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.
C
The elements are Lithium and Beryllium— just look at the mass number on the periodic table
Answer:
Let the mixture is X% by mass of CuSO
4
.5H
2
O and 100 - X % by mass of MgSO
4
.7H
2
O. 5.0 g of mixture will contain 0.05X g CuSO
4
.5H
2
O and 5.0 - 0.05X g MgSO
4
.7H
2
O
The molar masses of CuSO
4
.5H
2
O and MgSO
4
.7H
2
O are 249.7 g/mol and 246.5 g/mol respectively.
The number of moles of CuSO
4
.5H
2
O=
249.7
0.05X
=2.00×10
−4
X moles.
Explanation:
Pls mark it as branliest answere thanks
Answer:
A hydrocarbon containing a carbon - carbon double bond.
Explanation:
Alkene is hydrocarbon containing a
carbon - carbon double bond.
( Refer the attachment to understand more clearly )
Answer:
89.88 g
Explanation:
Atomic Mass of Ar: 39.948
Mass = moles * AM
Replacing moles = 2.25 and AM = 39.948 you get the mass of Ar:
Mass = 2.25 * 39.948
Mass = 89.88 g
