Answer:
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
The proton transfer reaction between Cyanide and water can be written as; X^- + H2O -----> HX + OH^-
<h3>What is a proton transfer reaction?</h3>
A proton transfer reaction is one in which a proton is moved from one chemical specie to another.It is in fact and acid - base reaction in the Brownstead - Lowry sense.
The proton transfer reaction between Cyanide and water can be written as(Let the cyanide ion be shown as X);
X^- + H2O -----> HX + OH^-
Learn more about proton transfer: brainly.com/question/861100?
The molality is 0.54 M when 1.34 moles of NaCl is present in 2.47 kg of solvent.
<u>Explanation:</u>
Molality is the measure of how much of amount of solute is dissolved in the solvent. So it is calculated as the ratio of moles of solute to the grams of solvent.
As in this case, the solute is NaCl and solvent is unknown. So the moles of solute is given as 1.34 moles and the mass of solvent is given as 2.47 kg.
Hence,
Thus, the molality is 0.54 M when 1.34 moles of NaCl is present in 2.47 kg of solvent.