Hbro dissociate as follows
HBro---> H+ + BrO-
Ka= (H+)(BrO-) / HBro
PH = -log (H+)
therefore (H+) = 10^-4.48= 3.31 x10^-5
ka is therefore= ( 3.31 x 10^-5)^2/0.55=1.99 x10^-9
Answer:
4.4×10² cm³
Explanation:
From the question given above, the following data were obtained:
Diameter (d) = 68.3 mm
Height (h) = 0.120 m
Volume (V) =?
Next, we shall convert the diameter (i.e 68.3 mm) to cm.
This can be obtained as follow:
10 mm = 1 cm
Therefore
68.3 mm = 68.3 mm / 10 mm × 1 cm
68.3 mm = 6.83 cm
Therefore, the diameter 68.3 mm is equivalent 6.83 cm.
Next, we shall convert the height (i.e 0.120 m) to cm. This can be obtained as follow:
1 m = 100 cm
Therefore,
0.120 m = 0.120 m/ 1 m × 100 cm
0.120 m = 12 cm
Therefore, the height 0.120 m is equivalent 12 cm.
Next, we shall determine the radius of the cylinder. This can be obtained as follow:
Radius (r) is simply half of a diameter i.e
Radius (r) = Diameter (d) /2
r = d/2
Diameter (d) = 6.83 cm
Radius (r) =?
r = d/2
r = 6.83/2
r = 3.415 cm
Finally, we shall determine the volume of the cylinder as follow:
Radius (r) = 3.415 cm
Height (h) = 12 cm
Volume (V) =?
Pi (π) = 3.14
V = πr²h
V = 3.14 × (3.415) ² × 12
V = 440 cm³
V = 4.4×10² cm³
Therefore, the volume of the cylinder is 4.4×10² cm³
Answer:
energy known as the latent heat of vaporization is required to break the hydrogen bonds. At 100 °C, 540 calories per gram of water are needed to convert one gram of liquid water to one gram of water vapour under normal pressure.
Explanation:energy known as the latent heat of vaporization is required to break the hydrogen bonds. At 100 °C, 540 calories per gram of water are needed to convert one gram of liquid water to one gram of water vapour under normal pressure.
Answer: H+ ia helyuim
explinanation: Hope this helped!!
