Answer:
0.297 °C
Step-by-step explanation:
The formula for the <em>freezing point depression </em>ΔT_f is
ΔT_f = iK_f·b
i is the van’t Hoff factor: the number of moles of particles you get from a solute.
For glucose,
glucose(s) ⟶ glucose(aq)
1 mole glucose ⟶ 1 mol particles i = 1
Data:
Mass of glucose = 10.20 g
Mass of water = 355 g
ΔT_f = 1.86 °C·kg·mol⁻¹
Calculations:
(a) <em>Moles of glucose
</em>
n = 10.20 g × (1 mol/180.16 g)
= 0.056 62 mol
(b) <em>Kilograms of water
</em>
m = 355 g × (1 kg/1000 g)
= 0.355 kg
(c) <em>Molal concentration
</em>
b = moles of solute/kilograms of solvent
= 0.056 62 mol/0.355 kg
= 0.1595 mol·kg⁻¹
(d) <em>Freezing point depression
</em>
ΔT_f = 1 × 1.86 × 0.1595
= 0.297 °C
The elements of same group are placed in same group as they have similar properties. So Calcium is the element which resembles Mangesium.
Calcium and Magnesium both are alkaline earth metals.
They have similar properties as they have similar outer or valence shell electronic configuration
The general electronic configuration of alkaline earht metal is ns2
The electronic configuration of Mg [atomic number 12] and Ca [atomic number 20] are
and
respectively.
Answer:
Amount of heat energy released by light bulb = 25 joules
Explanation:
Given:
Energy receive by light bulb = 100 Joules
Energy released by light bulb as light energy = 75 Joules
Find:
Amount of heat energy released by light bulb
Computation:
We know that, energy is neither be created nor destroys
So,
Using Law of conservation of energy
Energy receive by light bulb = Energy released by light bulb as light energy + Amount of heat energy released by light bulb
100 = 75 + Amount of heat energy released by light bulb
Amount of heat energy released by light bulb = 100 - 75
Amount of heat energy released by light bulb = 25 joules
A. Covalent bonds is the answer
Complete Question
The complete question is shown on the first uploaded image
Answer:
The solution to this question is shown on the second uploaded image
Explanation: