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2 years ago

Calculate the molarity of a solution prepared by diluting 7.0 mL of 4.0 M solution to a volume of 30 mL.

1 answer:

8 0

Considering the definition of dilution, the molarity of a solution prepared by diluting 7.0 mL of 4.0 M solution to a volume of 30 mL is 0.93 mL.

<h3>What is diluion</h3>

First of all, you have to know that when it is desired to prepare a less concentrated solution from a more concentrated one, it is called dilution.

Dilution is the process of reducing the concentration of solute in solution, which is accomplished by simply adding more solvent to the solution at the same amount of solute.

In a dilution the amount of solute does not change, but as more solvent is added, the concentration of the solute decreases, as the volume (and weight) of the solution increases.

A dilution is mathematically expressed as:

Ci×Vi = Cf×Vf

where

Ci: initial concentration
Vi: initial volume
Cf: final concentration
Vf: final volume

<h3>Molarity of the solution in this case</h3>

In this case, you know:

Ci= 4 M
Vi= 7 mL
Cf= ?
Vf= 30 mL

Replacing in the definition of dilution:

4 M× 7 mL= Cf× 30 mL

Solving:

(4 M× 7 mL)÷ 30 mL= Cf

In summary, the molarity of a solution prepared by diluting 7.0 mL of 4.0 M solution to a volume of 30 mL is 0.93 mL.

Learn more about dilution:

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Determine the mass of copper that has the same number of atoms as there are in 6.95 mg of potassium. Answer in units of mg.

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3 years ago

In Part B the given conditions were 1.00 mol of argon in a 0.500-L container at 18.0°C. You identified that the ideal pressure (

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Answer:

4,38%

small molecular volumes

Decrease

Explanation:

The percent difference between the ideal and real gas is:

(47,8atm - 45,7 atm) / 47,8 atm × 100 = 4,39% ≈ <em>4,38%</em>

This difference is considered significant, and is best explained because argon atoms have relatively <em>small molecular volumes. </em>That produce an increasing in intermolecular forces deviating the system of ideal gas behavior.

Therefore, an increasing in volume will produce an ideal gas behavior. Thus:

If the volume of the container were increased to 2.00 L, you would expect the percent difference between the ideal and real gas to <em>decrease</em>

<em />

I hope it helps!

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3 years ago

There are some data that suggest that zinc lozenges can significantly shorten the duration of a cold. If the solubility of zinc

zhuklara [117]

Answer:

$K_{sp}$ of $Zn(CH_{3}COO)_{2}$ is 0.0513

Explanation:

Solubility equilibrium of $Zn(CH_{3}COO)_{2}$ :

$Zn(CH_{3}COO)_{2}\rightleftharpoons Zn^{2+}+2CH_{3}COO^{-}$

Solubility product of $Zn(CH_{3}COO)_{2}$ ( $K_{sp}$ ) is written as- $K_{sp}=[Zn^{2+}][CH_{3}COO^{-}]^{2}$

Where $[Zn^{2+}]$ and $[CH_{3}COO^{-}]$ represents equilibrium concentration (in molarity) of $Zn^{2+}$ and $CH_{3}COO^{-}$ respectively.

Molar mass of $Zn(CH_{3}COO)_{2}$ = 183.48 g/mol

So, solubility of $Zn(CH_{3}COO)_{2}$ = $\frac{43.0}{183.48}M$ = 0.234M

1 mol of $Zn(CH_{3}COO)_{2}$ gives 1 mol of $Zn^{2+}$ and 2 moles of $CH_{3}COO^{-}$ upon dissociation.

so, $[Zn^{2+}]$ = 0.234 M and $[CH_{3}COO^{-}]$ = $(2\times 0.234)M=0.468M$

so, $K_{sp}=(0.234)\times (0.468)^{2}=0.0513$

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3 years ago

A solution is prepared by mixing 200.0 g of water, H2O, and 300.0 g of

VikaD [51]

Answer:

Mole Fraction (H₂O) = 0.6303

Mole Fraction (C₂H₅OH) = 0.3697

Explanation:

(Step 1)

Calculate the mole value of each substance using their molar masses.

Molar Mass (H₂O): 2(1.008 g/mol) + 15.998 g/mol

Molar Mass (H₂O): 18.014 g/mol

200.0 g H₂O 1 mole
--------------------- x ------------------ = 11.10 moles H₂O
18.014 g

Molar Mass (C₂H₅OH): 2(12.011 g/mol) + 6(1.008 g/mol) + 15.998 g/mol

Molar Mass (C₂H₅OH): 46.068 g/mol

300.0 g C₂H₅OH 1 mole
---------------------------- x -------------------- = 6.512 moles C₂H₅OH
46.068 g

(Step 2)

Using the mole fraction ratio, calculate the mole fraction of each substance.

moles solute
Mole Fraction = ------------------------------------------------
moles solute + moles solvent

11.10 moles H₂O
Mole Fraction = -------------------------------------------------------------
11.10 moles H₂O + 6.512 moles C₂H₅OH

Mole Fraction (H₂O) = 0.6303

6.512 moles C₂H₅OH
Mole Fraction = -------------------------------------------------------------
11.10 moles H₂O + 6.512 moles C₂H₅OH

Mole Fraction (C₂H₅OH) = 0.3697

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2 years ago

If 1. 3618 moles of AsF3 are allowed to react with 1. 0000 mole of C2Cl6, what would be the theoretical yield of AsCl3, in moles

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Answer:

AsF3:C2CI6

4:3

1.3618 moles: 1.02135 moles(1.3618÷4×3)

C2CI6 is the limting reagent

So the number of moles for AsCI3 is 0.817 moles( number of moles of the limting reagant) ÷3 ×4 (according to ratio by balancing chemical equation)=1.09 moles(3 s.f.)

Balanced equation

4AsF3 + 3C2Cl6 → 4AsCl3 + 3C2Cl2F4

Use stoichiometry to calculate the moles of AsCl3 that can be produced by each reactant.

Multiply the moles of each reactant by the mole ratio between it and AsCl3 in the balanced equation, so that the moles of the reactant cancel, leaving moles of AsCl3.

Explanation:

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2 years ago