Answer:
Exercise 1
a) 140 m
b) 100 m
c) 180 m
d) 140 m
Exercise 2
a) 20 yards
b) 30 yards
c) 20 yards
d) 55 yards
Exercise 3
a) 11 Kilometers
b) 7 Kilometers
Explanation:
Exercise 1
Distance from B to C = 140 m
Distance from to D = 100 m
Total distance = 100+40+40 = 180 m
Total displacement i.e. distance between A and D is 140 m
Exercise 2
a) Distance from B to C = 35 -15 = 20 yards
b) Distance from C to D = 5 + 35 = 30 yards
c) Distance from B to D = 5 + 15 = 20 yards
d) Displacement = 5 + 50 = 55 yards
Exercise 3
a) The total distance travelled = 5 + 2 + 4 = 11 Kilometers
b) Displacement = 5-2 + 4 = 7 Kilometers
In a) the final equation is AgNO3 + KCl = AgCl + KNO3, b) Ni(NO3)2 + Na2S = 2NaNO3 + NiS; c) CaCl2 + Na2CO3 = 2 NaCl + CaCO3. In 2) The total net equation is Ca 2+ + CO32- = CaCO3 (s).
Answer:
Given
mass of H2O (m) =35.6g
molarmass (mr) = H2O ), 1x2+16=18g/mol
moles of H2O (n) =?
sln
n=m/mr
n=35.6g/18g/mol
n=1.978moles
the moles of H2O are 1.978moles
Answer
CH3Cl is polar amoungst the choices
