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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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The British gold sovereign coin is an alloy of gold and copper having a total mass of 7.988 g, and is 22-karat gold.

azamat

Answer:

(a) $m_{gold}=7.322g$

(b)

$V_{gold}=0.379cm^3$

$V_{copper}=0.122cm^3$

Explanation:

Hello,

(a) In this case, with the given formula we easily compute the mass of gold contained in the sovereign as shown below:

$m_{gold}=\frac{m_{tota}*karats}{24}=\frac{7.988g*22}{24}=7.322g$

(b) Now, by knowing the density of gold and copper, 19.32 and 8.94 g/cm³ respectively, we compute each volume, by also knowing that the rest of the coin contains copper:

$V_{gold}=\frac{m_{gold}}{\rho_{gold}} =\frac{7.322g}{19.32g/cm^3}=0.379cm^3$

$m_{copper}=7.988g-7.322g=1.09g\\V_{copper}=\frac{m_{copper}}{\rho_{copper}}=\frac{1.09g}{8.94g/cm^3} \\\\V_{copper}=0.122cm^3$

$\rho _{coin}=\frac{m_{total}}{V_{gold}+V_{copper}}=\frac{7.988 g}{0.379cm^3+0.122cm^3}\\ \\\rho _{coin}=15.94g/cm^3$

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3 years ago

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Which of the following elements can form diatomic molecules held together by triple covalent bonds?

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Nitrogen can form a diatomic molecule held together by triple bonds.

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In the reaction below, how many grams of h2o(g) are produced when 2.1 grams o2(g) are consumed? c4h6(g) + o2(g) → co2(g) + h2o(g

amm1812

The balanced equation that illustrates the reaction is:
2C4H6 + 11O2 ......> 8CO2 + 6H2O

number of moles = mass / molar mass
number of moles of oxygen = 2.1 / 32 = 0.065625 moles

Now, from the balanced equation, we can note that:
11 moles of oxygen are required to produce 6 moles of water.
Therefore:
0.065625 moles of oxygen will produce:
(0.065625*6) / 11 = 0.03579 moles of water

number of moles = mass / molar mass
mass = number of moles * molar mass
mass of water = 0.03579 * 18 = 0.644 grams

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What is a form of potential energy that is due to relative positions of the particles within a materials.

Salsk061 [2.6K]

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Enter a balanced equation for the reaction between solid nickel(II)(II) oxide and carbon monoxide gas that produces solid nickel

velikii [3]

Answer: A balanced equation for the given reaction is $NiO(s) + CO \rightarrow Ni(s) + CO_{2}(g)$ .

Explanation:

The reaction equation will be as follows.

$NiO(s) + CO \rightarrow Ni(s) + CO_{2}(g)$

Number of atoms on the reactant side is as follows.

O = 2
C = 1

Number of atoms on the product side is as follows.

Ni = 1
O = 2
C = 1

Since number of atoms on both the reactant and product sides are equal. Hence, the reaction equation is balanced.

Thus, we can conclude that a balanced equation for the given reaction is $NiO(s) + CO \rightarrow Ni(s) + CO_{2}(g)$ .

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3 years ago