3.The simplest way of separating visible solid impurities that do not dissolve in water from water

1 answer:

3 0

The simplest way of separating visible solid impurities that do not dissolve in water from water is by using a sieve (option A).

<h3>What is sieving?</h3>

Sieving is a act of passing substances through a sieve, which is a device with a mesh bottom to separate, in a granular material, larger particles from smaller ones, or to separate solid objects from a liquid.

Sieving is a physical method of separation widely used to separate substances that cannot dissolve in a liquid from that liquid.

Therefore, the simplest way of separating visible solid impurities that do not dissolve in water from water is by using a sieve.

Learn more about sieving at: brainly.com/question/2760394

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What mass of butane in grams is necessary to produce 1.5×103 kj of heat what mass of co2 is produced?

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The heat of reaction (i.e. combustion) of butane ( $C_{4} H_{10}$ ) when reacted with oxygen ( $O_{2}$ )  is -2658 kJ/mol butane, and the chemical reaction is given by:

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Dividing this with the heat of reaction per mole of butane reacted would give the number of moles butane required. Then, multiplying the answer with the molar mass of butane which is 58 grams/mole, will give the mass of butane required.

Moles of butane = [(-1,500 kJ)/(-2658 kJ/mol butane)]
Moles of butane = 0.5643 moles butane

Mass of butane  = 0.5643 moles butane * 58 grams/mol butane
Mass of butane = 32.73 grams butane

The mass of carbon dioxide ( $CO_{2}$ ) can be determined by multiplying the moles of butane ( $C_{4} H_{10}$ ) with the mole ratio of ( $CO_{2}$ ) produced to the ( $C_{4} H_{10}$ ) reacted, and then with the molar mass of ( $CO_{2}$ ), which is 44 grams/mole.

Mass of carbon dioxide produced
= 0.5643 moles butane * [4 moles $CO_{2}$ / 1 mole $C_{4} H_{10}$ ] * 44 grams/mole $CO_{2}$

Mass of carbon dioxide produced
= 99.32 grams $CO_{2}$

Thus, the mass of butane required is 32.73 grams, and the mass of carbon dioxide produced from the reaction of this amount of butane is 99.32 grams.

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