Answer:
Since with LiBr no precipitation takes place. So, Ag+ is absent
When we add Li2SO4 to it, precipitation takes place.
Ca2+(aq) + SO42-(aq) ----> CaSO4(s) ...Precipitate
Thus, Ca2+ is present.
When Li3PO4 is added, again precipitation takes place.Reaction is:
Co2+(aq) + PO43-(aq)---->Co3(PO4)2(s) ... Precipitate
A. Ca2+ and Co2+ are present in solution
B. Ca2+(aq) + SO42-(aq) ----> CaSO4(s)
C. 3Co2+(aq) + 2PO43-(aq)---->Co3(PO4)2(s)
.............................
Answer:-
0.229 L
Explanation:-
Molar mass of AgBr = 107.87 x 1 + 79.9 x 1
=187.77 grams mol-1
Mass of AgBr = 150 grams
Number of moles of AgBr = 150 grams / 187.77 gram mol-1
= 0.8 mol
The balanced chemical equation is
NaBr (aq) + AgNO3 (aq)--> AgBr(s) + NaNO3(aq)
From the equation we can see that
1 mol of AgBr is produced from 1 mol of AgNO3.
∴ 0.8 mol of AgBr is produced from 1 x 0.8 / 1 = 0.8 mol of AgNO3.
Strength of AgNO3 = 3.5 M
Volume of AgNO3 required = Number of moles / strength
= 0.8 moles / 3.5
=0.229 L
Answer:
Various limitations of Mendeleev's periodic table are:-
Position of hydrogen - he couldn't assign a correct position to hydrogen as it showed properties of both alkali and halogens .
Position of isotopes - he considered that the properties of elements are a function of their atomic masses. Hence isotopes of a same element couldn't be placed.
In the d-block , elements with lower atomic number were placed before higher atomic number.
Explanation:
Hello!
To find the number of moles that are in the given amount, we need to divide the total number of atoms by Avogadro's number, which is 1 mole is equal to 6.02 x 10^23 atoms.
5.0 x 10^25 / 6.02 x 10^23 ≈ 83.0564
Therefore, there are about 83.06 moles of iron (sigfig: 83 moles).
