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ollegr [7]
3 years ago
5

What does the little ‘2’ in 2O2 mean?

Chemistry
2 answers:
Lady bird [3.3K]3 years ago
7 0
It means 2 oxygen atoms in the molecule
Andrews [41]3 years ago
4 0

Answer:200

Explanation:

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Answer:

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Explanation:

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Why is this mars landing a new era for space exploration
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Explanation:

3 0
3 years ago
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
4 years ago
2Fe(OH)3 Fe2O3+3H2O how many grams of Fe2O3 are produced if 10.7 grams of Fe (OH)3 react in this way
sasho [114]

Answer: 7.98 grams of Fe_2O_3 are produced if 10.7 grams of Fe(OH)_3 are reacted.

Explanation:

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)  

Putting values in equation 1, we get:

\text{Moles of} Fe(OH)_3=\frac{10.7g}{106.87g/mol}=0.100mol  

The chemical equation for the reaction is

2Fe(OH)_3\rightarrow Fe_2O_3+3H_2O

By Stoichiometry of the reaction:

2 moles of Fe(OH)_3 produce = 1 mole of Fe_2O_3

So, 0.100 moles of Fe(OH)_3 produce= \frac{1}{2}\times 0.100=0.05mol of Fe_2O_3  

Mass of Fe_2O_3 =moles\times {\text{Molar Mass}}=0.05mol\times 159.69g/mol=7.98g  

Hence 7.98 grams of Fe_2O_3 are produced if 10.7 grams of Fe(OH)_3 are reacted.

3 0
3 years ago
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