Given :
Compound A reacts with Compound B to form only one product, Compound C.
The usual percent yield of C in this reaction is 40%.
10.0 g of A are reacted with excess Compound B, and 6.4 g of Compound C
To Find :
The theoretical yield of C.
Solution :
We know, % yield is given by :
Putting given values , we get :
Therefore, theoretical yield of C is 16 g.
Hence, this is the required solution.
Answer:
1. HBr>HCl> H2S >BH3
2.K_a1 very large — H2SO4
K_a1= 1.7 x 10^−2 — H2SO3
K_a1 = 1.7 x 10^−7 — H2S
Explanation:
As one goes down a row in the Periodic Table the properties that determine the acid strength can be observed.
The atoms get larger in radius meaning that in strength, the strength of the bonds get weaker, conversely meaning that the acids get stronger.
For the halogen-containing acids above following the rows and periods, HBr has the strongest bond and is the strongest acid and others follow in this order.
HBr>HCl> H2S >BH3
Acid Dissociation Constant provides us with information known as the ionization constant which comes in handy to measure the acid's strength. The meaning of the proportions are thus, the higher the Ka value, the stronger the acid i.e. it liberates more number of hydrogen ions per mole of acid in solution.
In solution strong acids completely dissociate hence, the value of dissociation constant of strong acids is very high.
Following the cues above on Ka;
K_a1 very large — H2SO4
K_a1= 1.7 x 10^−2 — H2SO3
K_a1 = 1.7 x 10^−7 — H2S
Answer:
0.8 mol.
Explanation:
- The balanced equation for the reaction between Al and FeO is represented as:
<em>2Al + 3FeO → 3Fe + Al₂O₃,</em>
It is clear that 2 mol of Al react with 3 mol of FeO to produce 3 mol of Fe and 1 mol of Al₂O₃.
<em><u>Using cross multiplication:</u></em>
2 mol of Al needs → 3 mol of FeO, from stichiometry.
??? mol of Al needs → 1.2 mol of FeO.
∴<em> The no. of moles of Al are needed to react completely with 1.2 mol of FeO </em>= (2 mol)(1.2 mol)/(3 mol) = <em>0.8 mol.</em>
Mg3N2 is Magnesium nitride
The number of moles in 3.612 x 10²⁴ molecules of CaO is 6 moles.
<h3>
Number of moles in the molecules</h3>
The number of moles in 3.612 x 10²⁴ molecules of CaO is calculated as follows;
6.02 x 10²³ molecules = 1 mole
3.612 x 10²⁴ molecules = ?
= (3.612 x 10²⁴ ) / (6.02 x 10²³ )
= 6 moles
Thus, the number of moles in 3.612 x 10²⁴ molecules of CaO is 6 moles.
Learn more about number of moles here: brainly.com/question/15356425
