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Anni [7]
3 years ago
10

The concentration of commercially available concentrated nitric acid is 70.0 percent by mass, or 15.9 M. Calculate the density a

nd molality of the solution.
Chemistry
1 answer:
frosja888 [35]3 years ago
7 0

Answer:

Density = 1000 g/l

= 1.00 kg/l.

Molality = 0.037 mol/g

= 37 mol/kg.

Explanation:

Molar concentration of nitric acid = 15.9 M.

Mass of water in 100 g = 100 - 70

= 30 g

Mass of nitric acid = 70 g

Molar mass of HNO3 = 1 + 14 + (16*3)

= 63 g/mol

Number of moles = mass/molar mass

= 70/63

= 1.11 mol.

Molar concentration = number of moles/volume

Volume = 1.11/15.9

= 0.07 l

Density = mass/volume

Density =

= 70 * 0.07

= 1000 g/l

= 1.00 kg/l

Molality = number of moles of nitric acid/mass of solvent(water)

= 1.11/0.030

= 37 mol/kg

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The most common source of copper (cu) is the mineral chalcopyrite (cufes2). how many kilograms of chalcopyrite must be mined to
tigry1 [53]

Answer : 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.

Solution : Given,

Mass of Cu = 300 g

Molar mass of Cu = 63.546 g/mole

Molar mass of CuFeS_2 = 183.511 g/mole

  • First we have to calculate the moles of Cu.

\text{ Moles of Cu}=\frac{\text{ Given mass of Cu}}{\text{ Molar mass of Cu}}= \frac{300g}{63.546g/mole}=4.7209moles

The moles of Cu = 4.7209 moles

From the given chemical formula, CuFeS_2 we conclude that the each mole of compound contain one mole of Cu.

So, The moles of Cu = Moles of CuFeS_2 = 4.4209 moles

  • Now we have to calculate the mass of CuFeS_2.

Mass of CuFeS_2 = Moles of CuFeS_2 × Molar mass of CuFeS_2 = 4.4209 moles × 183.511 g/mole = 866.337 g

Mass of CuFeS_2 = 866.337 g = 0.8663 Kg         (1 Kg = 1000 g)

Therefore, 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.


3 0
3 years ago
If 4.0 g of helium gas occupies a volume of 22.4 L at 0 o C and a pressure of 1.0 atm, what volume does 3.0 g of He occupy under
WINSTONCH [101]

Answer:

the volume occupied by 3.0 g of the gas is 16.8 L.

Explanation:

Given;

initial reacting mass of the helium gas, m₁ = 4.0 g

volume occupied by the helium gas, V = 22.4 L

pressure of the gas, P = 1 .0 atm

temperature of the gas, T = 0⁰C = 273 K

atomic mass of helium gas, M = 4.0 g/mol

initial number of moles of the gas is calculated as follows;

n_1 = \frac{m_1}{M} \\\\n_1 = \frac{4}{4} = 1

The number of moles of the gas when the reacting mass is 3.0 g;

m₂ = 3.0 g

n_2 = \frac{m_2}{M} \\\\n_2 = \frac{3}{4} \\\\n_2 = 0.75 \ mol

The volume of the gas at 0.75 mol is determined using ideal gas law;

PV = nRT

PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\since, \ \frac{RT}{P} \ is \ constant,\  then;\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1n_2}{n_1} \\\\V_2 = \frac{22.4 \times 0.75}{1} \\\\V_2 = 16.8 \ L

Therefore, the volume occupied by 3.0 g of the gas is 16.8 L.

4 0
3 years ago
Most animals that live in soil live in the _______ layer.
labwork [276]

Answer:

Living organisms present in soil include archaea, bacteria, actinomycetes, fungi, algae, protozoa, and a wide variety of larger soil fauna, including springtails, mites, nematodes, earthworms, ants, insects that spend all or part of their life underground, and larger organisms such as burrowing rodents.

Explanation:

8 0
3 years ago
Read 2 more answers
Which of the following elements is a transition metal?
GenaCL600 [577]

Answer:

Scandium

Titanium

Vanadium

Chromium

Manganese

Iron

Cobalt

Nickel

Copper

Zinc

Yttrium

Zirconium

Niobium

Molybdenum

Technetium

Ruthenium

Rhodium

Palladium

Silver

Cadmium

Lanthanum

Hafnium

Tantalum

Tungsten

Rhenium

Osmium

Iridium

Platinum

Gold

Mercury

Actinium

Rutherfordium

Dubnium

Seaborgium

Bohrium

Hassium

Meitnerium

Darmstadtium

Roentgenium

Copernicium

Explanation:

all of those are transition metals lol

5 0
4 years ago
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A solid is crushed into a powder is chimacal change or pysical change
SSSSS [86.1K]

Answer:

Physical Change

Explanation:

A solid being crushed into a powder is a physical change because a chemical change is when substances combine to create new substances.

Hope this helps :)

4 0
3 years ago
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