Answer:
The magnesium atom loses 2 electron to the 2 atoms of chlorine. The 7 valency electrons of each chlorine atom will now be 8 to attain stable configuration. The final compound is written as MgCl2.
Explanation:
Ionic compounds are compound formed from the transfer of electron(s). One atom of the element loses electron(s) while the other atom gains electron(s).
The compound Magnesium chloride is an ionic compound . The bond between an atom of magnesium and 2 atoms of chlorine is an ionic bonding.
The valency electron of magnesium is 2 electron , for the atom of magnesium to attain octet rule, it will easily lose it 2 electrons to the chlorine atoms.
The chlorine atom on the other hand has 7 valency electrons, to attain octet configuration it will most likely gain 1 electron to become stable.
The magnesium atom loses 2 electron to the 2 atoms of chlorine. The 7 valency electrons of each chlorine atom will now be 8 to attain stable configuration. The final compound is written as MgCl2.
<u>Answer:</u> The correct answer is Option 1.
<u>Explanation:</u>
Neutralization reaction is defined as the reaction in which acid reacts with a base to produce a salt along with water.
Here, HCl is an acid and
is a base. When these two compounds react, the salt obtained is calcium chloride.
The equation for the above reaction is given by:

Hence, the correct answer is Option 1.
1) Chemical reaction
HCl + NaOH ---> NaCl + H2O
25.0 ml
0.150 M 0.250M
2) 50% completion => 0.025 l * 0.150 M * (1/2) = 0.001875 mol HCl consumed and 0.001875 mol HCl in solution
0.001875 mol HCl => 0.001875 mol H(+)
Volume = Volume of HCl solution + Volumen of NaOH solution added
Volume of HCl solution = 0.0250 l
Volume of NaOH = n / M = 0.001875 mol / 0.250M = 0.0075 l
Total volume = 0.0250 l + 0.0075 l = 0.0325 l
[H+] = 0.001875 mol / 0.0325 l = 0.05769 M
pH = - log [H+] = - log (0.05769) = 1.23
Answer: 1.23
3) Equivalence point
0.02500 l * 0.150 M = 0.250M * V
=> V = 0.02500 * 0.150 / 0.250 = 0.015 l
4) 1.00 ml NaOH added beyond the equivalence point
1.00 ml * 1 l / 1000 ml * 0.250 M = 0.00025 mol NaOH in excess
0.00025 mol NaOH = 0.00025 mol OH-
Volume of the solution = 0.02500 l + 0.015 l + 1.00/1000 l = 0.041 l
[OH-] = 0.00025 mol / 0.041 l = 0.00610 M
pOH = - log (0.00610) = 2.21
pH + pOH = 14 => pH = 14 - pOH = 14 - 2.21 = 11.76
Answer: 11.76
Answer:
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