6.02 x 1023 atoms weigh out 63.55 grams copper.
No. of Molecules in water = 3.5mole x (6.02 x 10^23) molecules/mole = 2.107 x 10^24 molecules of H2O
Use the PV = nRT equation T is in Kelvins = 31 + 273 = 304 K
P(0.5) = (2.91)(0.0821)(304)
P(0.5) = 72.6289
P = 145.25 atm or 1.45x10^2 atm
Answer:
q = - 2067.2 J of Heat is giving out when 85.0g of lead cools from 200.0 c to 10.0 c.
Explanation:
The Specific Heat capacity of Lead is 0.128 
This means, increase in temperature of 1 gm of lead by
will require 0.128 J of heat.
Formula Used :

q = amount of heat added / removed
m = mass of substance in grams = 85.0 g
c = specific heat of the substance = 0.128
= Change in temperature
= final temperature - Initial temperature
= 10 - 200
= -
put value in formula
q = - 
On calculation,
q = - 2067.2 J
- sign indicates that the heat is released in the process
I would say #3 I’m sorry if it’s wrong tho
Answer: D
Explanation:
Because velocity is speed