Decomposition of a potassium superoxide happens according to the scheme:
4KO₂(s) + 2H₂O(l) → 4KOH(aq) + 3O₂(g)
m(K₂O)=4M(K₂O)m(O₂)/{3M(O₂)}
m(K₂O)=4×71.1×6.5/{3×32.0}≈19.3 g
Answer:
Depends, but in most cases, 2.
It's best to use as many digits as possible to keep it accurate.
Explanation:
This varies between teachers, as most schools go with 2 decimal places.
This is something that depends in your situation.
You technically want as many decimals as possible to keep it as accurate, but most people stick with 2.
I personally do 3, and commonly do 5 sometimes.
The reaction of sugar with oxygen is as follows:
C₁₂H₂₂O₁₁ + 12 O₂ → 12 CO₂ + 11 H₂O
When in the presence of pure oxygen, sucrose (table sugar) will not form caramel, in fact it will combust into carbon dioxide and water like any other carbohydrate.
Caramel is actually formed by slowly heating sucrose to high temperatures of around 170 °C resulting in thermal decomposition. This essentially removes molecules of water from the sucrose which results in the compound isomerizing and eventually polymerizing to form caramel. The chemical make up of caramel is the same as sucrose, so it will still be composed of carbon, hydrogen and oxygen.
Answer:
You cannot make observations if you are 57 seconds late into the lab.
Explanation:
The atomic nucleus can split by decay into 2 or more particles as a result of the instability of its atomic nucleus due to the fact that radioactive elements possess an unstable atomic nucleus.
Now, the primary particles which are emitted by radioactive elements in order to make them decay are alpha, beta & gamma particles.
The half life equation is;
N_t = N₀(½)^(t/t_½)
Where:
t = duration of decay
t_½ = half-life
N₀ = number of radioactive atoms initially
N_t = number of radioactive atoms remaining after decay over time t
We are given;
t = 57 secs
N₀ = 100 g
Now, half life of Nitrogen-16 from online sources is 7.2 seconds. t_½ = 7.2
Thus;
N_t = 100(1/2)^(57/7.2)
N_t = 0.4139g
We are told that In order to make observations, you require at least .5g of material.
The value of N_t you got is less than 0.5g, therefore you cannot make observations if you are 57 seconds late.
Answer:
0.1127 grams of Mg is used to produce one gram of silver
Explanation:
Write a balanced equation
Mg(s) +2Ag + (aq) -->
(aq) + 2Ag(s)
The atomic mass of magnesium is 24.31 g/mol and the atomic mass of silver is 107.84g/mol
1 g Ag of silver is produced from [(1 mol Mg)(24.31 g/mol)] /[ (2mol Ag)(107.84g/mol)]
1 g Ag of silver is produced from 0.1127 grams of Mg
0.1127 grams of Mg is used to produce one gram of silver