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3 years ago

If 298 mL of 0.575 M CuCl2 is diluted into 750 mL, what is the concentration of new solution?

Chemistry

1 answer:

Sonja [21]3 years ago

4 0

Answer:

<h3>The answer is 0.228 M</h3>

Explanation:

To find the concentration of new solution we use the formula

$C_1V_1 = C_2V_2$

where

C1 is the concentration of the stock solution

V1 is the volume is the volume of the stock solution

C_2 is the concentration of the diluted solution

V2 is the volume of the diluted solution

Since we are finding the the concentration of new solution

$C_2 = \frac{C_1V_1}{V_2} \\$

From the question

C1 = 0.575 M

V1 = 298 mL

V2 = 750 mL

We have

$C_2 = \frac{0.575 \times 298}{750} \\ = 0.228466666...$

We have the final answer as

Hope this helps you

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Apples reproduce naturally by sexual reproduction.How does this explain why there are many types of apples be sure to use use me

kvasek [131]

Answer:

could be a that the apple produced a seed with genetic mutation in chromosome during the process meiosis.

Explanation:

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2 years ago

An ion of iron has an 26 protons, 30 neutrons, and 23 electrons. What are its atomic number, atomic mass, and net charge?

sleet_krkn [62]

Answer:

Atomic number 26,atomic mass 26+30=56 dalton and net charge is 3+

Explanation:

The total number of proton present in an atom is known as the atomic number of that atom.From that point of view the atomic number of iron ion is 26.

The total number proton and neutrons present in the nucleus of an atom or ion is termed as atomic mass.From that point of view the the atomic mass of iron ion is 26+30=56 dalton

According to the given question iron ion contain 3 more protons than electrons as a result the iron ion will contain 3 unit positive charge.

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3 years ago

Explain in your own words the process using the following key words: carbon-di-oxide, water, sugar, oxygen, thylakoid, chloropla

AleksAgata [21]

Answer:

The Photosynthesis process

Explanation:

Plants, algae, and some other organisms can transform the sunlight energy into chemical energy. The photosynthesis process occur thanks to the chloroplasts. The chloroplast is an organelle found in all green plants. Inside of the chloroplast you can find the thylakoids which are arranged in stacks named grana, they have membranes with chloropyll a photosynthetic pigment, also you can find the photosystems, they are functional and structural units of protein complexes. The thylakoids capture the light and allow the reactions to transform CO2. The set of reactions that occurs in the chloroplasts are known as the Calvin cycle.

The general equation of photosynthesis is:

$6 CO_{2} + 6 H_{2} O + Energy -> C_{6} H_{12} 12O_{6} + 6 O_{2}$

6 CO2 + 6 H2O + Energy -> C6H12O6 + 6 O2

Carbon Dioxide + water + Light -> Glucose (sugar) + Oxygen

After, this glucose is transformed into pyruvate, and it allowed the release of denosine triphosphate (ATP) by cellular respiration. The ATP is an organic chemical that is requires for the cell to perform any process (any kind or work).

5 0

3 years ago

You mix 285.0 mL of 1.20 M lead(II) nitrate with 300.0 mL of 1.60 M potassium iodide. The lead(II) iodide is insoluble. Which of

SIZIF [17.4K]

Answer:

D. The final concentration of NO3– is 0.821 M.

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For potassium iodide :

Molarity = 1.60 M

Volume = 300.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 300.0×10⁻³ L

Thus, moles of potassium iodide :

$Moles=1.60 \times {300.0\times 10^{-3}}\ moles$

Moles of potassium iodide = 0.48 moles

For lead(II) nitrate :

Molarity = 1.20 M

Volume = 285 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 285×10⁻³ L

Thus, moles of lead(II) nitrate :

$Moles=1.20\times {285\times 10^{-3}}\ moles$

Moles of lead(II) nitrate = 0.342 moles

According to the given reaction:

$2KI_{(aq)}+Pb(NO_3)_2_{(aq)}\rightarrow PbI_2_{(s)}+2KNO_3_{(aq)}$

2 moles of potassium iodide react with 1 mole of lead(II) nitrate

1 mole of potassium iodide react with 1/2 mole of lead(II) nitrate

0.48 moles potassium iodide react with 0.48/2 mole of lead(II) nitrate

Moles of lead(II) nitrate = 0.24 moles

Available moles of lead(II) nitrate = 0.342 moles

Limiting reagent is the one which is present in small amount. Thus, potassium iodide is limiting reagent.

Also, consumed lead(II) nitrate = 0.24 moles (lead ions precipitate with iodide ions)

Left over moles = 0.342 - 0.24 moles = 0.102 moles

Total volume = 300 + 285 mL = 585 mL = 0.585 L

So, Concentration = 0.102/0.585 M = 1.174 M

Statement A is correct.

The formation of the product is governed by the limiting reagent. So,

2 moles of potassium iodide gives 1 mole of lead(II) iodide

1 mole of potassium iodide gives 1/2 mole of lead(II) iodide

0.48 mole of potassium iodide gives 0.48/2 mole of lead(II) iodide

Mole of lead(II) iodide = 0.24 moles

Molar mass of lead(II) iodide = 461.01 g/mol

Mass of lead(II) chloride = Moles × Molar mass = 0.24 × 461.01 g = 111 g

Statement B is correct.

Potassium iodide is the limiting reagent. So all the potassium ion is with potassium nitrate . Thus,

2 moles of Potassium iodide on reaction forms 2 moles of potassium ion

0.48 moles of Potassium iodide on reaction forms 0.48 moles of potassium ion

Total volume = 300 + 285 mL = 585 mL = 0.585 L

So, Concentration = 0.48/0.585 M = 0.821 M

Statement C is correct.

Nitrate ions are furnished by lead(II) nitrate . So,

1 mole of lead(II) nitrate produces 2 moles of nitrate ions

0.342 mole of lead(II) nitrate produces 2*0.342 moles of nitrate ions

Moles of nitrate ions = 0.684 moles

So, Concentration = 0.684/0.585 M = 1.169 M

Statement D is incorrect.

4 0

2 years ago

A 10.21 mol sample of argon gas is maintained in a 0.7564 L container at 296.9 K. What is the pressure in atm calculated using t

soldi70 [24.7K]

Answer:

The pressure in atm calculated using the van der Waals' equation, is 337.2atm

Explanation:

This is the Van der Waals equation for real gases:

(P + a/v² ) ( v-b) = R .T

where P is pressure

v is Volume/mol

R is the gas constant and T, T° in K

a y b are constant for each gas, so those values are data, from the statement.

[P + 1.345 L²atm/mol² / (0.7564L/10.21mol)² ] (0.7564L/10.21mol - 3.219×10-2 L/mol ) = 0.082 L.atm/mol.K . 296.9K

[P + 1.345 L²atm/mol² / 5.48X10⁻³ L²/mol²] (0.074 L/mol - 3.219×10-2 L/mol ) = 0.082 L.atm/mol.K . 296.9K

(P + 245.05 atm) (0.04181L/mol) = 0.082 L.atm/mol.K . 296.9K

(P + 245.05 atm) (0.04181L/mol) = 24.34 L.atm/mol

0.04181L/mol .P + 10.24 L.atm/mol = 24.34 L.atm/mol

0.04181L/mol .P = 24.34 L.atm/mol - 10.24 L.atm/mol

0.04181L/mol. P = 14.1 L.atm/mol

P = 14.1 L.atm/mol / 0.04181 mol/L

P = 337.2 atm

4 0

2 years ago