15.5% by mass is
equivalent 15.5 g urea in 100 g solution or 155 g urea in 1 kg solution. <span>
<span>we know that molality = moles solute / kg solvent
<span>moles solute = 155 g / 60 g/mol = 2.58 moles urea
</span></span></span>
Since there are 155 g
urea in 1000g solution, hence the solvent is 845 g or 0.845 kg
So:<span>
<span>molality = 2.58 / 0.845 = 3.06 m</span></span>
There are 1.078 x 10²³ molecules
<h3>Further explanation</h3>
Given
4 dm³ = 4 L Nitrogen gas
Required
Number of molecules
Solution
Assumptions on STP (1 atm, 273 K), 1 mol gas = 22.4 L, so for 4 L :
mol = 4 : 22.4
mol = 0.179
1 mol = 6.02 x 10²³ particles(molecules, atoms)
For 0.179 :
= 0.179 x 6.02 x 10²³
= 1.078 x 10²³
Answer:
234 KJ ≡ 55.887 Kcal
Explanation:
∴ J ≡ Kg/m².s² ≡ N.m = 0.2389 cal
∴ cal = 4.187 J
⇒ 234 KJ * ( 1000 J / KJ ) * ( cal / 4.187 J ) * ( Kcal / 1000 cal ) = 55.887 Kcal
Specific heat capacity= Quantity of heat/massxΔT
Shc of iron (constant)= 0.4494J/³C for 1g
1.49kg=1490g
Q=1490x(22-155)x0.4494
Q=<span>89057.598J</span>
Answer:
480.06 g/mol, Thorium nitrate.
Explanation: