Answer:
Hey there!
Gold would have a greater mass, because density is mass/volume, and if the density is greater, so is the mass.
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Answer:
Most common oxidation state of the chalcogens is -2, most common oxidation state of the halogens is -1.
Explanation:
For atomic radii, the chalcogens have a larger atomic radii than the halogens
This is because atomic radii decreases across the period due to increase in nuclear charge.
For ionic radii the chalcogens also have larger ionic radii than the halogens. This is because the chalcogens always carry a -2 charge compared to halogens that carry a -1 charge. Since -2 is the most common oxidation state for chalcogens and -1 is the most common oxidation state for the halogens.
In terms of oxidation states, the halogens show a higher value of common oxidation state -1 while for chalcogens is -2 even though +2, +4 and +6 oxidation states are also well known.
First ionization energy of halogens is greater than that of the chalcogens due to greater effective nuclear charge.
The second ionization energy of chalcogens is greater than that of the halogens.
Answer:
Hydrogen: -141 kJ/g
Methane: -55kJ/g
The energy released per gram of hydrogen in its combustion is higher than the energy released per gram of methane in its combustion.
Explanation:
According to the law of conservation of the energy, the sum of the heat released by the combustion and the heat absorbed by the bomb calorimeter is zero.
Qc + Qb = 0
Qc = -Qb [1]
We can calculate the heat absorbed by the bomb calorimeter using the following expression.
Q = C . ΔT
where,
C is the heat capacity
ΔT is the change in the temperature
<h3>Hydrogen</h3>
Qc = -Qb = -C . ΔT = -(11.3 kJ/°C) . (14.3°C) = -162 kJ
The heat released per gram of hydrogen is:
<h3>Methane</h3>
Qc = -Qb = -C . ΔT = -(11.3 kJ/°C) . (7.3°C) = -82 kJ
The heat released per gram of methane is:
Answer:
7.41 × 10⁻⁵
Explanation:
Let's consider the basic dissociation reaction of trimethylamine (CH₃)N).
(CH₃)N + H₂O = (CH₃)NH⁺ + OH⁻
According to Brönsted-Lowry, in this reaction (CH₃)N is a base and (CH₃)NH⁺ is its conjugate acid. The pKb for (CH₃)N is 9.87. We can calculate the pKa of (CH₃)NH⁺ using the following expression.
pKa + pKb = 14
pKa = 14 - pKb = 14 - 9.87 = 4.13
Then, we can calculate the acid dissociation constant for (CH₃)NH⁺ using the following expression.
pKa = -log Ka
Ka = antilog - pKa = antilog -4.13 = 7.41 × 10⁻⁵