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3 years ago

Why is our water treated

Chemistry

1 answer:

DedPeter [7]3 years ago

8 0

Answer: For safety

Explanation: One of the biggest reason why the water we use are treated is so that it is safe for us to use.

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I really need help pls, will pay anything

Triss [41]

Explanation:

For the first part,

Reaction equation:

N₂ + 3H₂ → 2NH₃

Given:

Number of moles of NH₃ = 6 moles

Unknown:

Number of moles of N₂ = ?

Solution:

N₂ + 3H₂ → 2NH₃;

From the reaction above, we solve from the known specie to the unknown. Ensure that the equation is balanced;

2 moles of NH₃ is produced from 1 mole of N₂

6 moles of NH₃ will be produced from $\frac{6}{2}$ mole of N₂

= 3moles of N₂

The number of moles of N₂ is 3 moles

ii.

Given parameters:

Number of moles of sulfur = 2.4moles

Molar mass of sulfur = 32.07g/mol

Unknown:

Mass of sulfur = ?

Solution:

The number of moles of any substance can be found using the expression below;

Number of moles = $\frac{mass}{molar mass}$

Mass of sulfur = number of moles of sulfur x molar mass

Insert the parameters and solve;

Mass of sulfur = 2.4 x 32.07 = 76.97g

6 0

2 years ago

Can a solid defuse into both a liquid and a solid

MariettaO [177]

Diffusion of one state of matter into another: Solid can diffuse in liquid. When sugar is added to water, whole water becomes sweet without stirring it because of diffusion of sugar into water. ... Carbon dioxide and oxygen are the two gases in air which dissolves in water by diffusion.

5 0

2 years ago

Read 2 more answers

Give regents and mechanism

Kryger [21]

Answer:

Reagents: 1) $BH_3$ 2) $H_2_O2$ , $OH^-$

Mechanism: Hydroboration

Explanation:

In this case, we have a hydration of alkenes reaction. But, in this example, we have an anti-Markovnikov reaction. In other words, the "OH" is added in the least substituted carbon. Therefore we have to choose an anti-Markovnikov reaction: "hydroboration".

The first step of this reaction is the addition of borane ( $BH_3$ ) to the double bond. Then in the second step, we have the deprotonation of the hydrogen peroxide, to obtain the peroxide anion. In the third step, the peroxide anion attacks the molecule produced in the first step to produce a complex compound in which we have a bond " $BH_2-O-OH$ ". In step number 4 we have the migration of the C-B bond to oxygen. Then in step number 5, we have the attack of $OH^-$ on the $O-BH_2$ to produce an alkoxide. Finally, the water molecule produce in step 2 will protonate the molecule to produce the alcohol.