Answer:
The energy transferred between samples of matter because of a difference in their temperatures is called a. heat.
Explanation:
The first law of thermodynamics establishes that when two bodies with different temperatures are put in contact they will find thermic equilibrium to a final temperature by transferring heat. Thus the correct answer is (a).
Thermochemistry is the study of the transformations of heat energy on the chemical reactions. Chemical kinetics is the study of the rate of chemical reactions. And temperature is the measure of the heat.
Chemical equilibrium<span> is the state in which both reactants and products are present in concentrations which have no further tendency to change with time.
</span><span>Or, we can say that in chemical equilibrium the ratio between the concentration of the reactants and the products is constant.</span><span>
Chemical equilibrium is a result state when </span><span>the forward reaction proceeds at the same rate as the reverse reaction.
</span><span>Different reactions have different equilibrium.</span>
Explanation:
Defining law of definite proportions, it states that when two elements form more than one compound, the ratios of the masses of the second element which combine with a fixed mass of the first element will always be ratios of small whole numbers.
A. One of the oxides (Oxide 1) contains 63.2% of Mn.
Mass of the oxide = 100g
Mass of Mn = 63.2 g
Mass of O = 100 - 63.2
= 36.8 g
Ratio of Mn to O = 63.2/36.8
= 1.72
Another oxide (Oxide 2) contains 77.5% Mn.
Mass of oxide = 100 g
Mass of Mn = 77.5 g
Mass of O = 100 - 77.5
= 22.5 g
Ratio of Mn to O = 77.5/22.5
= 3.44
Therefore, the ratio of the masses of Mn and O in Oxide 1 and Oxide 2 is in the ratio 1.72 : 3.44, which is also 1 : 2. So the law of multiple proportions is obeyed.
B.
Oxide 1
Mass of Mn per 1 g of O = mass of Mn/mass of O
= 77.5/22.5
= 3.44 g/g of Oxygen.
Oxide 2
Mass of Mn per 1 g of O = mass of Mn/mass of O
= 77.5/22.5
= 3.44 g/g of Oxygen.
Answer:
Δ S = 93.8 J/mol-K
Explanation:
Given,
Boiling point of chloroform = 61.7 °C
= 273 + 61.7 = 334.7 K.
Enthalapy of vapourization = 31.4 kJ/mol.
Using Gibbs free energy equation
Δ G = Δ H - T (ΔS)
at equilibrium (when the liquid is boiling), Δ G = 0
so, 0 = ΔH - T (Δ S)
T (Δ S) = Δ H
and ΔS = ΔH / T
Δ S = (31400 J/mol.) / 334.7 K
Δ S = 93.8 J/mol-K
As we move down the group, the metallic bond becomes more stable and the formation of forming covalent bond decreases down the group due to the large size of elements.
Covalent and metallic bonding leads to higher melting points. Due to a decrease in attractive forces from carbon to lead there is a drop in melting point.
Carbon forms large covalent molecules than silicon and hence has a higher melting point than silicon.
Similarly, Ge also forms a large number of covalent bonds and has a smaller size as compared to that of Sn. Hence melting point decreases from Ge to Sn.
The order will be C>Si>Ge>Pb>Sn.
To learn more about the covalent bond, visit: brainly.com/question/10777799
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