Answer:
See explanation
Explanation:
The ionic radius of metal M decreases as the charge on the metal M increases. The ionic radius is generally defined as the distance between the nucleus and the outermost electron of the ion. Hence, ionic radius becomes much lesser as the magnitude of the positive charge increases.
It is obvious from the various formulae of metal chlorides in the question that the metal forms cations M^2+, M^3+ and M^4+ respectively. The order of decreasing ionic radius of the compounds is;
MX2 > MX3 > MX4
Answer:
option C. Addition of protons to the atomic nucleus.
Explanation:
Elements are fundamentally made up of atoms. Atoms in turn has three sub particles.
These particles are; neutrons, electrons and protons.
The sub particles responsible for the mass of an atoms is basically the protons and neutrons. Going through the options;
The correct option is option C. Addition of protons to the atomic nucleus.
Answer:
15.4 g of sucrose
Explanation:
Formula to be applied for solving these question: colligative property of freezing point depression. → ΔT = Kf . m
ΔT = Freezing T° of pure solvent - Freezing T° of solution
Let's replace data given: 0°C - (-0.56°C) = 1.86 C/m . m
0.56°C / 1.86 m/°C = m → 0.301 mol/kg
m → molality (moles of solute in 1kg of solvent)
Our mass of solvent is not 1kg, it is 150 g. Let's convert it from g to kg, to determine the moles of solute: 150 g. 1kg/1000g = 0.150 kg
0.301 mol/kg . 0.150kg = 0.045 moles.
We determine the mass of sucrose, by the molar mass:
0.045 mol . 342 g/1mol = 15.4 g
Answer:
Final temperature = T₂ = 155.43 °C
Explanation:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Given data:
Mass of coin = 4.50 g
Heat absorbed = 54 cal
Initial temperature = 25 °C
Specific heat of copper = 0.092 cal/g °C
Final temperature = ?
Solution:
Q = m.c. ΔT
ΔT = T₂ -T₁
Q = m.c. T₂ -T₁
54 cal = 4.50 g × 0.092 cal/g °C × T₂ -25 °C
54 cal = 0.414 cal/ °C × T₂ -25 °C
54 cal /0.414 cal/ °C = T₂ -25 °C
130.43 °C = T₂ -25 °C
130.43 °C + 25 °C = T₂
155.43 °C = T₂