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How are distillation and reverse osmosis alike?

Andrei [34K]

I'm having trouble with this one. At the very least, they both purify water. If I'm not mistaken, distillation is part of the process of purifying water.
Really sorry, I hope I helped at least a little! If you have any other questions I might be able to answer better, let me know..

5 0

3 years ago

What is the theoretical yield of Calcium if we begin with 12.6 grams of Aluminium?

AlladinOne [14]

The question is incomplete, here is the complete question:

The given chemical reaction is:

$2Al+3Cu(NO_3)_2\rightarrow 3Cu+2Al(NO_3)_3$

What is the theoretical yield of Calcium if we begin with 12.6 grams of Aluminium?

Answer: The theoretical yield of copper is 44.48 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of aluminium = 12.6 g

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

$\text{Moles of aluminium}=\frac{12.6g}{27g/mol}=0.467mol$

The given chemical equation follows:

$2Al+3Cu(NO_3)_2\rightarrow 3Cu+2Al(NO_3)_3$

By Stoichiometry of the reaction:

2 moles of aluminium produces 3 moles of copper

So, 0.467 moles of aluminium will produce = $\frac{3}{2}\times 0.467=0.7005mol$ of copper

Now, calculating the mass of copper from equation 1, we get:

Molar mass of copper = 63.5 g/mol

Moles of copper = 0.7005 moles

Putting values in equation 1, we get:

$0.7005mol=\frac{\text{Mass of copper}}{63.5g/mol}\\\\\text{Mass of copper}=(0.7005mol\times 63.5g/mol)=44.48g$

Hence, the theoretical yield of copper is 44.48 grams

4 0

3 years ago

Only 20 plant species, such as rice, corn,and barely, make up the main sources of food for the entire human population.

riadik2000 [5.3K]

Answer:

I believe it would be D, to enhance the survival rate.

Explanation:

7 0

3 years ago

Read 2 more answers

The raw water supply for a community contains 18 mg/L total particulate matter. It is to be treated by addition of 60 mg alum (A

s344n2d4d5 [400]

Solution :

Given :

The steady state flow = 8000 $m^3 /d$

$= 80 \times 10^5 \ I/d$

The concentration of the particulate matter = 18 mg/L

Therefore, the total quantity of a particulate matter in fluid $= 80 \times 10^5 \ I/d \times 18 \ mg/L$

$= 144 \times 10^6 \ mg/g$

$= 144 \ kg/d$

If 60 mg of alum $[Al_2(SO_4)_3.14 H_2O]$ required for one litre of the water treatment.

So Alum required for $80 \times 10^5 \ I/d$

$= 80 \times 15^5 \ I/d \times 60 \ mg \ alum /L$

$= 480 \times 10^6 \ mg/d$

or 480 kg/d

Therefore the alum required is 480 kg/d

1 mg of the alum gives 0.234 mg alum precipitation, so 60 mg of alum will give $= 60 \times 0.234 \text{ of alum ppt. per litre}$

$= 14.04$ mg of alum ppt. per litre

480 kg of alum will give = 480 x 0.234 kg/d

= 112.32 kg/d ppt of alum

Daily total solid load is $= 144 \ kg/d + 112.32 \ kg/d$

= 256.32 kg/d

So, the total concentration of the suspended solid after alum addition $= 18 \ mg/L + 60 \times 0.234$

= 32.04 mg/L

Therefore total alum requirement = 480 kg/d

b). Initial pH = 7.4

The dissociation reaction of aluminium hydroxide as follows :

$Al(OH)_3 \rightleftharpoons Al^{3+} + 3OH^{-}$

After addition, the aluminium hydroxide pH of water will increase due to increase in $OH^-$ ions.

Therefore, the pH of water will be acceptable range after the addition of aluminium hydroxide.

c). The reaction of $CO_2$ and water as follows :

$CO_2 (g) + H_2O (l) \rightarrow H_2CO_3$

For the atmospheric pressure :

$p_{CO_2} = 3.5 \times 10^{-4} \ atm$

And the pH is reduced into the range of 5.9 to 6.4

6 0

2 years ago

The compound trimethylamine, (CH3)3N, is a weak base when dissolved in water. Write the Kb expression for the weak base equilibr

Rina8888 [55]

Answer: The expression of $K_b$ is written below.

Explanation:

We are given a chemical compound which is trimethylamine that acts as a weak base when dissolved in water.

It accepts a proton from the water to form trimethylammonium ion and hydroxide ion.

The chemical equation for the reaction of trimethylamine in water follows:

$(CH_3)_3N+H_2O\rightleftharpoons (CH_3)_3NH^++OH^-$

The expression of $K_b$ for above equation follows:

$K_b=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}$

Hence, the expression of $K_b$ is written above.

5 0

3 years ago