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Anna [14]
2 years ago
10

Determine the name and formula of a hydrate with the following composition; 32.30%

Chemistry
1 answer:
Maslowich2 years ago
7 0

1. The formula of the hydrate is FePO₄•4H₂O

2. The name of the hydrate is Iron(iii) phosphate tetrahydrate.

1. Determination of the formula of the hydrate

  • H₂O = 32.3%
  • FePO₄ = 67.7%
  • Formula =?

The formula of the hydrate can be obtained as follow:

Divide by their molar mass

H₂O = 32.3 / 18 = 1.794

FePO₄ = 67.7 / 151 = 0.448

Divide by the smallest

H₂O = 1.794 / 0.448 = 4

FePO₄ = 0.448 / 0.448 = 1

Therefore, the formula of the hydrate is FePO₄•4H₂O

The name of the hydrate is Iron(iii) phosphate tetrahydrate

Learn more about empirical formula: brainly.com/question/16143068

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1. Do you think that the lightbulb and the Moon spheres are "to scale" compared to the real
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3 0
2 years ago
The reaction C4H8(g)⟶2C2H4(g) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ/mol.262 kJ/mol. At 600.0 K,600.0 K, the rate c
crimeas [40]

Answer : The rate constant at 785.0 K is, 1.45\times 10^{-2}s^{-1}

Explanation :

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

or,

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = rate constant at 600.0K = 6.1\times 10^{-8}s^{-1}

K_2 = rate constant at 785.0K = ?

Ea = activation energy for the reaction = 262 kJ/mole = 262000 J/mole

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 600.0K

T_2 = final temperature = 785.0K

Now put all the given values in this formula, we get:

\log (\frac{K_2}{6.1\times 10^{-8}s^{-1}})=\frac{262000J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{785.0K}]

K_2=1.45\times 10^{-2}s^{-1}

Therefore, the rate constant at 785.0 K is, 1.45\times 10^{-2}s^{-1}

7 0
3 years ago
Determine the AMOUNT OF NO2, LIMITING REACTANT, AND THE AMOUNT AND NAME OF EXCESS REACTANT.
zepelin [54]

Answer:

balanced equation mole ratio 5 2 mol NO/1 mol O2

10.00 g O2 3 1 mol O2/32.00 g O2 5 0.3125 mol O2

20.00 g NO 3 1 mol NO/30.01 g NO 5 0.6664 mol NO

actual mole ratio 5 0.6664 mol NO/0.3125 mol O2 5 2.132 mol NO/1.000 mol O2

Because the actual mole ratio of NO:O2 is larger than the balanced equation mole

ratio of NO:O2, there is an excess of NO; O2 is the limiting reactant.

Mass of NO used 5 0.3125 mol O2 3 2 mol NO/1 mol O2 5 0.6250 mol NO

0.6250 mol NO 3 30.01 g NO/1 mol NO 5 18.76 g NO

Mass of NO2 produced 5 0.6250 mol NO2 3 46.01 g NO2/1 mol NO2 5 28.76 g NO2

Excess NO 5 20.00 g NO 2 18.76 g NO 5 1.24 g N

Explanation:

3 0
2 years ago
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