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fgiga [73]
3 years ago
8

What is the final, balanced equation that is formed by combining these two half reactions?

Chemistry
1 answer:
lara31 [8.8K]3 years ago
7 0

Answer: Balanced equation that is formed by combining these two half reactions is Cu + NO^{-}_{3} + 2H^{+} \rightarrow Cu^{2+} + NO^{-}_{2} + H_{2}O.

Explanation:

A chemical equation which contains same number of atoms on both reactant and product side is called a balanced chemical equation.

For example, Cu \rightarrow Cu^{2+} + 2e^{-}

NO^{-}_{3} + 2e^{-} + 2H^{+} \rightarrow NO^{-}_{2} + H_{2}O

On cancelling the common species from both these half-reactions, the complete balanced equation will be as follows.

Cu + NO^{-}_{3} + 2H^{+} \rightarrow Cu^{2+} + NO^{-}_{2} + H_{2}O

Thus, we can conclude that balanced equation that is formed by combining these two half reactions is Cu + NO^{-}_{3} + 2H^{+} \rightarrow Cu^{2+} + NO^{-}_{2} + H_{2}O.

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A sample of cacl2⋅2h2o/k2c2o4⋅h2o solid salt mixture is dissolved in ~150 ml de-ionized h2o. the oven dried precipitate has a ma
Paraphin [41]

We are given that the balanced chemical reaction is:

cacl2⋅2h2o(aq) + k2c2o4⋅h2o(aq) ---> cac2o4⋅h2o(s) + 2kcl(aq) + 2h2o(l)

We known that the product was oven dried, therefore the mass of 0.333 g pertains only to that of the substance cac2o4⋅h2o(s). So what we will do first is to convert this into moles by dividing the mass with the molar mass. The molar mass of cac2o4⋅h2o(s) is molar mass of cac2o4 plus the molar mass of h2o.

molar mass cac2o4⋅h2o(s) = 128.10 + 18 = 146.10 g /mole

moles cac2o4⋅h2o(s) = 0.333 / 146.10 = 2.28 x 10^-3 moles

Looking at the balanced chemical reaction, the ratio of cac2o4⋅h2o(s) and k2c2o4⋅h2o(aq) is 1:1, therefore:

moles k2c2o4⋅h2o(aq) = 2.28 x 10^-3 moles

Converting this to mass:

mass k2c2o4⋅h2o(aq) = 2.28 x 10^-3 moles (184.24 g /mol) = 0.419931006 g

 

Therefore:

The mass of k2c2o4⋅<span>h2o(aq) in the salt mixture is about 0.420 g</span>

3 0
3 years ago
Read 2 more answers
Ethanol (C2H5OH) melts a - 144 oC and boils at 78 °C. The enthalpy of fusion of ethanol is 5.02 kj/mol, and its enthalpy of vapo
hammer [34]

<u>Answer:</u>

<u>For a:</u> The total heat required is 36621.5 J

<u>For b:</u> The total heat required is 58944.5 J

<u>Explanation:</u>

  • <u>For a:</u>

To calculate the heat required at different temperature, we use the equation:

q=mc\Delta T         .........(1)

where,

q = heat absorbed

m = mass of substance

c = specific heat capacity of substance

\Delta T = change in temperature

To calculate the amount of heat required at same temperature, we use the equation:

q=m\times \Delta H      ........(2)

where,

q = heat absorbed

m = mass of substance

\Delta H = enthalpy of the reaction

The processes involved in the given problem are:

1.)C_2H_5OH(l)(35^oC)\rightarrow C_2H_5OH(l)(78^oC)\\2.)C_2H_5OH(l)(78^oC)\rightarrow C_2H_5OH(g)(78^oC)

  • <u>For process 1:</u>

We are given:

Change in temperature remains the same.

m=42.0g\\c_l=2.3J/g.K\\T_2=78^oC\\T_1=35^oC\\\Delta T=[T_2-T_1]=[78-35]^oC=43^oC=43K

