Answer:
82.0 mL
Explanation:
Step 1: Given data
- Concentration of concentrated acid (C₁): 12.2 M
- Volume of concentrated acid (V₁): ?
- Concentration of dilute acid (C₂): 1.00 M
- Volume of dilute acid (V₂): 1.00 L
Step 2: Calculate the required volume of the concentrated acid
We want to prepare a dilute solution from a concentrated one. We can calculate the volume of the concentrated acid using the dilution rule.
C₁ × V₁ = C₂ × V₂
V₁ = C₂ × V₂ / C₁
V₁ = 1.00 M × 1.00 L / 12.2 M = 0.0820 L = 82.0 mL
If you have 58.93g of Co it means that you only have 1 mol (use a periodic table to find the answer, if you had more find it by proportion, it's easier).
There's 6.022 x 10^23 atoms per mol so you have 6.022 x 10^23 atoms of Co.
(once again if you had more mol, you could find the answer by proportions).
Answer:
b) 2.0 mol
Explanation:
Given data:
Number of moles of Ca needed = ?
Number of moles of water present = 4.0 mol
Solution:
Chemical equation:
Ca + 2H₂O → Ca(OH)₂ + H₂
now we will compare the moles of Ca and H₂O .
H₂O : Ca
2 : 1
4.0 : 1/2×4.0 = 2.0 mol
Thus, 2 moles of Ca are needed.
First, we need to get the value of Ka:
when Ka = Kw / Kb
we have Kb = 1.8 x 10^-5
and Kw = 3.99 x 10^-16 so, by substitution:
Ka = (3.99 x 10^-16) / (1.8 x 10^-5) = 2.2 x 10^-11
by using the ICE table :
NH4+ + H2O →NH3 + H+
intial 0.013 0 0
change -X +X +X
Equ (0.013-X) X X
when Ka = [NH3][H+] / [NH4+]
by substitution:
2.2 x 10^-11 = X^2 / (0.013 - X) by solving this equation for X
∴X = 5.35 x 10^-7
∴[H+] = X = 5.35 x 10^-7
∴PH = - ㏒[H+]
= -㏒(5.35 x 10^-7)
= 6.27
Answer:
B. Two electrons
Explanation:
Since the S orbital is closest to the nucleus and it only has one orbital, it can only hold 2 electrons
