The initial mass of sodium hydroxide is 3.3 g (answer C)
<u><em>calculation</em></u>
Step 1 : find the moles of iron (ii) hydroxide ( Fe(OH)₂
moles = mass÷ molar mass
from periodic table the molar mass of Fe(OH)₂ = 56 + [16 +1]2 = 90 g/mol
moles is therefore = 3.70 g÷ 90 g/mol = 0.041 moles
Step 2: use the mole ratio to calculate the moles of sodium hydroxide (NaOH)
from given equation NaOH : Fe(OH)₂ is 2 :1
therefore the moles of NaOH = 0.041 x 2 = 0.082 moles
Step 3: find mass of NaOH
mass = moles x molar mass
from the periodic table the molar mass of NaOH = 23 +16 +1 = 40 g/mol
mass = 0.082 moles x 40 g/mol = 3.3 g ( answer C)
Answer
Na OH reacts with H Cl and forms Na Cl and H₂O
NaOH + HCl → NaCl + H₂O
Here we can see that 1 mole of NaOH reacting with 1 mole of HCl and forming 1 mole of NaCl and 1 mole of H₂O
when NaOH and HCl are added together in equal amount then they will completely neutralize each other but NaOH is hygroscopic in nature which means it can absorb water from air so it will not be weighted accurately.
hence, for neutralization we will take extra NaOH.
PH= −log
10
[H
+
]
= −log
10
(0.001)
= −log
10
(10
−3
)
= −(−3)log
10
10
pH=3.
01
Metals of Group 1 donate 1 electron from its ns orbital to form ionic bond, where n is the no. of its outermost shell.
Metals of Group 2<span> donate 2 electrons from its ns orbital to form ionic bond, where n is the no. </span>of its <span>outermost shell. </span>