For the reaction HNO3 + Mg(OH)2→ Mg(NO3)2 + H2O, how many grams of magnesium nitrate are produced from 8.00 mol of nitric acid, HNO3? Balance the equation

tha mass of magnesium nitrate is 592g

Explanation:

from a balanced chemical equation

2HNO3 + Mg(OH)2→ Mg(NO3)2 + 2H2O

2 mol of nitric acid is equivalent to 1 mol of magnesium nitrate. then 8 mol of nitric acid will be equivalent to 4 mol of magnesium nitrate.

2HNO3 + Mg(OH)2 --> 2H2O + Mg(NO3)2

593.2 grams Mg(NO3)2

Explanation:

First, balance the equation

The first thing I notice is that there are 2 nitrates on the reactant side so I would put a 2 in front of HNO3. Now the problem is that there are 4 hydrogens on the reactants side, so add a 2 to H2O.

Next is stoichiometry

We are looking for Mg(NO3)2 so use the balanced equation to create a mole ratio and then convert to grams.

8 mol HNO3 * 1 mol Mg(NO3)2/2 mol HNO3 * 148.3 g Mg(NO3)2/1 mol Mg(NO3)2 = 593.2 grams Mg(NO3)2

i think its 100

i hope this !

(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