Lead(ii) nitrate and ammonium iodide react to form lead(ii) iodide and ammonium nitrate according to the reaction pb(no3)2(aq)+2nh4i(aq)⟶pbi2(s)+2nh4 no3(aq) what volume of a 0.610 m nh4i solution is required to react with 229 ml of a 0.640 m pb(no3)2 solution?

Answer : The volume of solution required is 482 mL

Explanation :

The given balanced chemical reaction is:

First we have to calculate the moles of .

Now we have to calculate the moles of

From the balanced chemical reaction we conclude that,

As, 1 mole of react to give 2 moles of

So, 0.147 mole of react to give moles of

Now we have to calculate the volume of

conversion used : (1 L = 1000 mL)

Therefore, the volume of solution required is 482 mL

the answer would be c

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied