Study the Lewis dot diagram for nitrogen. Explain what is correct and what is incorrect about the diagram. Use details to suppor
t your answer.
1 answer:
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At a particular temperature, the solubility of He in water is 0.080 M when the partial pressure is 1.7 atm. 4.25 atm is the partial pressure of He would give a solubility of 0.200 M.
<h3>What is Henry's Law ?</h3>Henry's Law is a gas law states that at a constant temperature the amount of gas that dissolved in a liquid is directly proportional to the partial pressure of that gas.
<h3>What is relationship between Henry's Law constant and Solubility ?</h3>The solubility of gas is directly proportional to partial pressure.
It is expressed as:
where,
= Solubility of gas
= Henry's Law constant
= Partial pressure of gas
Now put the values in above expression we get
0.080M = × 1.7 atm
= 0.047 M/atm
Now we have to find the partial pressure of He
0.200 M = 0.047 M/atm ×
= 4.25 atm
Thus from the above conclusion we can say that At a particular temperature, the solubility of He in water is 0.080 M when the partial pressure is 1.7 atm. 4.25 atm is the partial pressure of He would give a solubility of 0.200 M.
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<u>Answer:</u> The enthalpy change of the reaction is -27. kJ/mol
<u>Explanation:</u>
To calculate the mass of water, we use the equation:
Density of water = 1 g/mL
Volume of water = 25.0 mL
Putting values in above equation, we get:
To calculate the heat released by the reaction, we use the equation:
where,
q = heat released
m = Total mass = [1.25 + 25] = 26.25 g
c = heat capacity of water = 4.18 J/g°C
= change in temperature =
Putting values in above equation, we get:
To calculate the number of moles, we use the equation:
Given mass of ammonium nitrate = 1.25 g
Molar mass of ammonium nitrate = 80 g/mol
Putting values in above equation, we get:
To calculate the enthalpy change of the reaction, we use the equation:
where,
q = amount of heat released = -0.428 kJ
n = number of moles = 0.0156 moles
= enthalpy change of the reaction
Putting values in above equation, we get:
Hence, the enthalpy change of the reaction is -27. kJ/mol
The equation for carbon-14 emission by Radium-223 nuclei is given below:
Radioactivity is the spontaneous decay of a substance with emission of radiation.
The equation for carbon-14 emission by Radium-223 nuclei is given below:
In conclusion, the emission of carbon-14 by Radium-223 nuclei produces Lead-209 nuclei.
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