Answer:
59.8%
Explanation:
First find the Mr of manganese (III) nitrate.
Mr of Mn(NO₃)₃ = 54.9 + (14 × 3) + (16 × 3 × 3) = <u>240.9</u>
Since we have to find the percentage composition of oxygen, we need to find the Mr of oxygen in the compound, which is:
Mr of (O₃)₃ = (16 × 3) × 3 = <u>144</u>
Now we can find percentage composition / percentage by mass of oxygen.
% composition = 
× 100
% composition = 
× 100 = <u>59.776%</u>
∴ % compostion of oxygen in maganese(III)nitrate is 59.8% (to 3 significant figures).
Answer:
λ = 5.68×10⁻⁷ m
Explanation:
Given data:
Energy of photon = 3.50 ×10⁻¹⁹ J
Wavelength of photon = ?
Solution:
E = hc/λ
h = planck's constant = 6.63×10⁻³⁴ Js
c = 3×10⁸ m/s
Now we will put the values in formula.
3.50 ×10⁻¹⁹ J = 6.63×10⁻³⁴ Js × 3×10⁸ m/s/ λ
λ = 6.63×10⁻³⁴ Js × 3×10⁸ m/s / 3.50 ×10⁻¹⁹ J
λ = 19.89×10⁻²⁶ J.m / 3.50 ×10⁻¹⁹ J
λ = 5.68×10⁻⁷ m
Na₂CrO₄ + PbCl₂ → PbCrO₄ + 2 NaCl
<u>Explanation:</u>
In a double displacement reaction, the reactants which are involved in the reaction exchanging their ions thereby produces 2 new compounds. Here sodium chromate and lead chloride are undergoing double displacement reaction, the ions exchanges their position there by forming sodium chloride and lead chromate. So the double displacement reaction is given as,
Na₂CrO₄ + PbCl₂ → PbCrO₄ + 2 NaCl
Answer:
Halogens always form anions, alkali metals and alkaline earth metals always form cations. Most other metals form cations (e.g. iron, silver, nickel), whilst most other nonmetals typically form anions (e.g. oxygen, carbon, sulfur).
Explanation:
Examples: Sodium (Na+), Iron (Fe2+), Ammonium (NH4
I believe the correct answer is 11 g
