Answer:
-5.51 kJ/mol
Explanation:
Step 1: Calculate the heat required to heat the water.
We use the following expression.
where,
- c: specific heat capacity
- m: mass
- ΔT: change in the temperature
The average density of water is 1 g/mL, so 75.0 mL ≅ 75.0 g.
Step 2: Calculate the heat released by the methane
According to the law of conservation of energy, the sum of the heat released by the combustion of methane (Qc) and the heat absorbed by the water (Qw) is zero
Qc + Qw = 0
Qc = -Qw = -22.0 kJ
Step 3: Calculate the molar heat of combustion of methane.
The molar mass of methane is 16.04 g/mol. We use this data to find the molar heat of combustion of methane, considering that 22.0 kJ are released by the combustion of 64.00 g of methane.
Answer:
Explanation:
An electron in 4s is farther away from nucleus and it has higher energy when compared to electron from 1s.
Ca(OH)₂: strong base
pOH = a . M
a = valence ( amount of OH⁻)
M = concentration
Ca(OH)₂ ⇒ Ca²⁺ + 2OH⁻ (2 valence)
so:
pOH = 2 x 0.005
pOH = 0.01
pH = 14 - 0.01 = 13.99
Answer:
Mn2O3
Explanation:
Manga has a 3+ charge and oxygen has a 2- charge so to balance the charges there needs to be 3 oxygens for every 2 manga
Answer:
The sp³ orbital hybridization has 25% s-character and 75% p-character.
Explanation:
Hybridization refers to the process of mixing of a given number atomic orbitals to form equal number of new hybrid orbitals. The three types of hybrid orbitals are sp³, sp² and sp.
The<u> four sp³ hybridized orbitals</u> are formed by the mixing of <u>one s orbital and three p orbitals.</u>
Thus s-character in a sp³ hybridized orbital = (1÷4)×100 = 25%
and p-character in a sp³ hybridized orbital = (3÷4)×100 = 75%
<u>Therefore, the sp³ orbital hybridization has 25% s-character and 75% p-character.</u>
