Answer:
It is true answer. Ozone = O3
= 24.3
The average atomic mass of X is the <em>weighted average</em> of the atomic masses of its isotopes.
We multiply the atomic mass of each isotope by a number representing its <em>relative importance</em> (i.e., its % abundance).
Thus,
0.790 × 24 u = 18.96 u
0.100 × 25 u = 2.50 u
0.110 × 26 u = <u>2.86 u</u>
TOTAL = 24.3 u
∴ The relative atomic mass of X is 24.3.
Answer:
1.634 molL-1
Explanation:
The mol ration between NH3 and HCl is 1 : 1
Using Ca Va / Cb Vb = Na / Nb where a = acid and b = base
Na = 1
Nb = 1
Ca = 0.208 molL-1
Cb = ?
Va = 19.64 mL
Vb = 25.00mL
Solving for Cb
Cb = Ca Va / Vb
Cb = 0.208 * 19.64 / 25.0
Cb = 0.1634 molL-1 (Concentration of diluted ammonia solution)
Using the dilution equation;
C1V1 = C2V2
Initial Concentration, C1 = ?
Initial Volume, V1 = 25.00 mL
Final Volume, V2 = 250 mL
Final Concentration, C2 = 0.1634 molL-1
Solving for C1;
C1 = C2 * V2 / V1
C1 = 0.1634 * 250 / 25.00
C1 = 1.634 molL-1
Answer:
O₂; KCl; 33.3
Explanation:
We are given the moles of two reactants, so this is a limiting reactant problem.
We know that we will need moles, so, lets assemble all the data in one place.
2KCl + 3O₂ ⟶ 2KClO₃
n/mol: 100.0 100.0
1. Identify the limiting reactant
(a) Calculate the moles of KClO₃ that can be formed from each reactant
(i)From KCl
(ii) From O₂
O₂ is the limiting reactant, because it forms fewer moles of the KClO₃.
KClO₃ is the excess reactant.
2. Moles of KCl left over
(a) Moles of KCl used
(b) Moles of KCl left over
n = 100.0 mol - 66.67 mol = 33.3 mol
Answer:
P2= 125.26 Kpa
Explanation:
V1= 489.6 ml=0.4896L
V2= 750 ml= 0.750L
V1= 180 KPa= 180000 Pa
P2= ?
T1= 10 = 10 + 273.15 = 283.15K
T2= 28.7+273.15= 301.85K
180000Pa* 0.4896L/ 283.15K * 301.85K/0.75L
P2= 12526.28553
P2= 125.26 KPa