Q=McDeltaT
Q=731g×1.00cal/g°c×(83°c-35°c)
Q=35088cal
therefore, their are 35088 calories are required to heat up 731g of water.
Answer: its B , B is the answer.
Explanation:
Answer:
2.408 x 10^24molecules
Explanation:
no. of moles, n=4
avogadro's constant= 6.02 x 10^23
no. of molecules, N = ?
using
<em>N=</em><em>n</em><em> </em><em>x </em><em>l</em>
N= 4 x 6.02 x 10^23
N = 2.408 x 10^24molecules