Answer:
Explanation:
Hello there!
In this case, since the modelling of titration problems can be approached via the Henderson-Hasselbach equation to set up a relationship between pH, pKa and the concentration of the acid and its conjugate base, we can write:
![pH=pKa+log(\frac{[NO_2^-]}{[HNO_2]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5BNO_2%5E-%5D%7D%7B%5BHNO_2%5D%7D%20%29)
Whereas the pH is given as 3.14 and the concentrations are the same, that is why the pH would be equal to the pKa as the logarithm gets 0 (log(1)=0); thus, we can calculate the Ka via:
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The relation between density and mass and volume is
the dose required is 2.5 tsp
each tsp contain 5mL
So dose required in mL = 2.5 X 5 = 12.5 mL
the mass will be calculated using following formula
The mass of dose in grams will be 15.38 g
Answer:
it depends on the material of the bowl. if the material is a good heat conductor, then it is a bad insulator vice versa.
mass percent concentration = 15.7 %
molar concentration of glucose solution 1.03 M
Explanation:
To calculate the mass percent concentration of the solution we use the following formula:
concentration = (solute mass / solution mass) × 100
solute mass = 60.5 g
solution mass = solute mass + water mass
solution mass = 60.5 + 325 = 385.5 g (I used the assumption that the solution have a density of 1 g/mL)
concentration = (60.5 / 385.5) × 100 = 15.7 %
Now to calculate the molar concentration (molarity) of the solution we use the following formula:
molar concentration = number of moles / volume (L)
number of moles = mass / molecular weight
number of moles of glucose = 60.5 / 180 = 0.336 moles
molar concentration of glucose solution = 0.336 / 0.325 = 1.03 M
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molarity
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