Why is tin tin and aluminum the better choice for canned milk
1 answer:
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The answer has to depend on the mass of the copper that you're trying to melt.
But there is a formula which you can calculate the energy required if you have the information of the mass. This formula requires the value called specific latent heat of fusion, which is the energy required to melt or freeze a specific mass of copper without changing its temperature but change its state.
E is the total energy required, m is the mass, and lv is the specific latent heat of fusion of copper.
Usually, in questions, lv is given. But we can also look it up online which is around 205kJ kg^-1
this means, to melt or freeze a copper, we need to work the mass multiplied by 205kJ in order to calculate its total energy needed or released.
But there is a formula which you can calculate the energy required if you have the information of the mass. This formula requires the value called specific latent heat of fusion, which is the energy required to melt or freeze a specific mass of copper without changing its temperature but change its state.
E is the total energy required, m is the mass, and lv is the specific latent heat of fusion of copper.
Usually, in questions, lv is given. But we can also look it up online which is around 205kJ kg^-1
this means, to melt or freeze a copper, we need to work the mass multiplied by 205kJ in order to calculate its total energy needed or released.
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!