Atoms of Mg = mol * L
mol of Mg =
=
= 2 mol
∴ atoms of Mg = 2 mol * ( 6.02 x 10²³)
= 1.204 x 10²⁴
Bromine 35
Chlorine 17
Flourine 9
Iodine 53
Astatine 85
<span>100 kilo joules
There are several phases that this problem undergoes and the final answer is the sum of all the energy used for each phase.
Phase 1. Heating of solid ethanol until its melting point.
Phase 2. Melting of the ethanol until it's completely liquid.
Phase 3. Heating of the liquid ethanol until it reaches its boiling point.
Phase 4. Boiling the ethanol until it's completely vapor.
To make things more interesting, some of our constant are per gram and some others are per mole. So let's calculate how many moles of ethanol we have.
Atomic weight carbon = 12.0107
Atomic weight hydrogen = 1.00794
Atomic weight oxygen = 15.999
Molar mass ethanol = 2*12.0107 + 6*1.00794 + 15.999 = 46.06804 g/mol
Moles ethanol = 75g / 46.06804 g/mol = 1.628026719 mol
Phase 1. Use the specific heat of solid ethanol and multiply by the number of degrees we need to change by the mass we have. So
0.97 J/g*K * 75 g * (-114c - -120c)
= 0.97 J/g*K * 75 g * 6K
= 436.5 J
Phase 2: Time to melt. Just need the moles and the enthalpy of fusion. So:
1.628026719 mol * 5.02 kJ/mol = 8.172694128 kJ
Phase 3: Heat to boiling. Just like heating to melting, just a different specific heat and temperature
2.3J/g*K * 75g * (78c - -114c)
= 2.3J/g*K * 75g * 192 K
= 33120 J
Phase 4: Boil it to vapor. Need moles and enthalpy of vaporization. So
1.628026719 mol * 38.56 kJ/mol = 62.77671027 kJ
Now let's add them together:
436.5 J + 8.172694128 kJ + 33120 J + 62.77671027 kJ
= 0.4365 kJ + 8.172694128 kJ + 33.120 kJ + 62.77671027 kJ
=104.5059044 kJ
Since the least precise datum we have is 2 significant figures, round the result to 2 significant figures, giving 100 kilo joules.</span>
Which is the correct order of ease of carbon dioxide production by heating the group II metal carbonates?
A) 
B) 
C) 
D) 
Answer:
Explanation:
The correct option is A:
<h2>
</h2>

On the periodic table, carbonates required stronger heating as one does down the groupings in order for them to decompose. The implication is that the stability of compounds will increase from
to
.
So the carbonates will be stable according to the following order:
, thus making B the correct answer.
Cheers
Its the other way around actually. Benzoic acid is stronger than acetic acid because the electron-donating inductive effect (+I) by the alkyl group on acetic acid destabilise the conjugate base of acetic acid.
hope this helps :)