The quantity of NaOH required to reach the third equivalence point is 20mL.
Using the titration formula,
CaVa/CbVb = Na/Nb
Where,
Ca = concentration of citric acid (0.200 M)
Cb = concentration of NaOH (0.750 M)
Va = Volume of citric acid (25.0 mL)
Vb = volume of NaOH (x mL)
Na = number of reacting mole of citric acid (3)
Nb = number of reacting mole of NaOH (1)
Therefore Vb ( x mL) =CaVaNb/CbNa
= 0.2× 25×3/0.75 ×1
= 15/0.75
Vb ( x mL) = 20 mL
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Answer: Option (3) is the correct answer.
Explanation:
When there is a negative charge on an atom then we add the charge with the number of electrons. Whereas when there is a positive charge on an atom then we subtract the charge from the number of electrons.
Atomic number of chlorine is 17. So, number of electrons present in 
is 17 + 1 = 18 electrons.
Atomic number of cobalt is 27. So, number of electrons present in 
is 27 - 4 = 23 electrons.
Atomic number of iron is 26. So, number of electrons present in 
is 26 - 2 = 24 electrons.
Atomic number of vanadium is 23. So, number of electrons present in V is 23 electrons.
Atomic number of scandium is 21. So, number of electrons present in 
is 21 + 2 = 23 electrons.
Thus, we can conclude that out of the given species, has the greatest number of electrons.
Two electron pairs is the answer
Answer:
1.204 × 10²⁴ atoms
Explanation:
According to this question, one mole of aluminum (Al) atom contains 6.02 × 10²³ atoms.
If two moles of aluminum are given, this means that there will be 2 × 6.02 × 10²³ atoms of aluminum
2 × 6.02 × 10²3
= 12.04 × 10²³
= 1.204 × 10²⁴ atoms.
cl2>F2>H2
we can do this by molar mass
Hydrogen - 1
clorine - 35 x2 = 70
flourine- 18 x 2 = 36
flourine - 18
