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tatiyna
2 years ago
15

Some table salt substitutes have potassium chloride in them. If you consume potassium chloride, are you truly eating potassium a

nd chlorine atoms? Support your answer with chemical evidence.
Chemistry
1 answer:
4vir4ik [10]2 years ago
8 0

Answer:

You are consuming both

Explaination

2Na(s)+Cl two(g)

That gives us 2NaCl(s)

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Ilia_Sergeevich [38]

Missing question: Z:X = 7:1 A:Z = 2.5:1 A:Z = 2.2:1 Y:X = 11:1.

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Law of multiple proportions or Dalton's Law said that the ratios of the masses of the second element which combine with a fixed mass of the first element will be ratios of small whole numbers. 
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You would expect snow to fail at the peak or the top because the weather is coldest there.
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The chemical equation of rusting of iron is given
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2 years ago
If acetic acid is the only acid that vinegar contains (ka=1.8×10−5), calculate the concentration of acetic acid in the vinegar.
kicyunya [14]
Ethanoic (Acetic) acid is a weak acid and do not dissociate fully. Therefore its equilibrium state has to be considered here.

CH_{3}COOH \ \textless \ ---\ \textgreater \   H^{+} + CH_{3}COO^{-}

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^{+}]  = 10^{-2.88} =  0.001 moldm^{-3}

The change in Concentration Δ [CH_{3}COOH]= 0.001 moldm^{-3}


                                  CH3COOH          H+           CH3COOH    
Initial  moldm^{-3}                      x           0                     0
                                                                                                                       
Change moldm^{-3}        -0.001            +0.001           +0.001
                                                                                                       
Equilibrium moldm^{-3}      x- 0.001      0.001             0.001
                                                                              

Since the k_{a} value is so small, the assumption 
[CH_{3}COOH]_{initial} = [CH_{3}COOH]_{equilibrium} can be made.

k_{a} = [tex]= 1.8*10^{-5}  =  \frac{[H^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]} =  \frac{0.001^{2}}{x}

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! 



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