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Serhud [2]
3 years ago
12

In which of the following is concentration expressed in percent by volume?

Chemistry
2 answers:
nevsk [136]3 years ago
6 0

Answer:

Concentration expressed in percent by volume by volume is 10%(v/v).

Explanation:

  • w/w % or m/m% : The percentage mass or fraction of mass of the of solute present in total mass of the solution.

w/w\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100

  • w/v % or m/v: The percentage of mass of the of solute present in total volume of the solution.

w/v\%=\frac{\text{Mass of solute}}{\text{Volume of solution}}\times 100

  • v/v % : The percentage volume of the of solute present in total volumeof the solution.

v/v\%=\frac{\text{Volume of solute}}{\text{Volume of solution}}\times 100

Concentration expressed in percent by volume by volume is 10%(v/v).

alexira [117]3 years ago
4 0

Answer: 10% (v/v).

Explanation:

V - unit is litre

m/m - by mass concentration

m/v - unit of density

v/v - concentration by volume

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Estimate the percent ionic character of the co bond. (note that the electronegativity of c is 2.5 and that of o is 3.5.)
deff fn [24]

From the calcuation, the percent ionic character of the bond is  70%

<h3>What is percent ionic character?</h3>

The term percent ionic character has to do with the degree of ionic bonding that is contained in a compound. It can be estimated from the electronegativity of each element.

We can use the formula; 100(1 - e^(-ΔEN² / 4))

EN = χB − χA * 100/1

EN = 3.5 - 1.5 = 1

100(1 - e^(-1)^2/4)

= 70%

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5 0
2 years ago
If the mass of a material is 106 grams and the volume of the material is 23 cm3, what would the density of the material be? g/cm
den301095 [7]
Density is mass over volume
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7 0
3 years ago
What mass of solid sodium formate (of MW 68.01) must be added to 150 mL of 0.42 mol/L formic acid (HCOOH) to make a buffer solu-
Sergio [31]

Answer:

We need 4.28 grams of sodium formate

Explanation:

<u>Step 1:</u> Data given

MW of sodium formate = 68.01 g/mol

Volume of 0.42 mol/L formic acid = 150 mL = 0.150 L

pH = 3.74

Ka = 0.00018

<u>Step 2:</u> Calculate [base)

3.74 = -log(0.00018) + log [base]/[acid]

0 = log [base]/[acid]

0 = log [base] / 0.42

10^0 = 1 = [base]/0.42 M

[base] = 0.42 M

<u>Step 3:</u> Calculate moles of sodium formate:

Moles sodium formate = molarity * volume

Moles of sodium formate = 0.42 M * 0.150 L = 0.063 moles

<u>Step 4:</u> Calculate mass of sodium formate:

Mass sodium formate = moles sodium formate * Molar mass sodium formate

Mass sodium formate = 0.063 mol * 68.01 g/mol

Mass sodium formate = 4.28 grams

We need 4.28 grams of sodium formate

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3 years ago
A root is an example of a/an-<br><br> A: tissue <br> B: cell type<br> C:organ<br> D:organism
konstantin123 [22]

Answer: D an organism

Explanation:

7 0
3 years ago
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4. A 134.0 g sample of an unknown metal is heated to 91.0°C and then placed in 125 g of water at 25.0°C. The final temperature o
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