How much heat is required to increase the temperature of 198.5 grams of water from 25.0 oc to 88.5 oc? (the specific heat of water is 4.18 j/g x oc)

a.52, 700 j
b.199 j
c.265 j
d.73,400 j
e.20,700 j

Answer : The correct option is, (A) 52700 J

Solution :

Formula used :

Q = heat gained

m = mass of the substance = 198.5 g

c = heat capacity of water = 4.18 J/g°C      

Now put all the given values in the above formula, we get

Q = 52687.855 J ≈ 52700 J

Therefore, the amount of heat required is, 52700 J

under a microscope, cancer cells may look very different from normal cells. they often have different sizes and some may be larger than normal while others are smaller. cancer cells are often abnormally shaped and the control centre of the cell (the nucleus) may have an abnormal appearance.

i think? ?

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer