Answer:
HOFO = (0, 0, +1, -1)
Explanation:
The formal charge (FC) can be calculated using the following equation:
<u>Where:</u>
V: are the valence electrons
N: are the nonbonding electrons
B: are the bonding electrons
The arrange of the atoms in the oxyacid is:
H - O₁ - F - O₂
Hence, the formal charge (FC) on each of the atoms is:
H: FC = 1 - 0 - 1/2*(2) = 0
O₁: FC = 6 - 4 - 1/2*(4) = 0
F: FC = 7 - 4 - 1/2*(4) = +1
O₂: FC = 6 - 6 - 1/2*(2) = -1
We can see that the negative charge is in the oxygen instead of the most electronegative element, which is the F. This oxyacid is atypical.
I hope it helps you!
Your Question: {How many objects are in a mole?}
Helpful Knowledge: (We Know the amount in an object: 12g or C^12)
{A number of objects that are in a mole of objects?}
Well for the question it is pretty easy to answer because a number of objects in One mole would equal 6.02 × 10²³
Which 6.02 × 10²³ is an Avogadro's Number.
So it depends on how many objects you have.
So for every object you have, One mole would equal 6.02 × 10²³. Or 62,000,000,000,000,0000,000,000. Big Number am I right. So that's why we just use 6.02 × 10²³.
Anywho, your answer would be 6.02 x 10²³ x n.
N would equal the number of objects you're calculating.
Final Answer: 6.02 x 10²³ x (n) = (Your Answer)
Hope this helps! Have a great day. If you need anything else, feel free to hope right in my inbox. Or comment below. ↓
What are you asking? Id love to help but cant without proper information
The acid dissociation constant of benzoic acid is 6.5 x 10^-5. Therefore, the pH of the benzoic acid solution prior to adding sodium benzoate is:
pH = -log[Ka]
pH = -log (6.5 x 10^-5)
pH = 4.19
The pH of the benzoic acid solution is 4.19 which is acidic, but a weak acid.
First, we have to get:
1- The heat required to increase T of ice from -50 to 0 °C:
according to q formula:
q1 = m*C*ΔT
when m is the mass of ice = mol * molar mass
= 1 mol * 18 mol/g
= 18 g
and C is the specific heat capacity of ice = 2.09 J/g-K
and ΔT change in temperature = 0- (-50) = 50°C
by substitution:
∴q1 = 18 g * 2.09 J/g-K *50°C
= 1881 J = 1.881 KJ
2- the heat required to melt this mass of ice is :
q2 = n*ΔHfus
when n is the number of moles of ice = 1 mol
and ΔHfus = 6.01 KJ/mol
by substitution:
q2 = 1 mol * 6.01 KJ/mol
= 6.01 KJ
3- the heat required to increase the water temperature from 0°C to 60 °C is:
q3 = m*C*ΔT
when m is the mass of water = 18 g
C is the specific heat capacity of water = 4.18 J/g-K
ΔT is the change of Temperature of water = 60°C - 0°C = 60°C
by substitution:
∴q3 = 18 g * 4.18 J/g-K * 60°C
= 4514 J = 4.514 KJ
∴the total change of enthalpy = q1+q2+q3
= 1.881 KJ +6.01 KJ + 4.514 KJ
= 12.405 KJ
