Answer:
Those with 3 protons and 3 neutrons are known as the isotope lithium-6. Those with 3 protons and 4 neutrons are known as the isotope lithium-7.
Explanation:
Answer:-
The reaction of 2-bromopropane reacts with sodium iodide in acetone is an example of Sn2 reaction.
The I - attacks from backside to give the transition state for both.
If we compare the transition state for cyclobromopropane 2-bromopropane then we see in case of cyclobromopropane transition state, one of the H is very close to the incoming I -.
This results in steric strain and less stability of the transition state. Hence 2-bromopropane reacts with sodium iodide in acetone over 104 times faster than bromocyclopropane.
Answer:
=154.8 J
Explanation:
The rise in temperature is contributed by the change in temperature.
Change in enthalpy = MC∅, where M is the mass of the substance, C is the specific heat capacity and ∅ is the change in temperature.
Change in temperature = 100.0°C-20.0°C=80°C
ΔH=MC∅
The specific heat capacity of gold= 0.129 J/g°C
ΔH= 15.0g×0.129J/g°C×80°C
=154.8 J
To solve this problem, we must assume ideal gas behaviour so
that we can use Graham’s law:
vA / vB = sqrt (MW_B / MW_A)
where,
<span>vA = speed of diffusion of A (HBR)</span>
vB = speed of diffusion of B (unknown)
MW_B = molecular weight of B (unkown)
MW_A = molar weight of HBr = 80.91 amu
We know from the given that:
vA / vB = 1 / 1.49
So,
1/1.49 = sqrt (MW_B / 80.91)
MW_B = 36.44 g/mol
Since this unknown is also hydrogen halide, therefore this
must be in the form of HX.
HX = 36.44 g/mol , therefore:
x = 35.44 g/mol
From the Periodic Table, Chlorine (Cl) has a molar mass of
35.44 g/mol. Therefore the hydrogen halide is:
HCl
