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Tasya [4]
4 years ago
8

Calculate the number of moles of solute present and the number of grams

Chemistry
1 answer:
nekit [7.7K]4 years ago
6 0

Answer:

The answer is: 18 moles and 1341, 72 grams of KCl

Explanation:

The molarity is defined as the moles of solute ( in this case KCl) in 1 liter of solution:

1L solution-----3 moles of KCl

6L solution-----x= (6L solutionx 3 moles of KCl)/1 L solution= <em>18 moles of KCl</em>

<em></em>

We calculate the weight of 1 mol of KCl:

Weight 1 mol KCl= Weight K + Weight Cl= 39,09 g + 35, 45 g=74, 54 g/mol

1 mol KCl----- 74, 54 g

18 mol KCl----x= (18 mol KCl x 74, 54 g)/1 mol KCl=<em>1341, 72 g</em>

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7 0
3 years ago
Because of river discharge, large changes in ocean salinity often occur __.
Lorico [155]
I’m guessing it would be A because the river discharge tends to gather near coastlines where the river ends
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1. How is the modern periodic table organized?
Dimas [21]

1. How is the modern periodic table organized?  Increasing atomic number

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4. Would you expect Strontium (Sr) to be more like potassium (K) or bromine (Br)?  

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3 0
3 years ago
Which lists the elements in order from least conductive to most conductive?
Savatey [412]

Answer: Option (A) is the correct answer.

Explanation:

Nitrogen is a non-metal and it is known that non-metals do not conduct electricity. Thus, it will be least conductive out of the given options.

Whereas antimony (Sb) is a metalloid. Metalloid are the substance that show properties of both metals and non-metals. Thus, antimony will conduct electricity.

On the other hand, bismuth (Bi) is a metal hence, it will conduct electricity.

Thus, we can conclude that the order from least conductive to most conductive will be nitrogen (N), antimony (Sb), bismuth (Bi).

 

7 0
3 years ago
Read 2 more answers
You have 100 mL of a solution of benzoic acid in water; the amount of benzoic acid in the solution is estimated to be about 0.30
dimaraw [331]

Answer:

0.00370 g

Explanation:

From the given information:

To determine the amount of acid remaining using the formula:\dfrac{(final \ mass \ of \ solute)_{water}}{(initial \ mass \ of \ solute )_{water}} = (\dfrac{v_2}{v_1+v_2\times k_d})^n

where;

v_1 = volume of organic solvent = 20-mL

n = numbers of extractions = 4

v_2 = actual volume of water = 100-mL

k_d = distribution coefficient = 10

∴

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +20 \ ml \times 10})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +200 \ ml})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{1}{3})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = 0.012345

Thus, the final amount of acid left in the water = 0.012345 * 0.30

= 0.00370 g

3 0
3 years ago
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