Answer:
The atoms of noble gases already have complete outer shells, so they have no tendency to lose, gain, or share electrons. This is why the noble gases are inert and do not take part in chemical reactions. The table summarises the electronic configurations of elements in groups 1, 7 and 0.
Explanation:
The atoms of noble gases already have complete outer shells, so they have no tendency to lose, gain, or share electrons. This is why the noble gases are inert and do not take part in chemical reactions. The table summarises the electronic configurations of elements in groups 1, 7 and 0.
Answer:
35 amu
Explanation:
In an atom, only masses of protons and neutrons are relevant. electrons are so small in size that their masses are negligible.
The mass of 1 proton/neutron is 1amu.
HCl + NaHCO3 = NaCl + H2O + CO2
We have 0.033*0.2 = 0.0066 mol of Hcl
According to reaction we need the same amount of NaHCO3
M(NaHCO3) = 23+1+12+48=84g/mold
m = 0.0066mol * 84g/mol = 0.5544g
Answer:
The greenhouse effect is the process by which radiation from a planet's atmosphere warms the planet's surface to a temperature above what it would be without this atmosphere. Radiatively active gases in a planet's atmosphere radiate energy in all directions.
Answer:
The right answer is:
Replacing the powdered lead oxide with its large crystals
Removing lead (IV) oxide from the reaction mixture
Using 1.0 gram of lead (IV) oxide
Explanation:
Based on the given information this reaction is the catalytic decomposition of H₂O₂ into water and oxygen using Lead (IV) oxide as a catalyst.
- The catalyst surface area is directly proportional to the reaction rate
- So, Replacing the powdered lead oxide with its large crystals would decrease the reaction rate due to the has larger surface area than its large crystals.
2. Also, Removing lead (IV) oxide from the reaction mixture the reaction rate decreased because as the catalyst is removed.
3. Using 50 cm³ of hydrogen peroxide doesn't affect the rate because the concentration of the reactant doesn't change.
4. Using 1.0 gram of lead (IV) oxide would decrease the reaction rate because the amount of catalyst decreased
So, The right answer is:
Replacing the powdered lead oxide with its large crystals
Removing lead (IV) oxide from the reaction mixture
Using 1.0 gram of lead (IV) oxide
