Answer:
<h2>9.03 × 10²³ molecules</h2>
Explanation:
The number of molecules can be found by using the formula
N = n × L
where n is the number of moles
N is the number of entities
L is the Avogadro's constant which is
6.02 × 10²³ entities
From the question we have
N = 1.5 × 6.02 × 10²³
We have the final answer as
<h3>9.03 × 10²³ molecules</h3>
Hope this helps you
Answer:
Explanation:
3: Given data:
Number of moles of strontium nitrate = 3.00×10⁻³ mol
Number of atoms = ?
Solution:
There are 9 moles of atoms in 1 mole of Sr(NO₃)₂.
In 3.00×10⁻³ moles,
9 mol × 3.00×10⁻³
27.00×10⁻³ mol
Number of atoms in 3.00×10⁻³ mol of Sr(NO₃)₂:
27.00×10⁻³ mol × 6.022×10²³ atoms / 1mol
162.59×10²⁰ atoms
4)Given data:
Mass of calcium hydroxide = 4500 Kg (4500/1000 = 4.5 g)
Number of moles = ?
Solution:
Number of moles = mass in g/molar mass
by putting values,
Number of moles = 4.5 g/ 74.1 g/mol
Number of moles = 0.06 mol
5) Given data:
Number of atoms of silver nitrate = 1.06×10²³
Number of moles = ?
Solution:
1 mole of any substance contain 6.022×10²³ atoms .
1.06×10²³ atoms × 1 mol / 6.022×10²³ atoms
0.176 moles of silver nitrate
Moles KClO₃ = 0.239
<h3>Further explanation</h3>
Given
Reaction
2KClO₃(s) ⇒2KCl(s) + 3O₂(g)
P water = 23.8 mmHg
P tot = 758 mmHg
V = 9.07 L
T = 25 + 273 = 298 K
Required
moles of KClO₃
Solution
P tot = P O₂ + P water
P O₂ = P tot - P water
P O₂ = 758 - 23.8
P O₂ = 734.2 mmHg = 0.966 atm
moles O₂ :
n = PV/RT
n = 0.966 x 9.07 / 0.082 x 298
n = 0.358
From equation, mol ratio KClO₃ : O₂ = 2 : 3, so mol KClO₃ :
= 2/3 x mol O₂
= 2/3 x 0.358
= 0.239
According to the reaction equation:
and by using ICE table:
CN- + H2O ↔ HCN + OH-
initial 0.08 0 0
change -X +X +X
Equ (0.08-X) X X
so from the equilibrium equation, we can get Ka expression
when Ka = [HCN] [OH-]/[CN-]
when Ka = Kw/Kb
= (1 x 10^-14) / (4.9 x 10^-10)
= 2 x 10^-5
So, by substitution:
2 x 10^-5 = X^2 / (0.08 - X)
X= 0.0013
∴ [OH] = X = 0.0013
∴ POH = -㏒[OH]
= -㏒0.0013
= 2.886
∴ PH = 14 - POH
= 14 - 2.886 = 11.11
W=0.360
p=1.18g/mL
pH=2.12
v=14.0 L
M(HCl)=36.46 g/mol
v₀-?
1) pH=-lg[H⁺]
[H⁺]=c(HCl)=10^(-pH)
n(HCl)=v[H⁺]=v*10^(-pH)
2) n(HCl)=m(HCl)/M(HCl)=wv₀p/M(HCl)
3) v*10^(-pH)=wv₀p/M(HCl)
v₀=v*10^(-pH)M(HCl)/(wp)
v₀=14.0*10^(2.12)*36.46/(0.360*1.18)=9.115 mL
Approximately 9.1 mL of concentrated solution required.
