a) when adding to 24 mL of 0.19 M NaOH:first, we need to get moles of HCl = molarity * volume
= 0.19 M * 0.019 L
= 0.00361 moles
then, moles of NaOH = molarity * volume
= 0.19 M * 0.024 L
= 0.00456 moles
NaOH remaining = 0.00456 - 0.00361
= 0.00095 moles
when the total volume = 24mL+19 mL = 43 mL
∴ molarity of NaOH = moles / total volume
= 0.00095 / 0.043L
= 0.0221 M
when POH = -㏒[OH-]
= -㏒0.0221 M
= 1.66
∴PH = 14 - POH
= 14- 1.66
= 12.34
b) when adding to 29 mL of 0.24 M NaOH:when moles of HCl = 0.00361 moles
then, moles of NaOH = molarity * volume
= 0.24 M * 0.029 L
= 0.00696 moles
∴ NaOH remaining = 0.00696 - 0.00361
= 0.00335 moles
the total volume = 29 mL + 19 mL = 48 mL
molarity of NaOH = moles / total volume
= 0.00335 / 0.048L
= 0.0698 M
∴POH = -㏒[OH-]
= -㏒ 0.0698
= 1.16
∴ PH = 14- POH
= 14- 1.16
= 12.84
The answer is B. F=ma, so plug that into the equation (10.8=0.6a) and then solve algebraically, resulting in 18.
The correct answer is 12.044 × 10²³ molecules.
The molecular mass of H₂S is 34 gram per mole.
Number of moles is determined by using the formula,
Number of moles = mass/molecular mass
Given mass is 68 grams, so no of moles will be,
68/34 = 2 moles
1 mole comprises 6.022 × 10²³ molecules, therefore, 2 moles will comprise = 6.022 × 10²³ × 2
= 12.044 × 10²³ molecules.
The given image shows a reaction between an acid and a alcohol that results in the formation of ester and water. This reaction is an example of Esterification reaction.
Here it can be seen that a acetic acid reacts with ethanol that forms ethyl ethanoate, that is an ester and water molecule that is H₂O. So this reaction is an example of esterification reaction.