Answer:
Explanation:
a ) Total mixture = 4.656 g
Sand recovered = 2.775 g
percent composition of sand in the mixture
= (2.775 g / 4.656 g ) x 100
= 59.6 % .
b )
Total of sand and salt recovered = 2.775 g + .852 g = 3.627 g .
Total mixture = 4.656 g
percent recovery = (3.627 / 4.656 ) x 100
= 77.9 % .
Assume 1 tsp is approximately can hold 5 mL liquid.
Given the dose of medicine = 1.5 tsp
Converting 1.5 tsp to mL:
= 7.5 mL
Given the specific gravity of the medicine = 1.23
That means density of the medicine with respect to water will be 1.23
As the density of water is 1 g/mL
We can take density of the medicine to be 1.23 g/mL
Calculating the mass of medicine in grams:

9.225 g medicine is present in one dose.
Answer:
pH = 11.95≈12
Explanation:
Remember the reaction among aqueous acetic acid (
) and aqueous sodium hydroxide (NaOH)

First step. Need to know how much moles of the substances are present
= 0.0025 mol NaOH
0.003 mol NaOH *
/ 1 mol NaOH = 0.003 mol CH_3COOH[/tex]
NaOH is in excess. Now, how much?
0.003 mole NaOH - 0.0025 mole NaOH = 0.0005 mole NaOH
Then, that amount in excess would be responsable for the pH.
Third step. Know the pH
Remember that pH= -log[H+]
According to the dissociation of water equilibrium
Kw=[H+]*[OH-]= 10^(-14)
The dissociation of NaOH is
NaOH -> 
Now, concentration of OH^{-}[/tex] would be given for the excess of NaOH.
[OH-]= 0.0005 mole / 0.055 L = 0.00909 M
Careful: we have to use the total volumen
Les us to calculate pH
![pH= -log [H+]\\pH= -log \frac{K_w}{[OH-]} \\pH= 11.95](https://tex.z-dn.net/?f=pH%3D%20-log%20%5BH%2B%5D%5C%5CpH%3D%20-log%20%5Cfrac%7BK_w%7D%7B%5BOH-%5D%7D%20%5C%5CpH%3D%2011.95)
Pure sodium can severely harm your body (toxic) which can lead to death