Answer:
Final temperature is 34.2 °C
Explanation:
Given data:
mass of metal = 125 g
temperature of metal = 93.2 °C
mass of water v= 100 g
temperature of water = 18.3 °C
specific heat of meta is = 0.900 j/g. °C
specific heat of water is = 4.186 j/g. °C
final temperature of water and metal = ?
Solution:
Q = m . c . ΔT
ΔT = T2-T1
now we will put the values in equation
Q1 = m . c . ΔT
Q1 = 125 g. 0.900 j/g. °C .93.2°C - T2
Q1 = 112.5 (93.2°C - T2)
Q1 =10,485 - 112.5T2
Q2 = m . c . ΔT
Q2 = 100 . 4.186. (T2- 18.3)
Q2 = 418.6 . (T2- 18.3)
Q2 = 418.6T2 - 7660.38
10,485 - 112.5T2 = 418.6T2 - 7660.38
10,485 + 7660.38 = 418.6T2+ 112.5T2
18145.38 = 531.1 T2
T2 = 18145.38/531.1
T2 = 34.2 °C
The number of electrons in the outermost shell of an atom determines its reactivity. Noble gases have low reactivity because they have full electron shells. Halogens are highly reactive because they readily gain an electron to fill their outermost shell
Neutrons for example isotopes
Answer:
Greater. Inversely proportional
Explanation:
Scientific data-reproducible,reliable results are demanded and experiments must be precisely described while pseudoscience failures are ignored,hidden,no actual physical phenomena or processes ever found