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goldenfox [79]
3 years ago
15

What happens if I overshoot the endpoint in titration?

Chemistry
1 answer:
xxTIMURxx [149]3 years ago
5 0
Your solution in the conical flask will be acidic since you will have gone beyond the amount of acid that should completely neutralise the base completely. thax
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(b) If 45.0% of the ethanol that did not produce the ether reacts by the side reaction, what mass (g) of ethylene is produced?
Finger [1]

Ethylene is a hydrocarbon which has the formula C ₂H ₄ or H₂C=CH₂. It is a colourless flammable gas with a faint "sweet and musky" odour when pure. It is the simplest alkene.

Ethylene is widely used in the chemical industry, and its worldwide production exceeds that of any other organic compound.

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Medical: Ethylene is used as an anaesthetic. Metal Fabrication: Ethylene is used as oxy-fuel gas in metal cutting, welding and high velocity thermal spraying. Refining: Ethylene is used as refrigerant, especially in LNG liquefaction plants. Rubber & Plastics: Ethylene is used in the extraction of rubber.

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3 0
1 year ago
Place the following in order of decreasing radius.<br><br> Te2- F- O2-
notka56 [123]

Answer:

Option 2 is correct.

6 0
2 years ago
what is the approximate ph at the equivalence point of a weak acid-strong base titration if 25 ml of aqueous formic acid require
vivado [14]

Answer:

pH at equivalence point is 8.52

Explanation:

HCOOH+NaOH\rightarrow HCOO^{-}Na^{+}+H_{2}O

1 mol of HCOOH reacts with 1 mol of NaOH to produce 1 mol of HCOO^{-}

So, moles of NaOH used to reach equivalence point equal to number of moles HCOO^{-} produced at equivalence point.

As density of water is 1g/mL, therefore molarity is equal to molality of an aqueous solution.

So, moles of HCOO^{-} produced = \frac{29.80\times 0.3567}{1000}moles=0.01063moles

Total volume of solution at equivalence point = (25+29.80) mL = 54.80 mL

So, at equivalence point concentration of HCOO^{-} = \frac{0.01063\times 1000}{54.80}M=0.1940M

At equivalence point, pH depends upon hydrolysis of HCOO^{-}. So, we have to construct an ICE table.

HCOO^{-}+H_{2}O\rightleftharpoons HCOOH+OH^{-}

I: 0.1940                                   0                 0

C: -x                                          +x               +x

E: 0.1940-x                                x                  x

So, \frac{[HCOOH][OH^{-}]}{[HCOO^{-}]}=K_{b}(HCOO^{-})=\frac{10^{-14}}{Ka(HCOOH)}

species inside third bracket represent equilibrium concentrations

So, \frac{x^{2}}{0.1940-x}=5.56\times 10^{-11}

or,x^{2}+(5.56\times 10^{-11}\times x)-(1.079\times 10^{-11})=0

So, x=\frac{-(5.56\times 10^{-11})+\sqrt{(5.56\times 10^{-11})^{2}+(4\times 1.079\times 10^{-11})}}{2}

So, x=3.285\times 10^{-6}M

So, pH=14-pOH=14+log[OH^{-}]=14+logx=14+log(3.285\times 10^{-6})=8.52

5 0
3 years ago
Ok, I need helpies! .__.
yKpoI14uk [10]
It is in a large elliptical shape!
3 0
3 years ago
How many significant figures does 4,982cm have
Elodia [21]

Answer:

4 significant figures.

Explanation:

All non-zero numbers are significant figures.

8 0
3 years ago
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