Answer:
2) 25.0mL aliquots of the solution in problem 1 are titrated with EDTA to the calmagite end point. A blank containing a small measured amount of Mg2+ requires 2.12mL of the EDTA to reach the end point. An aliquot to which the same amount of Mg2+ is added requires 25.88mL of the EDTA to reach the end point.
a. How many mL of EDTA are needed to titrate the Ca2+ ion in the aliquot?
b. How many moles of EDTA are there in the volume obtained in a.?
c. What is the molarity of the EDTA solution?
Explanation:
Given that;
Volume of aliquot = 25mL
Blank reading = 2.12mL
2a)
Volume of EDTA used for Ca²⁺ ion
25.88mL - 2.12mL = 23.76mL
Therefore mL of EDTA needed to titrate the Ca²⁺ ion in the aliquot is 23.76mL
2b)
Molarity of Ca²⁺ ion is 0.0172M
Mole of EDTA =
2c)
Molarity of EDTA = mole of EDTA / Vol. of EDTA
Answer:
An alkali metal present in period 2 have larger first ionization energy.
Explanation:
Ionization energy:
The amount of energy required to remove the electron from the atom is called ionization energy.
Trend along period:
As we move from left to right across the periodic table the number of valance electrons in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases and ionization energy increases because it is very difficult to remove the electron from atom and more energy is required.
Trend along group:
As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.
As the size of atom increases the ionization energy from top to bottom also decreases because it becomes easier to remove the electron because of less nuclear attraction and as more electrons are added the outer electrons becomes more shielded and away from nucleus. Thus alkali metal present in period 2 have larger ionization energy because of more nuclear attraction as compared to the alkali metal present in period 4.
The answer is A, a new chemical substance has to be formed for there to be a chemical change. This is because chemical change happens when a phase of matter changes due to chemicals.
Answer:
The answer to your question is 1137.45 kJ
Explanation:
Data
mass = 424 g
T1 = 22.6°C
T2 = 149°C
Clw = 4.184 J/g°C
Cv = 2.078 j/g°C
Latent heat = 2257 J/g
Process
1.- Heat of liquid
Heat = (424)(4.184)(100 - 22.6)
Heat = 137308.8 J
2.- Calculate latent heat
Latent heat = (424)(2257)
Latent heat = 956968 J
3.- Heat of vapor
Heat = (424)(2.078)(149 - 100)
Heat = 43172.5 J
4.- Total heat
Heat = 137308.8 + 956968 + 43172.5
Heat = 1137449.3 J = 1137.45 kJ
Answer:
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