Answer:
Making oxygen
Oxygen can be made from hydrogen peroxide, which decomposes slowly to form water and oxygen:
hydrogen peroxide → water + oxygen
2H2O2(aq) → 2H2O(l) + O2(g)
The rate of reaction can be increased using a catalyst, manganese(IV) oxide. When manganese(IV) oxide is added to hydrogen peroxide, bubbles of oxygen are given off.
Apparatus arranged to measure the volume of gas in a reaction. Reaction mixture is in a flask and gas travels out through a pipe in the top and down into a trough of water. It then bubbles up through a beehive shelf into an upturned glass jar filled with water. The gas collects at the top of the jar, forcing water out into the trough below.
To make oxygen in the laboratory, hydrogen peroxide is poured into a conical flask containing some manganese(IV) oxide. The gas produced is collected in an upside-down gas jar filled with water. As the oxygen collects in the top of the gas jar, it pushes the water out.
Instead of the gas jar and water bath, a gas syringe could be used to collect the oxygen.
litmus paper
Because it will just tell if the solution is acidic or basic, it won't tell the pH
Answer:
potassium hydrogen phthalate KHP MOLAR MASS = 204.233 glmol
to get 1000 ml
Molar concentration = Mass concentration/Molar Mass
mass concentration = molar concentration x molar mass
mass concentration=0.1 M,
molar mass= 204.233 g/mol
so to get 1L
mass conc = 204.233 x 0.1
= 20.4233g for 1L or 1000 ml
to get 6.00 ml
if 20.4233g is for 1000ml
then to 6.00 ml
= 20.4233 x 6 / 1000
= 0.123g for 6.00 ml
according to the equation below
NaOH(aq) + KHC8H4O4(aq) --> KNaC8H4O4(aq) + H2O(l)
number of moles of NaOH is equal to that of KHP
so the same amount will be needed too, which is
= 0.123g
False. An increase in temperature is an exothermic reaction. However, when a temperature decreases this is known as an endothermic reactionz
Answer:
Formula: Na2S2O3
we get solubility.
Divide the mass of the compound by the mass of the solvent and then multiply by 100 g to calculate the solubility in g/100g .
Solution given:
mass of sodium thiosulphate [m1]=25.5g
mass of water [m2]=40g
at temperature [t]=25°C
we have
<u>solubility in g/dm^3</u> :
- =
- =63.75g /litre=63.75g/dm³
<u>solubility in g/dm^3 :63.75g/dm³</u>
<u>n</u><u>o</u><u>w</u>
solubility of the solute in mol/dm^3=:63.75g/dm³/178=0.4 mol/dm³
