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3 years ago

2 What is the correct sequence of steps for the preparation of a pure sample of copper(II) sulfate

Chemistry

1 answer:

Lina20 [59]3 years ago

4 0

Answer:

The correct answer is - B. dissolving → evaporation filtration → crystallisation

Explanation:

The method of the preparation of a pure sample of copper(II) sulfate from dilute sulfuric acid and copper II oxide is given as follows:

step 1. Adding dilute sulfuric acid into a beaker. Using bunsen burner heat the beaker.

step 2. Adding the copper (II) oxide into the beaker and give it a little time at a time to the warm dilute sulfuric acid and stir

step 3. Filtering the mixture into an evaporating vessel to remove the excess copper (II) oxide and water from the filtrate.

Step 4. leave the rest filtrate to crystallize.

Copper (II) Oxide {CuO (s)} + Dilute Sulfuric Acid {H2SO4 (aq)} → Copper (II) Sulphate {CuSO4 (s)} + Water {H2O}

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Potassium hydrogen phthalate, KHP, is a monoprotic acid often used to standardize NaOH solutions. If 0.212 g of KHP are dissolve

maxonik [38]

The molarity of the NaOH solution is 0.03 M

We'll begin by calculating the mole of the KHP

Mass = 0.212 g
Molar mass = 204.22 g/mol
Mole of KHP =?

Mole = mass /molar mass

Mole of KHP = 0.212 / 204.22

Mole of KHP = 0.001 mole

Next, we shall determine the molarity of the KHP solution

Mole of KHP = 0.001 mole
Volume = 50 mL = 50/1000 = 0.05 L
Molarity of KHP =?

Molarity = mole / Volume

Molarity of KHP = 0.001 / 0.05

Molarity of KHP = 0.02 M

Finally , we shall determine the molarity of the NaOH solution

KHP + NaOH —> NaPK + H₂O

From the balanced equation above,

The mole ratio of the acid, KHP (nA) = 1
The mole ratio of base, NaOH (nB) = 1

From the question given above, the following data were obtained:

Volume of acid, KHP (Va) = 50 mL
Molarity of acid, KHP (Ma) = 0.02 M.
Volume of base, NaOH (Vb) = 35 mL
Molarity of base, NaOH (Mb) =?

MaVa / MbVb = nA / nB

(0.02 × 50) / (Mb × 35) = 1

1 / (Mb × 35) = 1

Cross multiply

Mb × 35 = 1

Divide both side by 35

Mb = 1 / 35

Mb = 0.03 M

Thus, the molarity of the NaOH solution is 0.03 M

Complete question:

See attached photo

Learn more about titration: brainly.com/question/25866669

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2 years ago

A chemist added hydrochloric acid to a sample of baking soda in a beaker. Immediately, the mixture of the two chemicals began to

tester [92]

Answer: B

Explanation:

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3 years ago

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Given a fixed amount of gas held at constant pressure, calculate the volume it would occupy if a 2.00 L sample were cooled from

aliina [53]

Answer:

1.82 L

Explanation:

We are given the following information;

Initial volume as 2.0 L
Initial temperature as 60.0°C
New volume as 30.0 °C

We are required to determine the new volume;

From Charles's law;

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

Where, $V_1 and V_2$ are initial and new volume respectively, while $T_1 and T_2$ are initial and new temperatures respectively;

$T_1= 333 K$

$T_2=303K$

$V_1 =2.0L$

Rearranging the formula;

$V_2=\frac{V_1T_2}{T_1}$

$= \frac{(2.0L)(303K)}{333K} \\=1.820 L$

Therefore, the new volume that would be occupied by the gas is 1.82 L

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3 years ago

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Aliun [14]

Metallioids
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2 years ago

determine mass of water formed when 12.5 L NH3(at298K and 1.50atm) is reacted with 18.9L of O2 (at 323K and 1.1atm)

sasho [114]

The mass of water formed is

calculation

Use the ideal gas equation to calculate the moles of NH3 and O2

that is Pv= n RT

where; P= pressure,

V= volume,

n = number of moles,

R=gas constant = 0.0821 l .atm/ mol.K

make n the formula of the subject by diving both side by RT

n = PV /RT

The moles of NH3

n= (1.50 atm x 12.5 L) /( 0.0821 L. atm /mol.k x 298 K) =0.766 moles

The moles of O2

=(1.1 atm x 18.9 L) / ( 0.0821 L. atm/ mol.k x 323 K) = 0.784 moles

write the reaction between NH3 and O2

4 NH3 + 5 O2 →4 No +6H2O

from equation above 0.766 moles of NH3 reacted to produce

0.766 x 6/4 =1.149 moles of H2O

0.784 moles of O2 reacted to produce 0.784 x 6/5=0.9408 moles of H20

since O2 is totally consumed, O2 is the limiting reagent and therefore the moles of H2O produced= 0.9408 moles

mass of H2O = moles x molar mass

from periodic table the molar mass of H2O = (1 x2)+16= 18 g/mol

mass = 18 g/mol x 0.9408 moles= 16.93 grams

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3 years ago