Answer is: molality of urea is 5.84 m.
If we use 100 mL of solution:
d(solution) = 1.07 g/mL.
m(solution) = 1.07 g/mL · 100 mL.
m(solution) = 107 g.
ω(N₂H₄CO) = 26% ÷ 100% = 0.26.
m(N₂H₄CO) = m(solution) · ω(N₂H₄CO).
m(N₂H₄CO) = 107 g · 0.26.
m(N₂H₄CO) = 27.82 g.
1) calculate amount of urea:
n(N₂H₄CO) = m(N₂H₄CO) ÷ M(N₂H₄CO).
n(N₂H₄CO) = 27.82 g ÷ 60.06 g/mol.
n(N₂H₄CO) = 0.463 mol; amount of substance.
2) calculate mass of water:
m(H₂O) = 107 g - 27.82 g.
m(H₂O) = 79.18 g ÷ 1000 g/kg.
m(H₂O) = 0.07918 kg.
3) calculate molality:
b = n(N₂H₄CO) ÷ m(H₂O).
b = 0.463 mol ÷ 0.07918 kg.
b = 5.84 mol/kg.
I will solve this question assuming the reaction equation look like this:
<span>MnO2 + 4 HCl ---> MnCl2 + Cl2 + 2 H2O.
</span>
For every one molecule of MnO2 used, there will be one molecule of Cl2 formed. If the molecular mass of MnO2 is 87g/mol and molecular mass of Cl2 is <span> 73.0 g/mol, the mass of MnO2 needed would be:
Cl mass/Cl molecular mass * MnO2 molecular mass=
25g/ (73g/mol) * (87g/mol) * 1/1= 29.8 grams</span>
Answer:
100g/mol
Explanation:
Given parameters:
Mass of unknown gas = 2g
Volume of gas in flask = 500mL = 0.5dm³
Unknown:
Molar mass of gas = ?
Solution:
Since we know the gas is at STP;
1 mole of substance occupies 22.4dm³ of space at STP
Therefore,
0.5dm³ will have 0.02mole at STP
Now;
Number of moles =
Molar mass =
=
= 100g/mol
Answer:
The correct answer is option C, that is, ΔS and ΔSsurr for the process H2O (s) ⇒ H2O(l) are equal in magnitude and opposite in sign.
Explanation:
The temperature at which solid state of water get transformed into liquid state is termed as the melting point of 0 °C. It can be shown by the reaction:
H2O (s) ⇒ H2O (l)
The degree of randomness of a molecule is known as entropy. With the transformation of ice into liquid state, there is an increase in randomness. Thus, the value of entropy becomes positive as shown:
Entropy change (ΔSsys) = ΔSproduct - ΔSreactant
= (69.9 - 47.89) J mol/K
= 22.0 J mol/K
Therefore, the value of entropy change is positive.
Now the value of entropy for surrounding ΔSsurr will be,
ΔSsurr = -ΔHfusion/T
= -6012 j/mol/273
= -22.0 J/molK
Hence, the value of ΔSsurr and ΔSsys exhibit same magnitude with opposite sign.