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Ivenika [448]
3 years ago
8

A sample of hydrated cobalt (II) chloride has a mass of 5.22 g. After heating, it has a mass of 2.85 g. What is the percent by m

ass of water in the hydrate?
Chemistry
2 answers:
Anastasy [175]3 years ago
8 0

<u>Answer:</u> The percent by mass of water in the hydrate is 45.4 %

<u>Explanation:</u>

The chemical equation for the heating of hydrated cobalt (II) chloride follows:

CoCl_2.xH_2O\rightarrow CaCl_2+xH_2O

We are given:

Mass of hydrated cobalt (II) chloride = 5.22 g

Mass of anhydrous cobalt (II) chloride = 2.85 g

Mass of water released = (5.22 - 2.85)g = 2.37 g

To calculate the mass percentage of water in the hydrate, we use the equation:

\text{Mass percent of water}=\frac{\text{Mass of water}}{\text{Mass of hydrated cobalt (II) chloride}}\times 100

Mass of hydrated cobalt (II) chloride = 5.22 g

Mass of water = 2.37 g

Putting values in above equation, we get:

\text{Mass percent of water}=\frac{2.37g}{5.22g}\times 100=45.4\%

Hence, the percent by mass of water in the hydrate is 45.4 %

Tom [10]3 years ago
5 0
Answer:
mass of water in hydrate = 2.37 grams

Explanation:
The mass of the hydrated cobalt (III) chloride is the summation of the salt and the water it contains.
This means that:
Total mass of sample = mass of salt + mass of water

Now, we are given that:
total mass of sample = 5.22 grams
mass of salt = mass of sample after heating = 2.85 grams

Substitute to get the mass of water as follows:
5.22 = mass of water in hydrate + 2.85
mass of water in hydrate = 5.22 - 2.85
mass of water in hydrate = 2.37 grams

Hope this helps :)
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Ionic compounds are formed between oppositely charged ions.

A binary ionic compound is composed of ions of two different elements - one of which is a positive ion(metal), and the other is negative ion (nonmetal).

To write the empirical formula of binary ionic compound we must remember that one ion should be positive and other ion should be negative, then only the correct formula should be written. To write the empirical formula the charges of opposite ions should be criss-crossed.

First empirical formula of binary ionic compound is written betweenZn^{2+} (Positive ion)and F^{-} (Negative ion)

First Formula would be ZnF_{2}

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Second Formula would be Zn_{2}O_{2}

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Third empirical formula is between Ni^{4+}(Positive ion) and F^{-}(Negative ion)

Third Formula would be :NiF_{4}

Forth empirical formula is between Ni^{4+}(Positive ion)and O^{2-}(negative ion)

Forth Formula would be : Ni_{2}O_{4} or NiO_{2}

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Answer:

K(48.5°C) = 1.017 E-8 s-1

Explanation:

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at T1 = 25°C (298 K) ⇒ K1 = 3.32 E-10 s-1

at T2 = 48.5°C (321.5 K) ⇒ K2 = ?

Arrhenius eq:

  • K(T) = A e∧(-Ea/RT)
  • Ln K = Ln(A) - [(Ea/R)(1/T)]

∴ A: frecuency factor

∴ R = 8.314 E-3 KJ/K.mol

⇒ Ln K1 = Ln(A) - [Ea/R)*(1/T1)]..........(1)

⇒ Ln K2 = Ln(A) - [(Ea/R)*(1/T2)].............(2)

(1)/(2):

⇒ Ln (K1/K2) = (Ea/R)* (1/T2-1/T1)

⇒ Ln (K1/K2) = (116 KJ/mol/8.3134 E-3 KJ/K.mol)*(1/321.5 K - 1/298 K)

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⇒ K2 = (3.32 E-10 s-1)/0.0326

⇒ K2 = 1.017 E-8 s-1

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