All categories

3 years ago

Q6. A student mixes 50.0 mL of 1.00 M Ba(OH)2 with 86.4 mL of 0.494 M H2SO4. Calculate the

Chemistry

1 answer:

skelet666 [1.2K]3 years ago

6 0

Answer:

The correct answer should be 12.72 pH of the mixed solution and 9.94g mass of solid BASO4 formed.

Explanation:

The molecules of Ba(OH)2 are : 1.00M x 0.05L = 0.05 ( moles )

The molecules of H2SO4 are : 0.494M x 0.0864L = 0.0426816

Ba(OH)2 + H2SO4 ----> BaSO4 + 2 H2O

0.0426816<--- 0.0426816

The mixed solution is Ba(OH)2 with 0.05 - 0.0426816 = 0.0073184

The concentration of mixed solution is : 0.0073184 : ( 0.05 + 0.0864 ) = 0.054 M

The pH of mixed solution is 14 - -log[0.054] = 14 - 1.27 = 12.73 PH

And the mass of BaSO4 is 0.0426816 x ( 137 + 32 + 16 x 4 ) = 9.94 gams.

You might be interested in

What would you observe in the reaction of calcium with nitric acid

Marina CMI [18]

Answer:

reacts to form carbondioxide

6 0

3 years ago

Calculate the mass of 3 moles of clorine

Vitek1552 [10]

Moles Chlorine to grams = 35.453 grams.
moles Chlorine to grams = 70.906 grams.
moles Chlorine to grams = 106.359 grams.
moles Chlorine to grams = 141.812 grams.
moles Chlorine to grams = 177.265 grams.
moles Chlorine to grams = 212.718 grams.

4 0

3 years ago

A 33.0 mL sample of 1.15 M KBr and a 59.0 mL sample of 0.660 M KBr are mixed. The solution is then heated to evaporate water unt

Oksi-84 [34.3K]

Answer:

We need 13.06 grams of silver nitrate to precipitate out silver bromide in the final solution

Explanation:

Step 1: Data given

Sample 1: The 1.15 M sample has a volume of 33.O mL

Sample 2: The 0.660 M sample has a volume of 59.0 mL

Molar mass of KBr = 119 g/mol

Molar mass of AgNO3 = 169.87 g/mol

Step 2: Calculate number of moles for both samples

Number of moles = Molarity * Volume

Sample 1: 1.15 M * 33 *10^-3 L = 0.03795 moles

Sample 2: 0.660 M *59*10^-3 L = 0.03894 moles

Total mol KBr = 0.03795 + 0.03894 = 0.07689 moles

Step 3: Calculate total mass

mass = Number of moles * Molar mass

mass = 0.07689 moles * 119 g/moles = 9.15 grams ( in 55mL)

Step 4: Calculate moles of AgBr

AgNO3 reacts with KBr

KBr(aq) + AgNO3(aq) → AgBr(s) + KNO3(aq)

1 mole of KBr consumed, needs 1 mole of AgNO3 to produce 1 mole of AgBr and 1 mole of KNO3

So 0.07689 moles of KBr wll need 0.07689 moles of AgNO3

Step 5: Calculate mass of silver nitrate

mass of AgNO3 = Moles of AgNO3 * Molar mass of AgNO3

mass of AgNO3 = 0.07689 moles * 169.87 g/mol = 13.06 grams

We need 13.06 grams of silver nitrate to precipitate out silver bromide in the final solution

8 0

3 years ago

A tank holds 6.90 m3 of water. What is the volume in ft3? (3.28 ft = 1 m, Remember significant figures, but this answer does not

AleksAgata [21]

Volume of tank = $6.9 m^{3}$ (given)

Since, $1 m = 3.28 ft$

So,

$1 m^{3} = 35.287 ft^{3}$

For $6.9 m^{3}$ :

$6.9\times 35.287 ft^{3} = 243.4803 ft^{3}$

The significant rule for multiplication, states that the number of significant figures in the answer obtained by multiplication is determined by the value with the lowest number of significant digits.

Since, the minimum number of decimal places in the above multiplication operation is 1 so, the final result must be upto 1 decimal place only.

$243.4803 ft^{3} \simeq 243.5 ft^{3}$

Hence, volume in $ft^{3}$ is 243.5.

5 1

3 years ago

Read 2 more answers

When Lithium-7 loses 1 electron it

IrinaK [193]

If it loses an electron, it will become an ion.

8 0

3 years ago

Read 2 more answers