Answer:
A
C
A
Explanation:
Iam not sure if it's correct
I am pretty sure the answer is 1.5*10^25
Answer:
a) a space in an atom where an electron is most likely to be found
Explanation:
Atomic orbital is the mathematical function which describes wave-like behavior of the electrons present in the atom.
It is used to calculate probability of finding the electron of the atom in any region around nucleus of the atom. Atomic orbital is the physical region or a three dimensional space where the probability of finding the electron is more than 90% which also means that the space where an electron is most likely to be found. They are boundless space and have definite energy.
Option A best describes atomic orbital.
Answer:
Molarity of solution is 1.10x10⁻³ M
Explanation:
Solute NaOCl
7.4% by mass means, that in 100 grams of solution, we have 7.4 g of solute.
Molar mass of NaOCl = 74.45 g/m
Mol = Mass / Molar mass
7.4 g / 74.45 g/m = 0.099 moles
Density of solution = 1.12 g/mL
Density = Mass / volume
1.12g/mL = 100 g / volume
Volume = 100 g / 1.12g/mL = 89.3 mL
Molarity = mol /L
89.3 mL = 0.0893 L
0.099 moles / 0.0893 L = 1.10x10⁻³ M
Answer:
66 g of CO₂
Solution:
The Balance Chemical Reaction is as follow,
C₂H₂ + 5/2 O₂ → 2 CO₂ + H₂O
Or,
2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O ------- (1)
Step 1: Find out the limiting reagent as;
According to Equation 1,
56.1 g (2 mole) C₂H₂ reacts with = 160 g (5 moles) of O₂
So,
125 g of C₂H₂ will react with = X g of O₂
Solving for X,
X = (125 g × 160 g) ÷ 56.1 g
X = 356.5 g of O₂
It means for total combustion of Ethylene we require 356.5 g of O₂, but we are only provided with 60.0 g of O₂. Therefore, O₂ is the limiting reagent and will control the yield.
Step 2: Calculate Amount of CO₂ produced as;
According to Equation 1,
160 g (5 mole) O₂ produces = 176 g (4 moles) of CO₂
So,
60.0 g of O₂ will produce = X g of CO₂
Solving for X,
X = (60.0 g × 176 g) ÷ 160 g
X = 66 g of CO₂
