I believe it's webbed feet
Answer:
40.8g of sodium sulfate must be added
Explanation:
The reaction of barium nitrate, Ba(NO₃)₂ with sodium sulfate, Na₂SO₄ is:
Ba(NO₃)₂ + Na₂SO₄ → 2 NaNO₃ + BaSO₄(s)
That means, for a complete reaction of an amount of barium nitrate you must add the same amount in moles of sodium sulfate. To solve this problem we need to convert the mass of barium nitrate to moles = Moles of sodium sulfate that must be added:
<em>Moles Ba(NO₃)₂ -Molar mass: 261.3g/mol-:</em>
75g * (1mol / 261.3g) = 0.287 moles = Moles Na₂SO₄
<em>Mass Na₂SO₄ -Molar mass: 142.04g/mol-:</em>
0.287 moles * (142.04g / mol) =
<h3>40.8g of sodium sulfate must be added</h3>
Answer:
It is basic and has a pH of 9.8.
Explanation:
pOH = 4.2
we will determine its pH.
pOH + pH = 14
pH = 14 - pOH
pH = 14 - 4.2
pH = 9.8
According to pH scale the the pH lower than 7 is consider acidic while pH of seven is neutral and pH greater than 7 is basic.
The given solution has pH 9.8 it means it is basic.
1.70 × 10³ seconds
<h3>Explanation </h3>
+ 2 e⁻ → 
It takes two moles of electrons to reduce one mole of cobalt (II) ions and deposit one mole of cobalt.
Cobalt has an atomic mass of 58.933 g/mol. 0.500 grams of Co contains 
of Co atoms. It would take 
of electrons to reduce cobalt (II) ions and produce the 
of cobalt atoms.
Refer to the Faraday's constant, each mole of electrons has a charge of around 96 485 columbs. The 0.01697 mol of electrons will have a charge of 
. A current of 0.961 A delivers 0.961 C of charge in one single second. It will take 
to transfer all these charge and deposit 0.500 g of Co.
Answer:
The mass percent of Al(OH)₃ is 15.3%
Explanation:
The reaction is:
Al(OH)₃ + 3HCl = AlCl₃ + 3H₂O
The excess acid is neutralized with a solution of sodium hidroxide, in the reaction:
NaOH + HCl = NaCl + H₂O
The total moles of HCl is:
From the second titration, the moles of excess of HCl is:
The difference between the total and excess of HCl, it can be know the moles that reacts with the aluminum hydroxide, is:
The ratio between HCl and Al(OH)₃ is 3:1. The MW for aluminum hydroxide is 78 g/mol, thus:
The percentage of Al(OH)₃ is:
%
