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3 years ago

Calcium reacts with Hydrochloric acid to produce Calcium chloride and

Chemistry

1 answer:

hammer [34]3 years ago

8 0

Answer:

Mass = 2.77 g

Explanation:

Given data:

Mass of HCl = 2 g

Mass of CaCl₂ produced = ?

Solution:

Chemical equation:

2HCl + Ca → CaCl₂ + H₂

Number of moles of HCl:

Number of moles = mass / molar mass

Number of moles = 2 g/ 36.5 g/mol

Number of moles = 0.05 mol

now we will compare the moles of HCl with CaCl₂.

HCl : CaCl₂

2 : 1

0.05 : 1/2×0.05 = 0.025 mol

Mass of CaCl₂:

Mass = number of moles × molar mass

Mass = 0.025 mol × 110.98 g/mol

Mass = 2.77 g

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neonofarm [45]

Answer:

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Explanation:

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2 years ago

Four moles of hydrogen chloride (HCl) gas react with one mole of oxygen (O2) gas to produce two moles of chlorine gas (Cl2) and

Marianna [84]

Answer:

4HCl(g) + O₂(g) → 2Cl₂(g) + 2H₂O(g)

Explanation:

In order to find the equation we should state:

The reactants → Hydrogen chloride (HCl) and oxygen (O₂)

The products → Chlorine gas (Cl₂) and water gas (H₂O)

The balanced equation is:

4HCl + O₂ → 2Cl₂ + 2H₂O

It is a redox reaction, where the oxygen reduces to make water, and the chloride is oxidized to produce elemental chlorine.

7 0

2 years ago

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I performed an experiment and mixed copper nitrate and potassium iodide. When they reacted, they formed a precipitate, even thou

Alexus [3.1K]

Answer:

2Cu2^+ + 2I^- ----> 2Cu^+ + I2

Explanation:

The reaction performed in the experiment is;

2 Cu(NO3)2 + 4 KI → 2 CuI (s) + 4 KNO3 + I2

The iodide ions reduces Cu^2+ to Cu^+ which is insoluble in water hence the precipitate. This is so because iodine is a good oxidizing agent seeing that it requires one electron to fill its outermost shell. Potassium on the other hand is a good reducing agent since it easily looses its one electron.

The oxidation - reduction equation is as follows;

2Cu2^+ + 2e ----> 2Cu^+ reduction half equation

2I^- ----> I2 + 2e. Oxidation half equation

Balanced redox reaction equation;

2Cu2^+ + 2I^- ----> 2Cu^+ + I2

3 0

2 years ago

15 Celsius to kelvin

navik [9.2K]

Answer:

308K

Explanation:

293+15=308

7 0

3 years ago

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What volume is occupied by 18.4 g oxygen at 28.0°C and a pressure of 0.998 torr? (round to sig figs)

Alexxandr [17]

Answer:

11.0 dm³

Explanation:

From the question,

Applying

PV= nRT............... Equation 1

Where P = pressure of oxygen gas, V = volume of oxygen gas, n = number of moles of oxygen, R = molar constant, T = Temperature.

make V the subeject of the equation

V = nRT/P............. Equation 2

But,

Number of mole (n) = Mass of oxygen(m)/Molar mass of oxygen(m')

n = m/m'....................... Equation 3

Substitute equation 3 into equation 2

V = mRT/Pm'............. Equation 4

Given: T = 28°C = (28+273) = 301 K, P = 0.998 torr = (0.998×0.00131579) = 1.3132 atm, m = 18.4 g

Constant: R = 0.082 atm.dm³/K.mol, m' = 32 g/mol.

Substitute these values into equation 4

V = (301×18.4×0.082)/(32×1.3132)

V = 454.1488/42.0224

V = 10.81 dm³

V = 11.0 dm³

8 0

2 years ago