Explanation:
(a) potassium oxide with water
According to reaction,1 mole of potassium oxide reacts with 1 mole of water to give 1 mole of potassium hydroxide.
(b) diphosphorus trioxide with water
According to reaction,1 mole of diphosphorus trioxide reacts with 2 moles of water to give 2 moles of phosphorus acid.
(c) chromium(III) oxide with dilute hydrochloric acid,
According to reaction,1 mole of chromium(III) oxide reacts with 6 moles of hydrochloric acid to give 2 moles of chromium(III) chloride and 3 moles of water.
(d) selenium dioxide with aqueous potassium hydroxide
According to reaction,1 mole of selenium dioxide reacts with 2 moles of potassium hydroxide to give 1 mole of potassium selenite and 1 mole of water.
Answer:
Density is mass over volume.
Explanation:
We are given the chemical reaction and the amount of reactant used for the process. We use these data together to obtain what is asked. We do as as follows:
0.882 mol H2O2 ( 1 mol O2 / 2 mol H2O2 ) = 0.441 mol O2 produced
Hope this answers the question.
Answer: Heterogenous mixtures are the ones in which we can see the different components clearly. Air is considered to be homogenous, because we don't see the different noble gases' particles floating around. Salt water is also homogenous, because salt dissolves almost completely into water, becoming hard to see without the proper equipment salt particles in it. Steel is also homogenous, since the metals that compose it are mixed during the heating. The only heterogenous would be soil, because we can see different particles in it, with different sizes, shapes, colors, etc without having to use special equipment.
<span>The choices are as follows:
h2o + 2o2 = h2o2
fe2o3 + 3h2 = 2fe + 3h2o
al + 3br2 = albr3
caco3 = </span><span>cao + co2
The correct answers would be the second and the last option. The equations that are correctly balanced are:
</span> fe2o3 + 3h2 = 2fe + 3h2o
caco3 = cao + co2
To balance, it should be that the number of atoms of each element in the reactant and the product side is equal.
