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ankoles [38]
3 years ago
9

5. Given that a 40.14 g sample of hydrated NiSO4•XH2O is reduced in mass to 2.14 g upon heating.

Chemistry
1 answer:
Ahat [919]3 years ago
4 0
<h3>Answer:</h3>

Value of X is 7

<h3>Explanation:</h3>

We are given;

  • Mass of the Hydrated sample as 40.14 g
  • Mass of the anhydrous salt as 22.14 g (Missing in the question)

We are required to determine the value of X in the hydrated sample, NiSO4•XH2O .

  • First, we determine the mass of the anhydrous salt
  • Mass of water of crystallization = Mass hydrate sample - Mass of  anhydrous sample

                                      = 40.14 g - 22.14 g

                                      = 18.0 g

We can determine the value of X by determine simple number ratio of anhydrous compound and water molecules.

  • Determine the number of moles
  • Molar mass of NiSO₄ = 154.75 g/mol
  • Moles of NiSO₄ = 22.14 g ÷ 154.75 g/mol

                                  = 0.1431 mole

Molar mass of water = 18.02 g/mol

Moles of water = 18.0 g  ÷ 18.02 g/mol

                          = 1.00 mole

We can then determine the mole ratio.

Mole ratio = 0.1431 mole : 1.00 mole

                 = 0.1431 mole/0.1431 mole : 1.00 mole/0.1421 mole

                 = 1 : 7.03

                = 1 : 7

Therefore, the formula of the hydrated salt is NiSO₄•7H₂O

Thus, the value of X is 7

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3 years ago
a 125 g chunk of aluminum at 182 degrees Celsius was added to a bucket filled with 365 g of water at 22.0 degrees Celsius. Ignor
Diano4ka-milaya [45]
<h3>Answer:</h3>

32.98°C

<h3>Explanation:</h3>

We are given the following;

Mass of Aluminium as 125 g

Initial temperature of Aluminium as 182°C

Mass of water as 265 g

Initial temperature of water as 22°C

We are required to calculate the final temperature of the two compounds;

First, we need to know the specific heat capacity of each;

Specific heat capacity of Aluminium is 0.9 J/g°C

Specific heat capacity of water is 4.184 J/g°C

<h3>Step 1: Calculate the Quantity of heat gained by water.</h3>

Assuming the final temperature is X°C

we know, Q = mcΔT

Change in temperature, ΔT = (X-22)°C

therefore;

Q = 365 g × 4.184 J/g°C × (X-22)°C

    = (1527.16X-33,597.52) Joules

<h3>Step 2: Calculate the quantity of heat released by Aluminium </h3>

Using the final temperature, X°C

Change in temperature, ΔT = -(X°- 182°)C (negative because heat was lost)

Therefore;

Q = 125 g × 0.90 J/g°C × (182°-X°)C

  = (20,475- 112.5X) Joules

<h3>Step 3: Calculating the final temperature</h3>

We need to know that the heat released by aluminium is equal to heat absorbed by water.

Therefore;

(20,475- 112.5X) Joules = (1527.16X-33,597.52) Joules

Combining the like terms;

1639.66X = 54072.52

             X = 32.978°C

                = 32.98°C

Therefore, the final temperature of the two compounds will be 32.98°C

7 0
4 years ago
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xeze [42]

Answer:

can only be determined experimentally.

Explanation:

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These ions appear to have structures that defied accurate elucidation. However, by diligent laboratory investigation, Alfred Werner was able to accurately determine the structure of cobalt complexes. As a result of this, he is regarded as a pathfinder in coordination chemistry.

Hence, the structure of complex ions can only be determined experimentally.

6 0
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Answer:

The correct answer is A

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