Putting values in equation 1, we get:

q_1=42.0g\times 2.3J/g.K\times 43K\\\\q_1=4153.8J

  • <u>For process 2:</u>

We are given:

Conversion factor: 1 kJ = 1000 J

Molar mass of ethanol = 46 g/mol

m=42.0g\\\Delta H_{vap}=38.56kJ/mol=\frac{35.56kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=773.04J/g

Putting values in equation 2, we get:

q_2=42.0g\times 773.04J/g\\\\q_2=32467.7J

Total heat required = [q_1+q_2]

Total heat required = [4153.8J+32467.7J]=36621.5J

Hence, the total heat required is 36621.5 J

  • <u>For b:</u>

The processes involved in the given problem are:  

1.)C_2H_5OH(s)(-155^oC)\rightarrow C_2H_5OH(s)(-144^oC)\\2.)C_2H_5OH(s)(-144^oC)\rightarrow C_2H_5OH(l)(-144^oC)\\3.)C_2H_5OH(l)(-144^oC)\rightarrow C_2H_5OH(l)(78^oC)\\4.)C_2H_5OH(l)(78^oC)\rightarrow C_2H_5OH(g)(78^oC)

  • <u>For process 1:</u>

We are given:

Change in temperature remains the same.

m=42.0g\\c_s=0.97J/g.K\\T_2=-144^oC\\T_1=-155^oC\\\Delta T=[T_2-T_1]=[-144-(-155)]^oC=11^oC=11K

Putting values in equation 1, we get:

q_1=42.0g\times 0.97J/g.K\times 11K\\\\q_1=448.14J

  • <u>For process 2:</u>

We are given:

m=42.0g\\\Delta H_{fusion}=5.02kJ/mol=\frac{5.02kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=109.13J/g

Putting values in equation 2, we get:

q_2=42.0g\times 109.13J/g\\\\q_2=4583.5J

  • <u>For process 3:</u>

We are given:

Change in temperature remains the same.

m=42.0g\\c_l=2.3J/g.K\\T_2=78^oC\\T_1=-144^oC\\\Delta T=[T_2-T_1]=[78-(-144)]^oC=222^oC=222K

Putting values in equation 1, we get:

q_3=42.0g\times 2.3J/g.K\times 222K\\\\q_3=21445.2J

  • <u>For process 4:</u>

We are given:

m=42.0g\\\Delta H_{vap}=38.56kJ/mol=\frac{38.56kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=773.04J/g

Putting values in equation 2, we get:

q_4=42.0g\times 773.04J/g\\\\q_4=32467.7J

Total heat required = [q_1+q_2+q_3+q_4]

Total heat required = [448.14+4583.5+21445.2+32467.7]J=58944.5J

Hence, the total heat required is 58944.5 J

8 0
3 years ago
8SnCl4<br> How many atoms
Veronika [31]

Answer:

i think i like 14 or 8 atom

Explanation:

8 0
3 years ago
2. How many grams of water can be heated from 20.0 oC to 75oC using 12500.0 Joules?
kenny6666 [7]

Answer;

  = 0.054 kg or 54 g

Explanation;

Using the equation; Q = mcΔT where Q is the quantity of heat transferred, m is the mass, c is specific heat of the substance, ΔT is delta T, the change in temperature.  

ΔT = 75 - 20 = 55 C.  

Solve the equation for m  

m = Q/ cΔT

Mass = 12500 / (55 × 4200)

        = 0.054 kg or 54 g


4 0
3 years ago
Read 2 more answers
When the valves are opened and the gases are allowed to mix at a constant temperature, what is the distribution of atoms in each
jek_recluse [69]

Answer:

Equal number of atoms of each gas in each container

Explanation:

When the valves opened, the two contaienrs become one and the gases beging to mix by diffusion. This phenomenom is produced by the differeces of concentration of a gas between two points of the container.

The gases will continue diffunding util their concentration in both containers are equal.

7 0
3 years ago
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