you throw a rock with a mass of 3.6 kg. the ball leaves your hand with a speed of 10 m/s

DO NOT INCLUDE J IN THE ANSWER

1 answer:

ANTONII [103]3 years ago

7 0

it is 3.9 i know tgis because you throw a rock with a mass of 3.6 kg. the ball leaves your hand with a speed of 10

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A solution is prepared by mixing 93.0 mL of 5.00 M HCl and 37.0 mL of 8.00 M HNO3. Water is then added until the final volume is

Charra [1.4K]

Answer:

$[H^{+}] = 0.761 \frac{mol}{L}$

$[OH^{-}]=1.33X10^{-14}\frac{mol}{L}$

$pH = 0.119$

Explanation:

HCl and HNO₃ both dissociate completely in water. A simple method is to determine the number of moles of proton from both these acids and dividing it by the total volume of solution.

$n_{H^{+} } from HCl = [HCl](\frac{mol}{L}). V_{HCl}(L) \\ n_{H^{+} } from HNO_{3} = [HNO_{3}](\frac{mol}{L}). V_{HNO_{3}}(L)$

Here, n is the number of moles and V is the volume. From the given data moles can be calculated as follows

$n_{H^{+} } from HCl = (5.00)(0.093)$

$n_{H^{+} } from HCl = 0.465 mol$

$n_{H^{+} } from HNO_{3} = (8.00)(0.037)$

$n_{H^{+} } from HNO_{3} = 0.296 mol$

$n_{H^{+}(total) } = 0.296 + 0.465$

$n_{H^{+}(total) } = 0.761 mol$

For molar concentration of hydrogen ions:

$[H^{+}] = \frac{n_{H^{+}}(mol)}{V(L)}$

$[H^{+}] = \frac{0.761}{1.00}$

$[H^{+}] = 0.761 \frac{mol}{L}$

From dissociation of water (Kw = 1.01 X 10⁻¹⁴ at 25°C) [OH⁻] can be determined as follows

$K_{w} = [H^{+} ][OH^{-} ]$

$[OH^{-}]=\frac{Kw}{[H^{+}] }$

$[OH^{-}]=\frac{1.01X10-^{-14}}{0.761 }$

$[OH^{-}]=1.33X10^{-14}\frac{mol}{L}$

The pH of the solution can be measured by the following formula:

$pH = -log[H^{+} ]$

$pH = -log(0.761)$

$pH = 0.119$

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<span>Since P, M and T are constant, the equation becomes </span>
<span>V1/m1 = V2/m2 </span>
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