The weather behind a cold front is dominated by which of the following?

What is the molarity of 225 grams of Cu(NO2)2 in a total volume of 2.59 L?

Kay [80]

Answer:

The molarity is 0.56 $\frac{moles}{L}$

Explanation:

In a mixture, the chemical present in the greatest amount is called a solvent, while the other components are called solutes. Then, the molarity or molar concentration is the number of moles of solute per liter of solution.

In other words, molarity is the number of moles of solute that are dissolved in a given volume.

The Molarity of a solution is determined by:

$Molarity (M)=\frac{number of moles of solute}{Volume}$

Molarity is expressed in units ( $\frac{moles}{liter}$ ).

Then you must know the number of moles of Cu(NO₂)₂. For that it is necessary to know the molar mass. Being:

Cu: 63.54 g/mol
N: 14 g/mol
O: 16 g/mol

the molar mass of Cu(NO₂)₂ is:

Cu(NO₂)₂= 63.54 g/mol + 2*(14 g/mol + 2* 16 g/mol)= 155.54 g/mol

Now the following rule of three applies: if 155.54 g are in 1 mole of the compound, 225 g in how many moles are they?

$moles=\frac{225 g*1 mole}{155.54 g}$

moles= 1.45

So you know:

number of moles of solute= 1.45 moles
volume=2.59 L

Replacing in the definition of molarity:

$Molarity=\frac{1.45 moles}{2.59 L}$

Molarity= 0.56 $\frac{moles}{L}$

<u><em>The molarity is 0.56</em></u> $\frac{moles}{L}$ <u><em></em></u>

5 0

3 years ago

A chemistry graduate student is given of a chlorous acid solution. Chlorous acid is a weak acid with . What mass of should the s

DerKrebs [107]

Answer:

11.31g NaClO₂

Explanation:

<em> Is given 250mL of a 1.60M chlorous acid HClO2 solution. Ka is 1.110x10⁻². What mass of NaClO₂ should the student dissolve in the HClO2 solution to turn it into a buffer with pH =1.45? </em>

It is possible to answer this question using Henderson-Hasselbalch equation:

pH = pKa + log₁₀ [A⁻] / [HA]

<em>Where pKa is -log Ka = 1.9547; [A⁻] is the concentration of the conjugate base (NaClO₂), [HA] the concentration of the weak acid</em>

You can change the concentration of the substance if you write the moles of the substances:

[Moles HClO₂] = 250mL = 0.25L×(1.60mol /L) = <em>0.40 moles HClO₂</em>

Replacing in H-H expression, as the pH you want is 1.45:

1.45 = 1.9547 + log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]

-0.5047 = log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]

<em>0.3128 = </em>[Moles NaClO₂] / [0.40 moles HClO₂]

0.1251 = Moles NaClO₂

As molar mass of NaClO₂ is 90.44g/mol, mass of 0.1251 moles of NaClO₂ is:

0.1251 moles NaClO₂ ₓ (90.44g / mol) =