<span>Filtration, if its a precipitate that means its insoluble. </span>
Answer
5
Explanation:
We can go about this using the percentage compositions.
First, we calculate the percentage composition of the copper sulphate. This is obtainable by using the mass.
0.96/1.5 * 100 = 64%
Hence the percentage by mass of the water present is 36%
The molar mass of the anhydrous sulphate is 64 + 32 +4(16) = 160g/mol
The molar mass of the water is 2(1) + 16 = 18g/mol
Not forgetting that it is in multiples of x, the total molar mass of the water is 18x moles
The total mass of the copper sulphate hydrate is 160+ 18x
Now how do we get x? Like it is said earlier, the percentage composition is constant.
Hence, 64/100 * (160 + 18x) = 160
16000 = 64(160 + 18x)
16000 = 10,240 + 1152x
16,000 - 10,240 = 1152x
1152x = 5760
x = 5760/1152
x = 5
The law of conservation of mass states that mass is neither created nor destroyed. Since we have 2 g/mol of A and 3 g/mol of B then AB should be equal to the sum of their molar mass that is
2 g/mol + 3 g/mol = 5 g/mol AB
for the case of A2B3
A2 = 2 * 2 = 4 g/mol
B3 = 3 * 3 = 9 g/mol
therefore A2B3 = 13 g/mol
Answer:
A 3s orbital is at a greater average distance from the nucleus than a 2s orbital
Explanation:
As the principal quantum number n increases, the distance of the orbital from the nucleus increases. Hence if we consider the 2s and 3s orbitals, it is easy to see that the 3s orbital is at a greater distance from the nucleus than the 2s orbitals.
This is clearly seen when we plot the radial distribution against the distance from the nucleus. This enables us to visualize the region in space in which an electron may be found.
Answer:
Molecular formula of the compound = H₂CO₃
Explanation:
First, the empirical formula of the compound is determined
Percentage by mass of each element is given as shown below:
H = 3.3% ; C = 19.9%; O = 77.4%
Mole ratio of the elements= percentage mass/ molar mass
H = 3.3/ 1 = 3.3
C = 19.3/12 = 1.6
O = 77.4/16 = 4.8
whole number ratio is obtained by dividing through with the smallest ratio
H = 3.3/1.6; C = 1.6/1.6; O = 4.8/1.6
H : C : O = 2 : 1 : 3
Empirical formula = H₂CO₃
Molecular formula/mass = n(empirical formula/mass)
60 = n(2*1 + 12*1 + 16*3)
60 = n(62)
n = 60/62 = 0.96
n is approximately = 1
Therefore, molecular formula of the compound = (H₂CO₃) * 1
Molecular formula of the compound = H₂CO₃
