Answer;
K+ and NO3- ions
Explanation;
The main ions remaining are K+ and NO3- ions after pbi2 precipitation is complete.
However; There will always be tiny amounts of Pb2+ and I- ions, but most of them are in the solid precipitate.
Answer:
0.677 moles
Explanation:
Take the atomic mass of K = 39.1, O =16.0, P = 31.0
no. of moles = mass / molar mass
no. of moles of K3PO4 used = 4.79 / (39.1x3 + 31 + 16x4)
= 0.02256 mol
From the equation, the mole ratio of KOH : K3PO4 = 3 :1,
meaning every 3 moles of KOH used, produces 1 mole of K3PO4.
So, using this ratio, let the no. of moles of KOH required to be y.

y = 0.02256 x3
y = 0.0677 mol
If you don't find exactly 0.677 moles as one of the options, go for the closest one. A very slight error may occur because of taking different significant figures of atomic masses when calculating.
Answer:
a straight line passing from side to side through the center of a body or figure, especially a circle or sphere.
Explanation:
a straight line passing from side to side through the center of a body or figure, especially a circle or sphere.
Answer : The rate of effusion of sulfur dioxide gas is 52 mL/s.
Solution :
According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

or,
..........(1)
where,
= rate of effusion of nitrogen gas = 
= rate of effusion of sulfur dioxide gas = ?
= molar mass of nitrogen gas = 28 g/mole
= molar mass of sulfur dioxide gas = 64 g/mole
Now put all the given values in the above formula 1, we get:


Therefore, the rate of effusion of sulfur dioxide gas is 52 mL/s.