Answer:
ΔH°f P4O10(s) = - 3115.795 KJ/mol
Explanation:
- P4O10(s) + 6H2O(l) ↔ 4H3PO4(aq)
- ΔH°rxn = ∑νiΔH°fi
∴ ΔH°rxn = - 327.2 KJ
∴ ΔH°f H2O(l) = - 285.84 KJ/mol
∴ ΔH°F H3PO4(aq) = - 1289.5088 KJ/mol
⇒ ΔH°rxn = (4)(- 1289.5088) - (6)(- 285.84) - ΔH°f P4O10(s) = - 327.2 KJ
⇒ ΔH°f P4O10(s) = - 5158.035 + 1715.04 + 327.2
⇒ ΔH°f P4O10(s) = - 3115.795 KJ/mol
Answer:
I don't know and I don't understand
Explanation:
I don't know and I don't understand
sorry
sorry for not being able to answer your question
<u>Answer:</u> The number of electrons for n = 0, 1 and 2 are 2, 6 and 10 respectively.
<u>Explanation:</u>
Huckel's rule is used to determine the aromaticity in a compound. The number of delocalized
electrons are calculated by using the equation:

where,
n = 0 or any whole number
- Calculating the value of electrons for n = 0
Putting values in above equation, we get:

- Calculating the value of electrons for n = 1
Putting values in above equation, we get:

- Calculating the value of electrons for n = 2
Putting values in above equation, we get:

Hence, the number of electrons for n = 0, 1 and 2 are 2, 6 and 10 respectively.
Answer:
Empirical formula: CH₃O
Empirical formula mass = 31 g/mol
Explanation:
Data Given:
Molecular Formula = C₁₀H₃₀O₁₀
Empirical Formula = ?
Empirical Formula mass =
Solution
Empirical Formula:
Empirical formula is the simplest ration of atoms in the molecule but not all numbers of atoms in a compound.
So,
The ratio of the molecular formula should be divided by whole number to get the simplest ratio of molecule
As
C₁₀H₃₀O₁₀ Consist of 10 Carbon (C) atoms, 30 Hydrogen (H) atoms, and 10 Oxygen (O) atoms.
Now
Look at the ratio of these three atoms in the compound
C : H : O
10 : 30 : 10
Divide the ratio by two to get simplest ratio
C : H : O
10/10 : 30/10 : 10/10
1 : 3 : 1
So for the empirical formula the simplest ratio of carbon to hydrogen to oxygen is 1:3:1
So the empirical formula will be
Empirical formula of C₁₀H₃₀O₁₀ = CH₃O
Now
To find the empirical formula mass in g/mol
Formula mass:
Formula mass is the total sum of the atomic masses of all the atoms present in a formula unit.
**Note:
if we represent the molar mass of the empirical formula for one mol in grams then it is written as g/mol
So,
As the empirical formula of C₁₀H₃₀O₁₀ is CH₃O
Then Its empirical formula mass will be
CH₃O
Atomic Mass of C = 12
Atomic Mass of H = 3
Atomic Mass of O = 16
Total Molar mass of CH₃O
CH₃O = 12 + 3(1) + 16
CH₃O = 12 + 3 + 16
CH₃O = 31 g/mol
Answer:
3.6 × 10²⁴ molecules
Explanation:
Step 1: Given data
Moles of methane (n): 6.0 moles
Step 2: Calculate the number of molecules of methane in 6.0 moles of methane
In order to convert moles to molecules, we need a conversion factor. In this case, we will use Avogadro's number: there are 6.02 × 10²³ molecules of methane in 1 mole of molecules of methane.
6.0 mol × 6.02 × 10²³ molecules/1 mol = 3.6 × 10²⁴ molecules