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Yuri [45]
3 years ago
12

A solution is created by dissolving 13.5 grams of ammonium chloride in enough water to make 235 ml of solution. how many moles o

f ammonium chloride are present in the resulting solution?
Chemistry
1 answer:
erica [24]3 years ago
7 0
<span>Moles = 0.252 Molarity = 1.07 This question is badly worded. You're asking for moles and I suspect you really want molarity. The number of moles of ammonium chloride you have in the solution will remain constant regardless of the volume of the solution. However, the molarity of the solution will differ depending upon how concentrated it is. So I'll give you both the number of moles of ammonium chloride you have, and the molarity of the resulting solution. Please talk to your teacher if you're confused by the difference between moles and molarity. The formula for ammonium chloride is NH4Cl. So let's calculate it's molar mass. Start by looking up the associated atomic weights. Atomic weight nitrogen = 14.0067 Atomic weight hydrogen = 1.00794 Atomic weight chlorine = 35.453 Molar mass NH4Cl = 14.0067 + 4 * 1.00794 + 35.453 = 53.49146 g/mol Moles NH4Cl = 13.5 g / 53.49146 g/mol = 0.252376735 mol Molarity is defined as moles per liter, so let's divide the number of moles we have by the volume in liters. So: 0.252376735 mol / 0.235 l = 1.073943551 M Rounding to 3 significant figures gives: 0.252 moles, 1.07 molarity.</span>
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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! 



8 0
3 years ago
Why is acid added to water during eletrolysis
DedPeter [7]

Answer:

Pure water is a non conductor of electricity and dilute acids in their aqueous solutions form free ions, which conducts electricity. Thus when we need to electrolyse water, a dilute acid is added to increase its conductivity.

4 0
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