Answer:
Qm  = -55.8Kj/mole
Explanation:
NaOH(aq) + HNO₃(aq) => NaNO₃(aq) + H₂O(l)
Qm = (mc∆T)water /moles acid
Given => 100ml(0.300M) NaOH(aq) + 100ml(0.300M)HNO₃(aq)
=>	0.03mole NaOH(aq) + 0.03mole HNO₃(aq)
=>	0.03mole NaNO₃(aq) + 0.03mole H₂O(l)
ΔH⁰rxn = [(200ml)(1.00cal/g∙°C)(37 – 35)°C]water / 0.03mole HNO₃
= 13,333 cal/mole x 4.184J/cal = 55,787J/mol = 55.8Kj/mole (exothermic)*
Heat of reactions comes from formation of H-Oxy bonds on formation of water of reaction and heats the 200ml of solvent water from 35⁰C to 37⁰C. 
 
        
                    
             
        
        
        
Well, these particles happens to be small, like REALLY small. So microscopically small they aren't picked up or observed my the naked eye. also the vibrations are in an atomic scale which is also VERY tiny This goes for all solids too.
        
             
        
        
        
Answer:
Q = 3440Kj
Explanation:
Given data:
Mass of gold = 2kg
Latent heat of vaporization = 1720 Kj/Kg
Energy required to vaporize 2kg gold = ?
Solution:
Equation
Q= mLvap
It is given that heat required to vaporize the one kilogram gold is 1720 Kj thus, for 2 kg
by putting values,
Q= 2kg ×  1720 Kj/Kg
Q = 3440Kj
 
        
             
        
        
        
I think it would be minimize so u can have more friction