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Leya [2.2K]
3 years ago
15

Cobalt III hydroxide and nitric acid react according to the following balanced equation:

Chemistry
1 answer:
velikii [3]3 years ago
5 0
Q1)
molarity is defined as the number of moles in a volume of 1 L.
the molarity of the solution to be prepared is 0.42 M
The volume of the solution to be prepared is 5.2 L
the number of moles of Co(OH)₃ in 1 L solution - 0.42 mol
therefore number of moles required in 5.2 L - 0.42 mol/L x 5.2 L = 2.184 mol
molar mass of Co(OH)₃ - 110 g/mol
mass of Co(OH)₃ required to prepare the solution - 2.184 mol x 110 g/mol = 240.2 g
mass of Co(OH)₃ required - 240.2 g

Q2)
the balanced equation for the reaction between Co(OH)₃ and HNO₃ is;
Co(OH)₃ + 3HNO₃ --> Co(NO₃)₃ + 3H₂O
stoichiometry of Co(OH)₃ to HNO₃ is 1:3
the number of Co(OH)₃ moles present - 2.184 mol
the number of HNO₃ moles required - 2.184 x 3 = 6.552 mol
molarity of HNO₃ is 1.6 M
If 1.6 mol is in 1 L of solution 
then volume containing 6.552 mol is - 6.552 mol / 1.6 mol/L = 4.095 L
therefore volume of HNO₃required is 4095 mL 

Q3)
the balanced equation
Co(OH)₃ + 3HNO₃ --> Co(NO₃)₃ + 3H₂O
stoichiometry of Co(OH)₃ to H₂O is 1:3
the number of Co(OH)₃ moles reacted - 2.184 mol 
when 1 mol of Co(OH)₃ reacts - 3 mol of H₂O is formed
therefore when 2.184 mol of Co(OH)₃ moles react - 2.184 x 3 = 6.552 mol of H₂O formed
therefore number of H₂O moles formed - 6.552 mol
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<h3>What is a redox reaction?</h3>

A redox or oxidation-reduction reaction is a chemical reaction in which some of the atoms have their oxidation number changed.

In a chemical reaction that involves oxidation and reduction, the oxidation number of the involved ions either decreases or increases.

Therefore, the statement that identifies an oxidation-reduction reaction is a reaction in which oxidation numbers change.

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6 0
2 years ago
A sample of gas in a balloon has an initial temperature of 21 ∘C and a volume of 1310 L . If the temperature changes to 70 ∘C ,
bixtya [17]

Answer:

1528.3L

Explanation:

To solve this problem we should know this formula:

V₁ / T₁ = V₂ / T₂

We must convert the values of T° to Absolute T° (T° in K)

21°C + 273 = 294K

70°C + 273 = 343K

Now we can replace the data

1310L / 294K = V₂ / 343K

V₂ = (1310L / 294K) . 343K → 1528.3L

If the pressure keeps on constant, volume is modified directly proportional to absolute temperature. As T° has increased, the volume increased too

7 0
4 years ago
Which quantity is most closely related to the speed of gas particles?
max2010maxim [7]
The temperature of the gas, the hotter they get the faster they move. the colder they get, the slower they move
3 0
3 years ago
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Calculate the amount of heat released when 27.0 g H2O is cooled from a liquid at 314 K to a solid at 263 K. The melting point of
BlackZzzverrR [31]

Answer : The amount of heat released, 45.89 KJ

Solution :

Process involved in the calculation of heat released :

(1):H_2O(l)(314K)\rightarrow H_2O(l)(273K)\\\\(2):H_2O(l)(273K)\rightarrow H_2O(s)(273K)\\\\(3):H_2O(s)(273K)\rightarrow H_2O(s)(263K)

Now we have to calculate the amount of heat released.

Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+\Delta H_{fusion}+[m\times c_{p,s}\times (T_{final}-T_{initial})]

where,

Q = amount of heat released = ?

m = mass of water = 27 g

c_{p,l} = specific heat of liquid water = 4.184 J/gk

c_{p,s} = specific heat of solid water = 2.093 J/gk

\Delta H_{fusion} = enthalpy change for fusion = 40.7 KJ/mole = 40700 J/mole

conversion : 0^oC=273k

Now put all the given values in the above expression, we get

Q=[27g\times 4.184J/gK\times (314-273)k]+40700J+[27g\times 2.093J/gK\times (273-263)k]

Q=45896.798J=45.89KJ     (1 KJ = 1000 J)

Therefore, the amount of heat released, 45.89 KJ

7 0
3 years ago
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Which of the following effect increases the solubility of Ca(OH)2?
mash [69]

Answer:

adding an acidic solution

Explanation:

Strong acids dissociate completely in water and the H⁺ ions react with the OH⁻ ions to make H₂O. If we add an acid to our solution of Ca(OH)₂, the acid protons would react with some of the OH⁻ ions and drive the equilibrium to the right. More of the solid Ca(OH)₂ should dissolve.

4 0
3 years ago
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