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andreyandreev [35.5K]
3 years ago
11

In which situation are the individual molecules moving the fastest? A) in a bowl of hot soup B) in a glass of iced tea C) in a g

lass of tap water D) in a cup of boiling water\
Chemistry
2 answers:
Airida [17]3 years ago
5 0
<h2>Answer : Option D) In a cup of boiling water</h2><h3>Explanation :</h3>

The individual molecules of water while boiling will move fastest amongst the given choices. As the boiling of water will cause the liquid molecules to increase the intermolecular forces of attraction between molecules and get escaped into vapors. Here, there is a transformation of state that is about to occur from liquid to gaseous. Compared to other examples, where there are no change of state occurring.

tresset_1 [31]3 years ago
5 0

D) in a cup of boiling water

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<u>Satelites</u> are a type of space technology utilizes the electromagnetic spectrum to collect information.

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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
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