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2 years ago

What is the molarity of a solution that contains 0.082 mol KI in a 2.03 L solution?

Chemistry

1 answer:

astraxan [27]2 years ago

4 0

Answer:

molarity of the KI solution = 0.04 mol/L

Explanation:

Molarity (M) is the concentration of a solution expressed as the number of moles of solute per liter of solution.

The law that we can applied to calculate the M is:

M = n / V

n - number of moles

V- volume of the solution (liters)

Then insert in the equation the values from the question;

M = 0.082 mol / 2.03 L = 0.04 mol/L

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A buffer solution contains 0.120M acetic acid and 0.150M sodium acetate.

Dvinal [7]

A. M x L = moles.
b. CH3COOH + NaOH ==> CH3COONa + H2O 
I...6 mmols....0.......7.5 mmoles 
C... 0........0.51 mmols..0 
E...6-0.511 ....0.......7.5+0.511 

I stands for initial 
C stands for change. 
E stands for equilibrium. 
Just divide mmoles by 1000 to convert to moles. I work in mmoles because I get tired of writing those zeros. 

c. done as in b.

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2 years ago

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2 years ago

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Given a gas with an initial volume of 3.0 L and a temperature of 290 K, what is the final volume if the

statuscvo [17]

Answer:

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1 year ago

How many milliliters of an aqueous solution of 0.170 M ammonium carbonate is needed to obtain 16.1 grams of the salt

Citrus2011 [14]

There will be needed 982.35 mL of solution to obtain 16.1 grams of the salt.There will be needed mL of

Why?

In order to calculate how many milliliters are needed to obtain 16.1 grams of the salt given its concentration, we first need to find its chemical formula which is the following:

$(NH_{4})2CO_{3}$

Now that we know the chemical formula of the substance, we need to find its molecular mass. We can do it by the following way:

$N_{2}=14g*2=28g\\\\2H_{4}=2*1g*4=8g\\\\C=12.01g*1=12.01g\\\\O_{3}=15.99g*3=47.97g$

We have that the molecular mass of the substance will be:

$MolecularMass=\frac{28g+8g+12.01g+47.97g}{mol}=95.98\frac{g}{mol}$

Therefore, knowing the molecular mass of the substance, we need to calculate how many mols represents 16.1 grams of the same substance, we can do it by the following way:

$mol_{(NH_{4})2CO_{3}=\frac{mass_{(NH_{4})2CO_{3}}}{molarmass_{(NH_{4})2CO_{3}}}$

$mol_{(NH_{4})2CO_{3}=\frac{16.1g}{95.98\frac{g}{mol}}=0.167mol$

Finally, if we need to calculate how many milliliters are needed, we need to use the following formula:

$M=\frac{moles_{solute}}{volume_{solution}}$

$M=\frac{moles_{solute}}{volume_{solution}}\\\\volume_{solution}=\frac{moles_{solute}}{M}$

Now, substituting and calculating, we have:

$volume_{solution}=\frac{0.167mol}{0.170\frac{mol}{L}}\\\\volume_{solution}=0.982L=0.982L*1000=982.35mL$

Henc, there will be needed 982.35 mL of solution to obtain 16.1 grams of the salt.

Have a nice day!

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2 years ago