Na2SO4 is dissolved in water to make 450mL of a 0.2250 M solution. What is the molar mass of Na2SO4?
1 answer:
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Ethanoic (Acetic) acid is a weak acid and do not dissociate fully. Therefore its equilibrium state has to be considered here.
In this case pH value of the solution is necessary to calculate the concentration but it's not given here so pH = 2.88 (looked it up)
pH = 2.88 ==>![[H^{+}]](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D)
= 
= 0.001 
The change in Concentration Δ![[CH_{3}COOH]](https://tex.z-dn.net/?f=%5BCH_%7B3%7DCOOH%5D)
= 0.001
CH3COOH H+ CH3COOH
Initial 
0 0
Change -0.001 +0.001 +0.001
Equilibrium - 0.001 0.001 0.001
Since the
value is so small, the assumption
![[CH_{3}COOH]_{initial} = [CH_{3}COOH]_{equilibrium}](https://tex.z-dn.net/?f=%5BCH_%7B3%7DCOOH%5D_%7Binitial%7D%20%3D%20%5BCH_%7B3%7DCOOH%5D_%7Bequilibrium%7D)
can be made.
![k_{a} = [tex]= 1.8*10^{-5} = \frac{[H^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]} = \frac{0.001^{2}}{x}](https://tex.z-dn.net/?f=%20k_%7Ba%7D%20%3D%20%5Btex%5D%3D%201.8%2A10%5E%7B-5%7D%20%20%3D%20%20%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5BCH_%7B3%7DCOO%5E%7B-%7D%5D%7D%7B%5BCH_%7B3%7DCOOH%5D%7D%20%3D%20%20%5Cfrac%7B0.001%5E%7B2%7D%7D%7Bx%7D%20)
Solve for x to get the required concentration.
note: 1.)Since you need the answer in 2SF don&t round up values in the middle of the calculation like I've done here.
2.) The ICE (Initial, Change, Equilibrium) table may come in handy if you are new to problems of this kind
Hope this helps!
In this case pH value of the solution is necessary to calculate the concentration but it's not given here so pH = 2.88 (looked it up)
pH = 2.88 ==>
The change in Concentration Δ
CH3COOH H+ CH3COOH
Initial
Change
Equilibrium
Since the
Solve for x to get the required concentration.
note: 1.)Since you need the answer in 2SF don&t round up values in the middle of the calculation like I've done here.
2.) The ICE (Initial, Change, Equilibrium) table may come in handy if you are new to problems of this kind
Hope this helps!
Answer:
126.73 mL
Explanation:
The total pressure of the gas mixture is the sum of the vapor pressure of its constituents. So, the vapor pressure of N₂O(p) can be calculated:
750 = 18.85 + p
p = 750 - 18.85
p = 731.15 torr
It means that for 731.15 torr, N₂O occupied 130 mL. For the general gas equation, we know that
Where <em>p</em> is the pressure, <em>V</em> is the volume, <em>T</em> is the temperature, 1 is the initial state, and 2 the final state. For the same temperatue (21ºC), the equation results on Boyle's law:
p1V1 = p2V2, so:
731.15x130 = 750xV2
750V2 = 95049.5
V2 = 126.73 mL
Hello!
The exercise doesn't say the final temperature, but we are going to assume that the increase is of 20 °C (The procedure is the same, just change the final temperature) To calculate the heat absorbed by water, we need to use the equation for Heat Transfer, in the following way (assuming that the heat increase is 20°C):
So, the Heat absorbed by water is 1297,04 J
Have a nice day!
The exercise doesn't say the final temperature, but we are going to assume that the increase is of 20 °C (The procedure is the same, just change the final temperature) To calculate the heat absorbed by water, we need to use the equation for Heat Transfer, in the following way (assuming that the heat increase is 20°C):
So, the Heat absorbed by water is 1297,04 J
Have a nice day!