Explanation:
The dimensions of a standard backpack is 30cm x 30cm x 40cm
The mass of an average student is 70 kg
We know that, the density of gold is 19.3 g/cm³.
Let m be the mass of the backpack. So,
An average student has a mass of 70 kg. If we compare the mass of student and mass of backpack, we find that the backpack is 10 times of the mass of the student.
Answer:
-1
Explanation:
The relation between Kp and Kc is given below:
Where,
Kp is the pressure equilibrium constant
Kc is the molar equilibrium constant
R is gas constant , 0.082057 L atm.mol⁻¹K⁻¹
T is the temperature in Kelvins
Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)
For the first equilibrium reaction:
<u>Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants) = (2+1)-(2+2) = -1 </u>
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Answer:
Molar mass of solute: 300g/mol
Explanation:
<em>Vapor pressure of pure benzene: 0.930 atm</em>
<em>Assuming you dissolve 10.0 g of the non-volatile solute in 78.11g of benzene and vapour pressure of solution was found to be 0.900atm</em>
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It is possible to answer this question based on Raoult's law that states vapor pressure of an ideal solution is equal to mole fraction of the solvent multiplied to pressure of pure solvent:
Moles in 78.11g of benzene are:
78.11g benzene × (1mol / 78.11g) = <em>1 mol benzene</em>
Now, mole fraction replacing in Raoult's law is:
0.900atm / 0.930atm = <em>0.9677 = moles solvent / total moles</em>.
As mole of solvent is 1:
0.9677× total moles = 1 mole benzene.
Total moles:
1.033 total moles. Moles of solute are:
1.033 moles - 1.000 moles = <em>0.0333 moles</em>.
As molar mass is the mass of a substance in 1 mole. Molar mass of the solute is:
10.0g / 0.033moles = <em>300g/mol</em>