Answer:- C. Hafnium.
Solution:- Mass of the sample is 46.0 g and it's volume is 
.
From mass and volume, we can calculate it's density using the formula:
On the basis of the density, this substance could either be mercury or hafnium. Since the substance is a solid at room temperature where as mercury is liquid. So, it can't be mercury.
The right choice is C) Hafnium.
Answer:
Option A.
Lower air pressure results in a lower boiling point
Explanation:
This is because in an open system, the lower the pressure the lesser the energy that will be required for boiling point. The is little or no collision of air molecules with the surface of the liquid
But if there is increase in pressure, more energy will be required to get to boiling point because there will be strong collision between air molecules and surface of the liquid.
Answer:
C
Explanation:
The oxidation number of Sulphur in SO4^2- is;
x + 4(-2) = -2
x - 8 = -2
x = -2 + 8
x = 6
Now,
the oxidation number of sulphur in H2SO3 is
2 (1) + x + 3(-2) = 0
2 + x -6 = 0
-4 + x = 0
x = 4
Hence, the oxidation number of sulphur changed from +6 to +4 which signifies gain of two electrons as shown in option C.
Answer:
The correct answer to the question is
The standard heat of reaction for the reaction is
a. 216.8 kJ released per mole
Explanation:
The heat of reaction is given by [Heat of formation of products] - [Heat of formation of reactants]
In the question we have, heat of formation of the products Zn+2 (aq) = -152.4 kJ/mole and the heat of formation of the reactants = 64.4 kJ/mole
Therefore, the heat of formation of the reaction = (-152-64.4) kJ/mole or
-216.8 kJ/mole released
