Answer:
There are 3 steps of this problem.
Explanation:
Step 1.
Wet steam at 1100 kPa expands at constant enthalpy to 101.33 kPa, where its temperature is 105°C.
Step 2.
Enthalpy of saturated liquid Haq = 781.124 J/g
Enthalpy of saturated vapour Hvap = 2779.7 J/g
Enthalpy of steam at 101.33 kPa and 105°C is H2= 2686.1 J/g
Step 3.
In constant enthalpy process, H1=H2 which means inlet enthalpy is equal to outlet enthalpy
So, H1=H2
H2= (1-x)Haq+XHvap.........1
Putting the values in 1
2686.1(J/g) = {(1-x)x 781.124(J/g)} + {X x 2779.7 (J/g)}
= 781.124 (J/g) - x781.124 (J/g) = x2779.7 (J/g)
1904.976 (J/g) = x1998.576 (J/g)
x = 1904.976 (J/g)/1998.576 (J/g)
x = 0.953
So, the quality of the wet steam is 0.953
The answer is: A
C-14 is not stable and this is the reason why it goes through radioactive decay.
Answer:
the molarity is 3.68 moles/L
Explanation:
the molality of the solution of sucrose is
m= moles of glucose / Kg of solvent (water)= 6.81 ,
since the molecular weight of glucose is 180.156 gr/mole , then per each kilogram of solvent there is
6.81 moles*180.156 gr/mole + 1000 gr of water = 2226.86 gr of solution
from the density
volume of solution = mass of solution/density = 2286.86 gr / 1.2 gr/ml = 1855.71 ml
therefore there is 1000 gr of water in 1855.71 ml
then the molarity M is
M= moles of glucose / L of solution = (moles of glucose / Kg of solvent) * (Kg of solvent/L of solution) = 6.81 moles/Kg * 1Kg/1.85 L = 3.68 moles/L
M= 3.68 moles/L
Note:
- Would be wrong in this case to assume density of water = 1 Kg/L since the solution is heavily concentrated in glucose and therefore the density of water deviates from its pure value.
1. C : Mg(CN)2
2. B : N2O5
3. D : Ti(ClO4)3
4. A : Ni(NO3)3
5. D : Cobalt (III) Acetate
6. B : Nickel(II) sulfate
7. C : Dinitrogen Tetrafluoride
8. A : Phosphorus pentachloride<em />
9. C : Metallic <em>(<!> This is the only one I'm not 100% sure of)</em>
10. A : Ionic
11. C : Metallic
12. B : Covalent
Answer:
15 ml
Explanation:
Volume = mass / density.
So our answer is 15 / 3 = 15 mL
