Answer:
Possible lowest volume = 0.19 cm
Possible highest volume = 0.21 cm
Explanation:
given data
volumetric pipette uncertainty = 0.01 cm³
total volume = 0.20 cm³
solution
we will get here Possible lowest and highest volume that is express as
Possible lowest volume = total volume - uncertainty .....................1
Possible highest volume = total volume + uncertainty ....................2
put here value in both equation and we get
Possible lowest volume = 0.20 cm - 0.01 cm
Possible lowest volume = 0.19 cm
and
Possible highest volume = 0.20 cm + 0.01 cm
Possible highest volume = 0.21 cm
Answer:
An aquifer is a layer of porous substrate that contains and transmits groundwater. ... The upper level of this saturated layer of an unconfined aquifer is called the water table or phreatic surface. Below the water table, where in general all pore spaces are saturated with water, is the phreatic zone.
Answer:
a) positive feedback
Explanation:
positive as in increasing
Answer:
See explanation below
Explanation:
In this case, HCl is a strong acid, therefore, it dissociates completely in solution.
To know the quantity of water we need to add, we first need to know the concentration of the acid with pH = 6:
[H⁺] = antlog(-pH)
[H⁺] = antlog(-6) = 1x10⁻⁶ M
This means that the concentration is being diluted.
Now, even if we add great quantities of water, and the concentration and volume change, there is one time that do not change despite the quantity of water added; this is the moles. So, all we have to do, is calculate the moles of the acid in the 1 mL of water, and then, the volume of the acid when it's dilluted:
moles HCl = 0.1 * (1/1000) = 1x10⁻⁴ moles
Now that we have the moles, we can calculate the volume which the acid with the lowest concentration has:
V = mol/M
V = 1x10⁻⁴ / 1x10⁻⁶
V = 100 L
This means that we need to add 99.999 mL of water
Answer:
It takes 83 mL of a 0.45 M NaOH solution to neutralize 235 mL of an HCl ... 2). You are titrating an acid into a base to determine the concentration of the base. The endpoint of the neutralization is reached but the stopcock on the buret sticks slightly ... 5). It takes 12.5 mL of a 0.30 M HCl solution to neutralize 285 mL of NaOH ...
Explanation:
