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butalik [34]
2 years ago
12

How can the requirements of Na (sodium) solve the requirements of Cl (chloride) instability achievement?

Chemistry
1 answer:
Paul [167]2 years ago
8 0

Na releases 1 electron to be stable

Cl requires 1 electron to be stable

both are ionic bonded to be stable

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For the aqueous solution containing 75 mg of compound C in 1.0 mL of water, what will be the total amount of the solute C that w
Sindrei [870]

Answer:

75 mg

Explanation:

We can write the extraction formula as

x = m/[1 + (1/K)(Vaq/Vo)], where

x = mass extracted

m = total mass of solute

K = distribution coefficient

Vo = volume of organic layer

Vaq = volume of aqueous layer

Data:

m = 75 mg

K = 1.8

Vo = 0.90 mL

Vaq = 1.00 mL

Calculations:

For each extraction,

1 + (1/K)(Vaq/Vo) = 1  + (1/1.8)(1.00/0.90) = 1 + 0.62 = 1.62  

x = m/1.62 = 0.618m

So, 61.8 % of the solute is extracted in each step.

In other words, 38.2 % of the solute remains.

Let r = the amount remaining after n extractions. Then  

r = m(0.382)^n.

If n = 7,

r = 75(0.382)^7 = 75 × 0.001 18 = 0.088 mg

m = 75 - 0.088 = 75 mg

After seven extractions, 75 mg (99.999 %) of the solute will be extracted.

5 0
3 years ago
I can't leave till I answer please help!!​
vovikov84 [41]
I think the answer would be 2
3 0
3 years ago
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What is the only intermolecular force present in nonpolar compounds?
Gnom [1K]

The intermolecular force of attraction between non-polar molecules is called as dispersion force . Another trem for dispersion force is London dispersion force.

6 0
3 years ago
When 6.0 mol Al react with 13 mol HCl, what is the limiting reactant, and how many moles of H2 can be formed?
Ad libitum [116K]

You're looking for the number of moles of H2, and you have 6.0 mol Al and 13 mol HCL.

For the first part, you have to make your way from 6.0 mol of Al to mol of H2, right? For that to happen, you need to make a conversion factor that will cancel the mol Al, in such case use the 2 moles of Al from your equation to cancel them out. At the top of the equation, you can use the number of moles of H2 from the equation and find the moles that will be produced for the H2.

6.0mol Al x 3 mol H2/2 mol Al = 9 mol H2

For the second part, you have to make the same procedure, make a conversion factor that will cancel the mol of HCL and for that you need to use the 6 mol HCL from your equation, and at the numerator you can put the 3 mol of H2 from the equation so that you can find the number of moles of H2 that will be produced.

13 mol HCL x 3 mol H2/6 mol HCL = 6.5 mol H2

As it can be seen, HCL produces the less amount of H2 moles. Therefore, the reaction CANNOT produce more than 6.5 mol H2, in that case 6.5 mol will be the maximum number of moles that will be produced at the end because HCL does not have enough to produce more than 6.5 mol.

In that case HCL is the limiting reactant because it limits that will be produced, and so the answer is B!

6 0
3 years ago
What is the hypothesis
Ulleksa [173]
A hypothesis is an educated guess, most likely in an "If... and then..." statement.
So, this paper is asking you to make a hypothesis about this experiment?
So: "If this happened, then this is why." Just add in your own details. :)
6 0
3 years ago
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