in ancient times, people didn't have electronic or even effective weight scales. so to weigh things like gold or precious metals to sell was near impossible. the water displacement weighing method (created by Greek mathmetition Archemaidies) allowed people to effectively get an approximated guess of how much the metals were worth... That's a mouthful!
The dissolution of borax in water is a temperature dependent reaction. With the higher temperature, the salt dissolve quickly.
<h3>What is borax?</h3>
Borax is the hydrate salt of boric acid. It is white and widely used in cleaning and in laundry detergent.
Borax is a salt that will dissolve in water at almost any temperature, with the exception of steam and ice.
However, as with any salt, the higher the temperature, the faster the salt dissolves, so speed is dependent on temperature. It will dissolve in cold water, but it will take longer.
Thus, the dissolution of borax in water is a temperature dependent reaction.
Learn more about borax
brainly.com/question/14724418
#SPJ4
We are given with the initial concentrations of both reactants, nitrous oxide and chlorine of 0.16 M and the value of the rate constant equal to M. The initial rate constant is required in this problem. The rate is expressed in k NO^2 *Cl2. Substituting 0.16 to both components, the initial rate is 0.12288.
Answer:
It is a technique of collection and storage of rain water
- <u>The reaction that takes place is:</u>
Hg(NO₃)₂(ac) + Na₂S(ac) → HgS(s) + 2Na⁺ + 2NO₃⁻
Now we calculate the moles of each reagent -using the molecular weights-, in order to determine the limiting reactant:
- Moles of mercury (II) nitrate = 85.14 g * =0.2622 moles.
- Moles of sodium sulfide = 14.334 g *=0.1837 moles.
Because the stoichiometric ratio between the reactants is 1:1, we compare the number of moles of each one upfront.
moles Hg(NO₃)₂ > moles Na₂S
<u>Thus Na₂S is the limiting reagent.</u>
So in order to find the mass of solid precipitate, we must calculate it using the moles of Na₂S:
The mass of the solid precipitate is 42.760 g.
- In order to calculate the grams of the reactant in excess that will remain after the reaction, we convert the moles that reacted into mass and substract them from the original mass:
Mass of Hg(NO₃)₂ remaining =
The mass of the remaning reactant in excess is 25.49 g.
- Because we assume complete precipitation, there are no more Hg⁺² or S⁻² ions in solution. The moles of NO₃⁻ and Na⁺ in solution remain the same during the reaction, so the number is calculated from the number added in the reactant:
Hg⁺²: 0 mol
NO₃⁻:
Na⁺:
S²⁻: 0 mol