All categories

3 years ago

What is the concentration of a solution that has a volume of 2.5L and contains exactly 2.125 moles of calcium phosphate

Chemistry

1 answer:

Iteru [2.4K]3 years ago

7 0

Answer:

0.85 mol/L.

Explanation:

Molarity is defined as the no. of moles of solute dissolved in a liter of the solution.

M = (no. of moles of solute)/(Volume of the solution (L))

no. of moles of calcium phosphate = 2.125 mol.

Volume of the solution = 2.5 L.

∴ M of calcium phosphate = (2.125 mol)/(2.5 L) = 0.85 mol/L.

You might be interested in

Cell is to a tissue as brick is to _______. Cement Clay Bricklayer wall

svet-max [94.6K]

Answer:

bricklayer

Explanation:

is the correct answer

6 0

3 years ago

Características de la báscula granataria porfi

saveliy_v [14]

Answer:

Lowkey dont know the language

Explanation:

3 0

4 years ago

- What is the concentration of hydrogen ions if the pH is 12?

ELEN [110]

Answer:

1 x 10^-12 mol dm^-3

Explanation:

[H+] = 10 ^ -pH

Therefore, pH 12 = 1 x 10^-12 hydrogen ion concentration

4 0

3 years ago

You used a calorimeter in the heat transfer lab. Explain how the calorimeter works, and how to calculate the heat given off or a

Butoxors [25]

A calorimeter contains reactants and a substance to absorb the heat absorbed. The initial temperature (before the reaction) of the heat absorbent is measured and then the final temperature (after the reaction) is also measured. The absorbent's specific heat capacity and mass are also known. Given all of this data, the equation:
Q = mcΔT
To find the heat released.

3 0

3 years ago

How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos

Alexeev081 [22]

Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of $H_3PO_4$ = 54.219 g

Number of atoms of $Mg$ = $7.179\times 10^{23}$

Molar mass of $H_3PO_4$ = 98 g/mol

First we have to calculate the moles of $H_3PO_4$ and $Mg$ .

$\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}$

$\text{Moles of }H_3PO_4=\frac{54.219g}{98g/mol}=0.553mol$

and,

$\text{Moles of }Mg=\frac{7.179\times 10^{23}}{6.022\times 10^{23}}=1.19mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2$

From the balanced reaction we conclude that

As, 3 mole of $Mg$ react with 2 mole of $H_3PO_4$

So, 0.553 moles of $Mg$ react with $\frac{2}{3}\times 0.553=0.369$ moles of $H_3PO_4$

From this we conclude that, $H_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $Mg$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2$

From the reaction, we conclude that

As, 3 mole of $Mg$ react to give 3 mole of $H_2$

So, 0.553 mole of $Mg$ react to give 0.553 mole of $H_2$

Now we have to calculate the volume of $H_2$ gas at STP.

As we know that, 1 mole of substance occupies 22.4 L volume of gas.

As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas

So, 0.553 mole of hydrogen gas occupies $0.553\times 22.4=12.4L$ volume of hydrogen gas

Therefore, the volume of hydrogen gas produced will be, 12.4 L

4 0

3 years ago