<span>A) This solution was not basic when it was heated in part 3. ( in part 3 i convertedCu(OH)2 to CuO).
Incorrectly low, because not all copper compounds will precipitate out
B) A slightly blue solution was decanted from Cu in part V. (in part 5 i reduced Cu(H20)6 ions with zink)
Incorrectly low, because some copper were thrown away
C) In part 5 the water in the beaker boiled away, exposing the evaporating dish to excess heat (same as above).
incorrectly high, because other compounds might be present as well </span>
Protons, neutrons, and electrons.
Answer:
T2 =21.52°C
Explanation:
Given data:
Specific heat capacity of sample = 1.1 J/g.°C
Mass of sample = 385 g
Initial temperature = 19.5°C
Heat absorbed = 885 J
Solution:
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
ΔT = Final temperature - initial temperature
885J = 385 g× 1.1 J/g.°C×(T2 - 19.5°C )
885 J = 423.5 J/°C× (T2 - 19.5°C )
885 J / 423.5 J/°C = (T2 - 19.5°C )
2.02°C = (T2 - 19.5°C )
T2 = 2.02°C + 19.5°C
T2 =21.52°C
Answer:
1(a) N = 3
(b) N = 0
(c) N = 5
(d) N = -2
(2) Molecular formula for benzene is C6H6
Explanation:
1(a) N02 1-
N + (2×-2) = -1
N-4 = -1
N = -1+4 = 3
(b) N2
2(N) = 0
N = 0/2 = 0
(c) NO2Cl
N + ( 2×-2) + (-1) = 0
N - 4 - 1 = 0
N - 5 = 0
N = 0+5 = 5
(d) N2H4
2(N) + (4×1) = 0
2N + 4 = 0
2N = 0 - 4 = -4
N = -4/2 = -2
(2) Molcular mass of benzene = 78g/mole = (6×12g of carbon) + (6×1g of hydrogen) = 72+6 = 78g/mole
Therefore, molecular formula for benzene is C6H6
Answer:
Equation of reaction:
a) 2HCl + Ba(OH)2 ==> CaCl2 + 2H2O
b) Molarity of base = 0.042 M.
Explanation:
Using titration equation
CAVA/CBVB = NA/NB
Where NA is the number of mole of acid = 2
NB is the number of mole of base = 1
CA is the molarity of acid =0.15M
CB is the molarity of base = to be calculated
VA is the volume of acid = 25 ml
VB is the volume of base = 44.45mL
Substituting
0.15×25/CB×44.45 = 2/1
Therefore CB =0.15×25×1/44.45×2
CB = 0.042 M.
