We will use boiling point formula:
ΔT = i Kb m
when ΔT is the temperature change from the pure solvent's boiling point to the boiling point of the solution = 77.85 °C - 76.5 °C = 1.35
and Kb is the boiling point constant =5.03
and m = molality
i = vant's Hoff factor
so by substitution, we can get the molality:
1.35 = 1 * 5.03 * m
∴ m = 0.27
when molality = moles / mass Kg
0.27 = moles / 0.015Kg
∴ moles = 0.00405 moles
∴ The molar mass = mass / moles
= 2 g / 0.00405 moles
= 493.8 g /mol
For 7A(17) :
Electronic configuration 
So, there are 5 unpaired electrons present in group 7A(17).
<h3>
What are Unpaired Electrons?</h3>
- An unpaired electron is an electron that doesn't form part of an electron pair when it occupies an atom's orbital in chemistry.
- Each of an atom's three atomic orbitals, designated by the quantum numbers n, l, and m, has the capacity to hold a pair of two electrons with opposing spins.
- Unpaired electrons are extremely uncommon in chemistry because an object carrying an unpaired electron is typically quite reactive. This is because the production of electron pairs, whether in the form of a chemical bond or as a lone pair, is frequently energetically advantageous.
- They play a crucial role in describing reaction pathways even though they normally only appear momentarily during a reaction on a thing called a radical in organic chemistry.
To learn more about unpaired electrons with the given link
brainly.com/question/14356000
#SPJ4
Answer:
a - the position of equilibrium lies far to the right, with products being favored.
Explanation:
Answer:
The volume of CO2 produced is 6.0 L (option D)
Explanation:
Step 1: Data given
Volume of oxygen = 3.0 L
Carbon monoxide = CO = in excess
Step 2: The balanced equation
2 CO (g) + O2 (g) → 2 CO2 (g)
Step 3: Calculate moles of O2
1 mol of gas at STP = 22.4 L
3.0 L = 0.134 moles
Step 3: Calculate moles of CO2
For 2 moles CO we need 1 mol of O2 to produce 2 moles of CO2
For 0.134 moles O2 we'll have 2*0.134 = 0.268 moles CO2
Step 4: Calculate volume of CO2
1 mol = 22.4 L
0.268 mol = 22.4 * 0.268 = 6.0 L
The volume of CO2 produced is 6.0 L
