Answer:2.86x10^-7m
Explanation:E=hc/^
E=6.94x10^-19J
c = 2.9979x10^8m/s
h= 6.626x10^-34Js
^ =( 6.626x10^-34)x( 2.9979x 10^8)/ 6.94x10^-19
= 2.86x10^-7m
Given reaction represents dissociation of bromine gas to form bromine atoms
Br2(g) ↔ 2Br(g)
The enthalpy of the above reaction is given as:
ΔH = ∑n(products)Δ
- ∑n(reactants)Δ
where n = number of moles
Δ
= enthalpy of formation
ΔH = [2*ΔH(Br(g)) - ΔH(Br2(g))] = 2*111.9 - 30.9 = 192.9 kJ/mol
Thus, enthalpy of dissociation is the bond energy of Br-Br = 192.9 kJ/mol
1. What do they have in common?
As mentioned in the problem, these gases are present in equal amounts. So, that would infer that they are common in terms of their mass. Also, it is specified that the temperature is 25°C. Connected to that is the average kinetic energy, which is directly proportional. Hence, they are also common in temperature and average kinetic energy.
2. What are the differences?
They differ in type, of course. Also, they differ in average velocities which is a factor of temperature of molar mass. Since they are 3 different types of gases with different molar masses, they would also differ in their average velocities.
Answer:
1.123x10⁻⁴ moles of alanine
Explanation:
In order to convert grams of alanine into moles, <em>we need to know its molecular weight</em>:
The formula for alanine is C₃H₇NO₂, meaning <u>its molecular weight would be</u>:
- 12*3 + 7*1 + 14 + 16*2 = 89 g/mol
Then we <u>divide the sample mass by the molecular weight</u>, to do the conversion:
- 1.0x10⁻² g ÷ 89 g/mol = 1.123x10⁻⁴ moles
