Assume that 300 J of heat is added to 3.00g of water originally at 23°C. what would be the final temperature of the water?  The specific heat capacity of water is 4.184

47C

Explanation:

Recall the equation for specific heat:

q = mcΔT

300 = 3 x 4.184 x (T - 23)

T = 47

the formula mass of a molecule or a compound is the sum of the atomic weights of the atoms in its empirical formula

(that's what i know about formula mass)

(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