Answer: The mass of the sample will be 1417.7 grams.
Explanation:
We are given:

This means that 1 mole of NaCl has an enthalpy of fusion of 30.2 kJ
1 mole of NaCl has a mass of 58.44 grams.
So, 30.2 kJ of heat is require for a mass 58.44 grams of NaCl
So, 732.6 kJ of heat will be required for =
= 1417.65 grams of NaCl.
Hence, the mass of NaCl sample will be 1417.7 grams.
Answer : The correct option is, Mass
Explanation :
As we know that there are 3 states of matter :
Solid state : It is a state in which the particles are closely packed and does not have any space between them. This state have a definite shape and volume.
Liquid state : It is a state in which the particles are present in random and irregular pattern. The particles are closely arranged but they can move from one place to another. This state have a definite volume but does not have a fixed shape.
Gaseous state : It is a state in which the particles are loosely arranged and have a lot of space between them. This state have indefinite volume as well as shape.
If we are taking 100 grams of ice then after melting its mass remains same but its shape, volume and temperature will be changed and after evaporation its mass remains same but its shape, volume and temperature will be changed.
Hence, the mass will stay constant, no matter if the substance is in the solid, liquid, or gas state.
Answer:
Explanation:
From the information given:


no of moles of
= 0.01 L × 0.0010 mol/L
no of moles of
= 
no of moles of
= 0.01 L × 0.00010 mol/L
no of moles of
= 
Total volume = 0.02 L
![[Ca^{2+}}] = \dfrac{1\times10^{-5} \ mol}{0.02 \ L} \\ \\ \\ \[[Ca^{2+}}] = 0.0005 \ mol/L](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%7D%5D%20%3D%20%5Cdfrac%7B1%5Ctimes10%5E%7B-5%7D%20%5C%20mol%7D%7B0.02%20%5C%20L%7D%20%5C%5C%20%5C%5C%20%20%5C%5C%20%20%5C%5B%5BCa%5E%7B2%2B%7D%7D%5D%20%3D%200.0005%20%5C%20mol%2FL)
![[F^{-}] = \dfrac{(1\times 10^{-6} \ mol)}{0.02 \ L}](https://tex.z-dn.net/?f=%5BF%5E%7B-%7D%5D%20%3D%20%5Cdfrac%7B%281%5Ctimes%2010%5E%7B-6%7D%20%5C%20mol%29%7D%7B0.02%20%5C%20L%7D)
![[F^{-}] = 5 \times 10^{-5} \ mol/L](https://tex.z-dn.net/?f=%5BF%5E%7B-%7D%5D%20%3D%205%20%5Ctimes%2010%5E%7B-5%7D%20%20%5C%20mol%2FL)
![Q = [Ca^{2+}][F^-]^2 \\ \\ Q = 0.0005 \times (5\times 10^{-5})^2 \\ \\ Q = 1.25 \times 10^{-12}](https://tex.z-dn.net/?f=Q%20%3D%20%5BCa%5E%7B2%2B%7D%5D%5BF%5E-%5D%5E2%20%5C%5C%20%5C%5C%20Q%20%3D%200.0005%20%5Ctimes%20%285%5Ctimes%2010%5E%7B-5%7D%29%5E2%20%5C%5C%20%5C%5C%20Q%20%3D%201.25%20%5Ctimes%2010%5E%7B-12%7D)
Since Q<ksp, then there will no be any precipitation of CaF2
The answer is A . Saturated solution