Given:
P1 = 13.0 atm
T1 = 20 °C
T2 = 102 °C
Required:
P2 of oxygen
Solution:
At constant volume,
we can apply Gay-Lussac’s law of pressure and temperature relationship
P1/T1=P2/T2
(13.0 atm) / (20 °C)
= P2 / (102 °C)
P2 = 66.3 atm
The answer is not in the choices given.
Answer:
ΔH = 2.68kJ/mol
Explanation:
The ΔH of dissolution of a reaction is defined as the heat produced per mole of reaction. We have 3.15 moles of the solid, to find the heat produced we need to use the equation:
q = m*S*ΔT
<em>Where q is heat of reaction in J,</em>
<em>m is the mass of the solution in g,</em>
<em>S is specific heat of the solution = 4.184J/g°C</em>
<em>ΔT is change in temperature = 11.21°C</em>
The mass of the solution is obtained from the volume and the density as follows:
150.0mL * (1.20g/mL) = 180.0g
Replacing:
q = 180.0g*4.184J/g°C*11.21°C
q = 8442J
q = 8.44kJ when 3.15 moles of the solid react.
The ΔH of the reaction is:
8.44kJ/3.15 mol
= 2.68kJ/mol
2)soil it's soil because soil is made out of <span>weathered rock and organic matter</span>
Answer:
To prepare 1.00 L of 2.0 M urea solution, we need to dissolve 120 g of urea in enough water to produce a total of 1.00 L solution
Explanation:
Molarity of a solute in a solution denotes number of moles of solute dissolved in 1 L of solution.
So, moles of urea in 1.00 L of a 2.0 M urea solution = 2 moles
We know, number of moles of a compound is the ratio of mass to molar mass of that compound.
So, mass of 2 moles of urea = 
Therefore to prepare 1.00 L of 2.0 M urea solution, we need to dissolve 120 g of urea in enough water to produce a total of 1.00 L solution
So, option (C) is correct.
