Answer:
74mL
Explanation:
Given parameters:
Molar mass of citric acid = 192g/mol
Molar mass of baking soda = 84g/mol
Concentration of citric acid = 0.8M
Mass of baking powder = 15g
Unknown parameters:
Volume of citric acid = ?
Solution
Equation of the reaction:
C₆H₈O₇ + 3NaHCO₃ → Na₃C₆H₅O₇ + 3H₂O + 3CO₂
Procedure:
- We work from the known parameters to the unknown. From the statement of the problem, we can approach the solution from the parameters of the baking powder.
- From the baking powder, we can establish a molar relationship between the two reactants. We employ the mole concept in this regard.
- We find the number of moles of the baking powder that went into the reaction using the expression below:
Number of moles =
Number of moles = = 0.179mole
- From the equation of the reaction, we can find the number of moles of the citric acid:
3 moles of baking powder reacted with 1 mole of citric acid
0.179 moles of baking powder would react with :
This yields 0.059mole of citric acid
- To find the volume of the citric acid, we use the mole expression below:
Volume of citric acid =
Volume of citric acid = = 0.074L
Expressing in mL gives 74mL
Answer: With power and energy, power is units of energy divided by time. The same difference as distance and velocity. The units of power are watts, the units of energy are joules.
Explanation:
Answer:
i) for NaOH (or KOH) = 13.75 mL
ii) for = 13.50 mL.
Explanation:
Phenolphthalein is an indicator which shows change in color when the conditions are highly basic.
Methyl orange is an indicator which shows change in color in presence of highly acidic medium.
For titration of NaOH and HCl we can use phenolphthalein but for sodium carbonate (a weak base) with HCl we use mehtyl orange.
Now in case of mixture of given strong base and weak base the reading or end point obtained from phenolphthalein, shows the neutralization of NaOH only and half neutralization of sodium carbonate.
While the reading or end point of methyl orange shows the neutralization of both the base present.
a) The volume of HCl used for phenolphthalein end point = 20.50 mL
Let us say the volume of HCl used for NaOH = V1
The volume of HCl used for half neutralization of sodium carbonate = V2
(1)
b) the volume of HCl used for methyl organe end point = 27.25
This volume of HCl used for both NaOH and Na₂CO₃
(2)
Equating equation 1 and 2
i) Thus the volume of Acid used for NaOH (Or KOH if present in place of NaOH) = 20.50-6.75= 13.75 mL
ii) the volume of acid used for sodium carbonate =
D.) Atoms are always in motion
Answer:
1.08 g/mL
Explanation:
Density=Mass/Volume therefore you would do the mass which is 7.481 g divided by the volume which is 6.9 mL and get 1.08 g/mL since you combine the two labels!