<span>Radioactive isotopes have been used commercially in all these applications.
The last option (D) is your answer
</span>
Answer:
- Question 19: the three are molecular compounds.
Explanation:
<em>Question 19.</em>
All of them are the combination of two kinds of different atoms in fixed proportions.
- C₂H₄: two carbon atoms per four hydrogen atoms
- HF: one hydrogen atom per one fluorine atom
- H₂O₂: two hydrogen atoms per two oxygent atoms
Thus, they all meet the definition of compund: a pure substance formed by two or more different elements with a definite composition.
Molecular compounds are formed by covalent bonds and ionic compounds are formed by ionic bonds.
Two non-metal elements, like H-F, C - C, C - H, H-O, H - H, and O - O will share electrons forming covalent bonds to complete their valence shell. Thus, the three compounds are molecular and not ionic.
<em>Question 20. </em>Formula of copper(II) sulfate hydrate with 36.0% water.
Copper(II) sulfate is CuSO₄. Its molar mass is 159.609g/mol
Water is H₂O. Its molar mass is 18.015g/mol
Calling x the number of water molecules in the hydrate, the percentage of water is:

From which we can solve for x:

Thus, there are 5 molecules of water per each unit of CuSO₄, and the formula is:
Given information : H = -92 KJ/mol and S = -0.199 KJ/(mol.K)
At equilibrium G = 0
We have to find the Temperature at which reaction would be spontaneous.
For spontaneous reaction : 
For non-spontaneous reaction : 
We can find the temperature using the formula for Gibbs free energy which is:

Where, G = Gibbs free energy ,
H = Enthalpy
S = Entropy
T = Temperature
By plugging the value of G , H and S in the above formula we can find 'T'

Since reaction should be spontaneous that means
should be negative , so the above formula can be written as :

On rearranging the above formula we get :




For the reaction to be spontaneous , T should be less than 462.3 K, so out of given option , C is correct which is 400 K.
Answer:
-3.617 °C
Explanation:
Step 1: Given data
Mass of water (m): 210.0 g
Energy released in the form of heat (Q): -3178 J (the minus sign corresponds to energy being released)
Specific heat of water (c): 4.184 J/g.°C
Temperature change (ΔT): ?
Step 2: Calculate the temperature change
We will use the following expression.
Q = c × m × ΔT
-3178 J = 4.184 J/g.°C × 210.0 g × ΔT
ΔT = -3.617 °C
34g -------- 100g H₂O
Xg ---------- 500g H₂O
X = (500×34)/100
<u>X = 170g KCl
</u>:)