All categories

3 years ago

Find the volume in m3 of 52.6 lbm of iron:

Chemistry

1 answer:

balandron [24]3 years ago

8 0

Answer: The volume of iron is $3.031\times 10^{-3}m^3$

Explanation:

To calculate volume of a substance, we use the equation:

$\text{Density of a substance}=\frac{\text{Mass of a substance}}{\text{Volume of a substance}}$

We are given:

Mass of iron = 52.6 lbm = 23.86 kg (Conversion factor: 1 kg = 2.205 lbm)

Density of iron = $7873kg/m^3$

Putting values in above equation, we get:

$7873kg/m^3=\frac{23.86kg}{\text{Volume of iron}}\\\\\text{Volume of iron}=3.031\times 10^{-3}m^3$

Hence, the volume of iron is $3.031\times 10^{-3}m^3$

You might be interested in

If you have 16 g of manganese (II) nitrate tetrahydrate, how much water is required to prepare 0.16 M solution from this amount

Schach [20]

Answer: The amount of water required to prepare given amount of salt is 398.4 mL

Explanation:

To calculate the volume of solution, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Molarity of solution = 0.16 M

Given mass of manganese (II) nitrate tetrahydrate = 16 g

Molar mass of manganese (II) nitrate tetrahydrate = 251 g/mol

Putting values in above equation, we get:

$0.16M=\frac{16\times 1000}{251\times \text{Volume of solution}}\\\\\text{Volume of solution}=\frac{16\times 1000}{251\times 0.16}=398.4mL$

Volume of water = Volume of solution = 398.4 mL

Hence, the amount of water required to prepare given amount of salt is 398.4 mL

4 0

3 years ago

What happens to the density of group 1 when you go down the group

OLga [1]

the density increases down the group.

4 0

2 years ago

Needs to be answered ASAP What group of protists does Euglena belong to? Why?

Korvikt [17]

Euglena belongs to the Kingdom Protista. It is a genus of single-celled flagellates. Euglena are the best known organisms in the Euglenophyta Phylum.Euglena belongs to the Kingdom Protista. It is a genus of single-celled flagellates. Euglena are the best known organisms in the Euglenophyta Phylum.

8 0

3 years ago

A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli

Diano4ka-milaya [45]

Answer:

$\large \boxed{109.17 \, ^{\circ}\text{C}}$

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

ρ(EG) = 1.11 g/mL

ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

$\Delta T_{b} = iK_{b}b$

i is the van’t Hoff factor — the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

$\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}$

2. Kilograms of water

$m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}$

3. Molal concentration of EG

$b = \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}$

4. Increase in boiling point

$\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}$

5. Boiling point

$\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}$

7 0

3 years ago

A 25.0 ml sample of 0.150 m benzoic acid is titrated with a 0.150 m naoh solution. what is the ph at the equivalence point? the

Ad libitum [116K]

Solution:

At the equivalence point, moles NaOH = moles benzoic acid

HA + NaOH ==> NaA + H2O where HA is benzoic acid

At the equivalence point, all the benzoic acid ==> sodium benzoate

A^- + H2O ==> HA + OH- (again, A^- is the benzoate anion and HA is the weak acid benzoic acid)

Kb for benzoate = 1x10^-14/4.5x10^-4 = 2.22x10^-11

Kb = 2.22x10^-11 = [HA][OH-][A^-] = (x)(x)/0.150

x^2 = 3.33x10^-12

x = 1.8x10^-6 = [OH-]

pOH = -log [OH-] = 5.74

pH = 14 - pOH = 8.26

5 0

3 years ago

Read 2 more answers