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Wittaler [7]
4 years ago
14

A mixture consists of saw dust, salt, and pebbles. Which of the following instruments or materials would not be useful in separa

ting this mixture?
A.magnet
B.filter paper
C.screen
D.water
Chemistry
1 answer:
andriy [413]4 years ago
8 0

The screen will not be useful in separating the mixture of saw dust, salt and pebbles.

Answer: Option C

<u>Explanation: </u>

The separation of individual elements from the mixture can be done easily if the size of the elements in the mixture is varying significantly. Just like in the present case, the size of sawdust and salt is very small compared to that of pebbles.

So by using sieves having pores of smaller sizes, pebbles can be separated from the mixture as pebbles will not pass through the minute pores. Even depending upon the size of pores, the saw dust can also be separated using it.

Similarly, we can use water also to separate this mixture as sawdust have least density it will float in water surface and pebbles being the highest density will sink and settle in the bottom of the container.

Remaining salt will get dissolved in water which can be obtained by heating the water in order to precipitate the salt. Even filter paper with different mess size will also help in separating this mixture. But as this mixture is not permeable through screen, the screen cannot be used to separate this mixture.

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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
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Answer:

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

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

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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

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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 }

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

pH = -log[H^{+} ]

pH = -log(0.761)

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