Rutherford used gold for his scattering experiment because gold is the most malleable metal and he wanted the thinnest layer as possible. The goldsheet used was around 1000 atoms thick. Therefore, Rutherford selected a Gold foil in his alpha scatttering experiment.
Answer:
Q = 30355.2 J
Explanation:
Given data:
Mass of ice = 120 g
Initial temperature = -5°C
Final temperature = 115°C
Energy required = ?
Solution:
Specific heat capacity of ice is = 2.108 j/g.°C
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Q = m.c. ΔT
ΔT = T2 -T1
ΔT = 115 - (-5°C)
ΔT = 120 °C
Q = 120 g × 2.108 j/g.°C × 120 °C
Q = 30355.2 J
Answer:
Calcium for 2+ charge and Fluorine forms 1- charge
Explanation:
PH + pOH = 14.
So, subtract 3.73 from 14.
The pOH is 10.27 :)
The number of atoms of phosphorus in copper (II) phosphate [Cu3 (PO4)2] is determined by multiplying first 9.20 by 2 since there are two phosphorus atoms in the compound and multiplying again with Avogadro's number. The answer is 1.108 x 10^25 atoms.
