This is false, they have intermediate properties between suspension and true solution
Moles of ZnSO4 = Mass/Mr
= 2/(65.4 + 32.1 + (16 x 4))
= 0.012383.... mol
Since there is no chemical equation provided, I will assume that the ratio of ZnSO4:Li2CO3 is 1:1
Therefore there are 0.012383 mil of Li2SO4 as well
So the mass would be Moles x Mr = 0.012383.... x ((6.9 x 2) +32.1 + (4x16))
= 1.360 g of Li2SO4 produced
25 drops of acid is required to neutralize the 50.0 ml of 0.010m of NaOH in the experiment.
The equation of the reaction is;
NaOH(aq) + HCl(aq) ---------> NaCl(aq) + H2O(l)
We can use the titration formula;
CAVA/CBVB = NA/NB
CA= concentration of acid
VA = volume of acid
CB = concentration of base
VB = volume of base
NA = number of moles of acid
NB = number of moles of base
CB = 0.010 M
VB = 50.0 ml
CA = 0.50 M
VA = ?
NA = 1
NB = 1
Substituting values;
CAVANB = CBVBNA
VA = 0.010 × 50.0 × 1/ 0.50 × 1
VA = 1 ml
Since the total volume of acid used is 1 ml and each drop contains 0.040 ml
The number of drops required is 1ml/0.040 ml = 25 drops
Learn more: brainly.com/question/1527403
Entropy change of vaporization is simply the ratio of
enthalpy change and the temperature in Kelvin.
Temperature = 64 + 273.15 = 337.15 K
Hence,
δsvap = (32.21 kJ / mole) / 337.15 K
<span>δsvap = 0.0955 kJ / mole K = 95.5 J / mole K</span>
+2, because Beryllium is in the Group II of the periodic table.
Hope this helps!
