Answer:
It’s is a simple
Explanation:the prokaryotic cell involves a simple process which intern involves one chromosom
Explanation:
Let us assume that the ratio for the given reaction is 1:1.
Therefore, we will calculate the moles of 
as follows.
Moles of solution = molarity × volume (L)
= 0.0440 M × 0.014 L
= 0.000616 moles
Moles of excess EDTA = 0.000616 moles
Also, the initial moles of EDTA will be calculated as follows.
Total initial moles of EDTA = 0.0600 M × 0.025 L
= 0.0015
Therefore, moles of EDTA reacted with 
will be as follows.
= 0.0015 - 0.000616
= 0.00088 moles
Since, we have supposed a 1 : 1 ratio between and EDTA
.
So, moles of = 0.00088 moles
Now, we will calculate the molarity of as follows.
Molarity of solution = 
= 
= 0.015 M
Thus, we can conclude that the original concentration of the solution is 0.015 M.
1) H2O is able to dissolve both polar molecules and non polar ones
2) due to its extreme polarity it can even dissolve some I onic compounds
3 the h2o molecule itself is small in size
Answer:
ones analog and ones digital.
Explanation:
The difference between the two are that one loses quality but is easier to transfer, however digital takes longer to transfer, however this makes the quality the same.
The question is incomplete. The complete question is:
At 25◦C and atmospheric pressure the volume change of mixing of binary liquid mixtures of species 1 and 2 is given by the equation:
ΔV = x1x2(45x1 + 25x2)
Where ΔV is in cm3-mol-1. At these conditions, the molar volumes of pure liquid 1 and 2 are V1= 110 and V2= 90 cm3-mol-1. Determine the partial molar volumes 1VE and 2VE in a mixture containing 40 mole percent of specie 1.
Answer:
1VE = 117.92 cm³.mol⁻¹, 2VE = 97.92 cm³.mol⁻¹
Explanation:
In the equation given, x represents the molar fraction of each substance, thus x1 = 0.4 and x2 = 0.6. Because of the mixture, the molar partial volume of each substance will change by a same amount, which will be:
ΔV = 0.4*0.6(45*0.4+ 25*0.6)
ΔV = 7.92 cm³.mol⁻¹
1VE - V1 = 7.92
1VE = 7.92 + 110
1VE = 117.92 cm³.mol⁻¹
2VE - V2 = 7.92
2VE = 7.92 + 90
2VE = 97.92 cm³.mol⁻¹
