Answer:
both
Explanation:
id say that it could occur but also not as much. the moon would be smaller and further from the earth to where we would barely be able to see it. if the full moon is barely visible then im sure the total solar eclipse wouldn't be as noticeable as it is now. but thats just my opinion
Answer:
Fe(CN)₂, FeCO₃, Pb(CN)₄, Pb(CO₃)₂
Explanation:
Cations (positively charged ions) can only form ionic bonds with anions (negatively charged ions). However, you can't just simply put one cation and one anion together to form a compound. Each compound needs to been neutral, or have an overall charge of 0. When cations and anions do not have charges that perfectly cancel, you need to modify the amount of each ion in the compound.
1.) Fe(CN)₂
-----> Fe²⁺ and CN⁻
-----> +2 + (-1) + (-1) = 0
2.) FeCO₃
-----> Fe²⁺ and CO₃²⁻
-----> +2 + (-2) = 0
3.) Pb(CN)₄
-----> Pb⁴⁺ and CN⁻
-----> +4 + (-1) + (-1) + (-1) + (-1) = 0
4.) Pb(CO₃)₂
-----> Pb⁴⁺ and CO₃²⁻
-----> +4 +(-2) + (-2) = 0
Volume of the metal = change in volume reading
Volume = 37.4 - 33
Volume = 4.4 ml
Density = mass / volume
Density = 7.101 / 4.4
Density = 1.61 g/ml
Answer:
25.35%
Explanation:
Again let me restate the the equation of the reaction;
H2O (ℓ) + 2 MnO4 - (aq) + 3 CN- (aq) → 2 MnO2 (s) + 3 CNO- (aq) + 2 OH- (aq)
Amount of potassium permanganate reacted = 10.2/1000 * 0.08035 = 8.1957 * 10^-4 moles
If 2 moles of MnO4 - reacts with 3 moles of CN-
8.1957 * 10^-4 moles of MnO4 - reacts with 8.1957 * 10^-4 * 3/2
= 1.229 * 10^-3 moles of CN-
Mass of CN- reacted = 1.229 * 10^-3 moles of CN- * 26.02 g/mol
= 0.03 g
Hence, percentage of the cyanide = 0.03 g/0.1183 g * 100
= 25.35%
