Would question 1B be the same set up and answer as 1A? Please help!

21) What is the volume (inL) of 3470 g of liquid glycerol, which has a density of 1.26 g/?

Illusion [34]

Answer:

V = 2.75 L

Explanation:

Given that,

Mass of liquid glycerol, m = 3470 g

The density of liquid glycerol, d = 1.26 g/cm³

We need to find the volume of liquid glycerol.

We know that,

Density=mass/volume

$V=\dfrac{m}{d}\\\\V=\dfrac{3470\ g}{1.26\ g/cm^3}\\\\V=2753.96\ cm^3$

We know that, 1 cm³ = 0.001 L

V = 2.75 L

So, the volume of liquid glycerol is 2.75 L.

6 0

3 years ago

Worksheet Ch.3: % Composition, Molarity, and Other Units for

Ann [662]

The percentage composition by mass of the elements in the compound are:

Ai. The percentage composition by mass of Na in 3.65 g of NaF is 54.8%

Aii. The percentage composition by mass of F in 3.65 g of NaF is 45.2%

Bi. The percentage composition by mass of Zinc in the compound is 67.1%

Bii. The percentage composition by mass of Sulfur in the compound is 32.9%

Percentage composition by mass of an element in a compound can be obtained by using the following formula: $Percentage = \frac{mass of Element}{mass of compound} * 100$

Ai. Determination of the percentage of Na in 3.65 g of NaF

Mass of Na = 2 g

Mass of NaF = 3.65 g

<h3>Percentage of Na =? </h3>

$Percentage = \frac{mass of element }{mass of compound} * 100\\\\Percentage of Na = \frac{2}{3.65} * 100$

<h3>Percentage of Na = 54.8%</h3>

Aii. Determination of the percentage of F in 3.65 g of NaF

Mass of F = 1.65 g

Mass of NaF = 3.65 g

<h3>Percentage of F =? </h3>

$Percentage = \frac{mass of element }{mass of compound } * 100\\\\Percentage of F = \frac{1.65}{3.65} * 100\\\\$

<h3>Percentage of F = 45.2%</h3>

Bi. Determination of the percentage composition of Zn in 48.72 g of the compound

Mass of Zn = 32.69 g

Mass of compound = 48.72 g

<h3>Percentage of Zn =? </h3>

$Percentage = \frac{mass of element}{mass of compound} * 100\\\\Percentage of Zn = \frac{32.69}{48.72} *100\\\\$

<h3>Percentage of Zn = 67.1%</h3>

Bii. Determination of the percentage composition of Sulphur in 48.72 g of the compound

Mass of Zn = 32.69 g

Mass of compound = 48.72 g

Mass of S = 48.72 – 32.69 g

Mass of S = 16.03 g

<h3>Percentage of S =? </h3>

$Percentage = \frac{mass of element}{mass of compound} * 100\\\\Percentage of S = \frac{16.03}{48.72} * 100\\\\$

<h3>Percentage of S = 32.9%</h3>

Learn more: brainly.com/question/1350382

3 0

3 years ago

5) Calculate the molality of 0.210 mol of KBr dissolved in 0.075kg pure<br> water?

Margaret [11]

Answer:

$\boxed {\boxed {\sf 2.8 \ m }}$

Explanation:

The formula for molality is:

$m=\frac{moles \ of \ solute}{kg \ of \ solvent}$

There are 0.210 moles of KBr and 0.075 kilograms of pure water.

$moles= 0.210 \ mol \\kilograms = 0.075 \ kg$

Substitute the values into the formula.

$m= \frac{ 0.210 \ mol }{0.075 \ kg}$

Divide.

$m= 2.8 \ mol/kg= 2.8 \ m$

The molality is <u>2.8 moles per kilogram</u>

5 0

3 years ago

A 11.630 g milk chocolate bar is found to contain 7.815 g of sugar.How many milligrams of sugar does the milk chocolate bar cont

zzz [600]

3.815 that's the awnser......................................

4 0

3 years ago

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Which type of bond is present in hydrogen sulfide (H2S)? The table of electronegativities is given

LuckyWell [14K]

Bonds formed between atoms can be classified as ionic and covalent

Ionic bonds are formed between atoms that have a high difference in the electronegativity values.

In contrast, bonds formed between atoms that have a difference in electronegativity lower than the ionic counterparts are polar covalent bonds. If the atoms have very similar electronegativities, they form non-polar covalent bonds.

In H2S, the S atom is bonded to 2 H atoms. The electronegativity of H = 2.2 and S= 2.56. Since the difference is not high the bond formed will be covalent (polar covalent).

7 0

4 years ago

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