Answer:
14.2L at STP
Explanation:
Based on the problem, 2 moles of NH3 produce 6 moles of HF. To solve this question we have to convert the mass of NH3 to moles. With the chemical equation find the moles of HF and using PV = nRT find the liters of HF:
<em>Moles NH3 -Molar mass: 17.031g/mol-</em>
3.6g NH3 * (1mol / 17.031g) = 0.211 moles NH3
<em>Moles HF:</em>
0.211 moles NH3 * (6mol HF / 2mol NH3) = 0.634 moles HF
<em>Volume HF</em>
PV = nRT; V = nRT/P
<em>Where V is volume in liters, n are moles of the gas = 0.634 moles, R is gas constant = 0.082atmL/molK, T is absolute temperature = 273.15K at STP and P is pressure = 1atm at STP.</em>
Replacing:
V = 0.634moles*0.082atmL/molK*273.15K / 1atm
V = 14.2L at STP
Elements are the same type of atoms, so there is nothing else to separate. Hydrogen has hydrogen atoms, and no other kind. Therefore nothing can be taken out of it anymore. Chemical x is a mixture, made by y and z. Nothing can make up y and z but themselves.
Answer:
a. Vₐ = 111.5282 + 1.29396m
b. For m = 0.100m; Vₐ = 111.6576
Explanation:
The partial molar volume of compound A in a mixture of A and B is defined as
:

Where V is volume and n are moles of a.
a. As molality is proportional to moles of substance, partial molar volume of glucose can be defined as:
Vₐ = dV / dm = d(1001.93 + 111.5282m + 0.64698m²) / dm
<em>Vₐ = 111.5282 + 1.29396m</em>
b. Replacing for m = 0.100m:
Vₐ = 111.5282 + 1.29396×0.100
<em>Vₐ = 111.6576</em>
<em></em>
I hope it helps!
Answer:
CaF2 will not precipitate
Explanation:
Given
Volume of Ca(NO3)2
ml
Molar concentration of Ca(NO3)2 
Volume of NaF
ml
Molar concentration of NaF 
Ksp for CaF2 
CaF2 will precipitate if Q for the reaction is greater than ksp of CAF2
Moles of calcium ion

![[Ca2+] = \frac{0.01}{10 + 10} \\= \frac{0.01}{20} \\= 5 * 10^{-4}](https://tex.z-dn.net/?f=%5BCa2%2B%5D%20%3D%20%5Cfrac%7B0.01%7D%7B10%20%2B%2010%7D%20%5C%5C%3D%20%5Cfrac%7B0.01%7D%7B20%7D%20%5C%5C%3D%205%20%2A%2010%5E%7B-4%7D)
Moles of F- ion

![[F-] = \frac{0.001}{10 + 10} \\= \frac{0.001}{20} \\= 5 * 10^{-5}](https://tex.z-dn.net/?f=%5BF-%5D%20%3D%20%5Cfrac%7B0.001%7D%7B10%20%2B%2010%7D%20%5C%5C%3D%20%5Cfrac%7B0.001%7D%7B20%7D%20%5C%5C%3D%205%20%2A%2010%5E%7B-5%7D)
![Q = [Ca2+] [F-]^2\\= (5 * 10^{-4}) * (0.5* 10^-4)\\= 1.25 * 10^{-12}](https://tex.z-dn.net/?f=Q%20%3D%20%5BCa2%2B%5D%20%5BF-%5D%5E2%5C%5C%3D%20%285%20%2A%2010%5E%7B-4%7D%29%20%2A%20%280.5%2A%2010%5E-4%29%5C%5C%3D%201.25%20%2A%2010%5E%7B-12%7D)
Q is lesser than Ksp value of CaF2. Hence it will not precipitate