Which substance is a homogeneous mixture?

Calculate the number of milliliters of 0.710 M Ba(OH)2 required to precipitate all of the Mn2+ ions in 161 mL of 0.796 M KMnO4 s

USPshnik [31]

<u>Answer:</u> The volume of barium hydroxide is 183 mL.

<u>Explanation:</u>

To calculate the moles of cadmium nitrate, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

Molarity of $MnSO_4$ = 0.796 M

Volume of $MnSO_4$ = 161 mL = 0.161 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.796mol/L=\frac{\text{Moles of }MnSO_4}{0.161L}\\\\\text{Moles of }MnSO_4=0.13mol$

The chemical equation for the reaction of manganese sulfate and barium hydroxide follows:

$MnSO_4(aq.)+Ba(OH)_2(aq.)\rightarrow Mn(OH)_2(s)+BaSO_4(aq.)$

By Stoichiometry of the reaction:

1 mole of manganese sulfate reacts with 1 mole of barium hydroxide.

So, 0.13 moles of manganese sulfate will react with = $\frac{1}{1}\times 0.13=0.13mol$ of barium hydroxide

Now, calculating the volume of barium hydroxide by using equation 1, we get:

Moles of barium hydroxide = 0.13 moles

Molarity of barium hydroxide = 0.710 M

Putting values in equation 1, we get:

$0.710mol/L=\frac{0.13mol}{\text{Volume of barium hydroxide}}\\\\\text{Volume of barium hydroxide}=0.183L$

Converting this into milliliters, we use the conversion factor:

1 L = 1000 mL

So, $0.183L=0.183\times 1000=183mL$

Hence, the volume of barium hydroxide is 183 mL.

7 0

3 years ago

What is the energy change if 84.0 g of calcium oxide (CaO) reacts with excess water in the following reaction?

pochemuha

The energy change if 84.0 g of CaO react with excess water is 98KJ of heat is released.

calculation
heat = number of moles x delta H

delta H = - 65.2 Kj/mol

first find the number of moles of CaO reacted

moles = mass/molar mass
the molar mass of CaO = 40 + 16= 56 g/mol
mass = 84 g
moles therefore = 84 g/56 g/mol =1.5 moles

Heat is therefore = 1.5 moles x -65.2 = - 97.8 Kj = -98 Kj

since sign is negative the energy is released

6 0

3 years ago

How many grams are in 11.9 moles of chromirum

faust18 [17]

Answer:

Explanation:

Method 1 proportion

1 mole of chromium is 52 grams

11.9 moles = x grams

1/11.9 = 52/x Cross multiply

x = 11.9 * 52

x = 618.8 grams

Now I have used an approximate mass for Chromium. The answer you get here is expected to reflect the weigth given on your periodic table Use that to get your answer. You should give a number very close to mine. Round to 3 places as in 619.

Method Two Formula

mols = given mass / molecular mass

11.9 = given mass / 51.9961 Multiply both sides by 51.9961

11.9 *51.9961 = given mass

given mass = 618.75

given mass = 619

3 0

3 years ago

If 200 mg of HCl is added to water to achieve a final volume of 1L, what is the final pH?

Anvisha [2.4K]

hope this helps ..................

8 0

2 years ago

Read 2 more answers

9.05 mol of oxygen gas to g of oxygen gas

NARA [144]

Molar mass of oxygen gas:

O₂ = 16 * 2 = 32.0 g/mol

1 mole O₂ -------------- 32.0
9.05 mole O₂ ---------- ?

Mass = 9.05 * 32.0

Mass = 289.6 g of O₂

hope this helps!

7 0

2 years ago