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3 years ago

If the pressure of a gas is quadrupled (4x), how

Chemistry

1 answer:

umka2103 [35]3 years ago

4 0

Answer:

Yes. The volume would be 1/4 of the initial volume.

Explanation:

At constant temperature, the pressure of a gas is inversely proportional to the volume of the gas. Hence;

P1V1 = P2V2

where P1 is the initial pressure, V1 is the initial volume, P2 is the final pressure, and V2 is the final volume.

If the pressure of a gas is quadrupled;

P2 = 4P1, the equation becomes

P1V1 = 4P1 x V2

Making V2 the subject:

V2 = P1V1/4P1

V2 = V1/4

This means that the volume would change by being reduced to 1/4 of the initial volume.

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the purpose of the ____ system prevent foreign substances from entering the body and to attack foreign substances that are in th

Ratling [72]

Immune System is the answer.

7 0

3 years ago

(58 g)/ (4L) reduce units to one

katen-ka-za [31]

Answer is 14.5 g L⁻¹.

Explanation;

Here, the question says reduce the units as one.

The presented units are g/L. To reduce the units to one, what we can do is take L to the upper side.

This can be done according to the rules of indices;

1 / aˣ = a⁻ˣ

Like that, we can write 1 / L as L⁻¹.

Hence, the reduced unit is g L⁻¹.

But remember to keep a space between when writing two different units.

Actually, this is an unit for density.

3 0

3 years ago

In this experiment we will be using a 0.05 M solution of HCl to determine the concentration of hydroxide (OH-) in a saturated so

gulaghasi [49]

Answer: The moles of hydroxide ions present in the sample is 0.0008 moles

Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is HCl.

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ca(OH)_2$

We are given:

$n_1=1\\M_1=0.05M\\V_1=16mL\\n_2=2\\M_2=?M\\V_2=36.0mL$

Putting values in above equation, we get:

$1\times 0.05\times 16=2\times M_2\times 36\\\\M_2=\frac{1\times 0.05\times 16}{2\times 36}=0.011M$

To calculate the moles of hydroxide ions, we use the equation used to calculate the molarity of solution:

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

Molarity of solution = 0.011 M

Volume of solution = 36.0 mL

Putting values in above equation, we get:

$0.011=\frac{\text{Moles of }Ca(OH)_2\times 1000}{36}\\\\\text{Moles of }Ca(OH)_2=\frac{0.011\times 36}{1000}=0.0004mol$

1 mole of calcium hydroxide produces 1 mole of calcium ions and 2 moles of hydroxide ions.

Moles of hydroxide ions = (0.0004 × 2) = 0.0008 moles

Hence, the moles of hydroxide ions present in the sample is 0.0008 moles

8 0

3 years ago

The solubility of copper chloride at 20 °C is 73 g/100 g of water. Kiera adds 100 g of copper chloride to 100 g of water and sti

OLEGan [10]

Answer:

$m_{undissolved}=27g$

Explanation:

Hello there!

In this case, according to the given information of the solubility of copper chloride, as the maximum amount of this salt one can dissolve without having a precipitate, we infer that since just 73 grams are actually dissolved, the following amount will remain solid as a precipitate:

$m_{undissolved}=100g-73g\\\\m_{undissolved}=27g$

Best regards!

3 0

3 years ago

A mixture contains only nacl and al2(so4)3. a 1.45 g sample of the mixture is dissolved in water, and an excess of naoh is added

Aleks [24]

When mixture of NaCl and Al₂(SO₄)₃ is allowed to react with excess NaOH, only Al₂(SO₄)₃ reacts with it and NaCl does not react with NaOH due to presence of common ion (Na⁺). On reaction gelatinous precipitate of aluminium hydroxide [Al(OH)₃] is produced. The balanced chemical reaction is represented as-

Al₂(SO₄)₃ + 6NaOH → 2Al(OH)₃ + 3Na₂SO₄

On this reaction, 0.495 g = 0.495/78 moles =6.346 X 10⁻³ moles of Al(OH)₃.

As per balanced reaction, two moles of Al(OH)₃ is produced from one mole Al₂(SO₄)₃. So, 6.346 X 10⁻³ moles of Al(OH)₃ is produced from (6.346 X 10⁻³)/2 moles=3.173 X 10⁻³ moles of Al₂(SO₄)₃= 3.173 X 10⁻³ X 342 g of Al₂(SO₄)₃=1.085 g of Al₂(SO₄)₃.

So, mass percentage of Al₂(SO₄)₃ is= (amount of Al₂(SO₄)₃/total amount of mixture)X100 = $\frac{1.085 X 100}{1.45}$ =74.8 %.

7 0

3 years ago