Answer:
after 45 days 9 g left
Explanation:
Given data:
Half life Na-24 = 15 days
Mass of sample = 72 g
Mass remain after 45 days = ?
Solution:
Number of half lives passed:
Number of half lives = time elapsed / half life
Number of half lives = 45 days / 15 days
Number of half lives = 3
At time zero total amount = 72 g
After first half life = 72 g/ 2= 36 g
At 2nd half life = 36 g/2 = 18 g
At 3rd half life = 18 g/2 = 9 g
Thus after 45 days 9 g left.
The one that will dissolve most quickly is : D. a tablespoon of ultrafine sugar
Unlike any other type of sugar, ultrafine sugar tend to have smaller crystal size. Which means that this substance will dissolve easier if it put into liquid such as tea or water
hope this helps
First you need to know the molecular weight of sugar (C6H12O6) which is 180.156g/mol
You have half a mole so you have 90.078g
If you wanted to make 1L of a 1.2M solution of glucose you would need 180.156*1.2=216.1872g
But you only have 90.078g
So you need to figure out how much this 90.078g will make if the solution must be 1.2M:
90.078g/216.1872g=xL/1L
solve for the X and you get 0.416666666...
so 416.7ml or 0.417L
Answer:
(3) 5.36
Explanation:
Since this is a titration of a weak acid before reaching equivalence point, we will have effectively a buffer solution. Then we can use the Henderson-Hasselbalch equation to answer this question.
The reaction is:
HAc + NaOH ⇒ NaAc + H₂O
V NaOH = 40 mL x 1 L/1000 mL = 0.040 L
mol NaOH reacted with HAc = 0.040 L x 0.05 mol/L = 0.002 mol
mol HAC originally present = 0.050 L x 0.05 mol/L = 0.0025 mol
mol HAc left after reaction = 0.0025 - 0.002 = 0.0005
Now that we have calculated the quantities of the weak acid and its conjugate base in the buffer, we just plug the values into the equation
pH = pKa + log ((Ac⁻)/(HAc))
(Notice we do not have to calculate the molarities of Ac⁻ and HAc because the volumes cancel in the quotient)
pH = -log (1.75 x 10⁻⁵) + log (0.002/0.0005) = 5.36
THe answer is 5.36
