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Natali [406]
3 years ago
13

How many mL of 50% (v/v) alcohol should be mixed with 10% alcohol to obtain 50mL of 35% (v/v) alcohol? Round to the hundredths p

lace.
Chemistry
1 answer:
Furkat [3]3 years ago
8 0

Answer:

The volume of 50% alcohol used to make the mixture = 31.25 L

The volume of 10% alcohol used to make the mixture = 18.75 L

Explanation:

Let the volume of 50% alcohol used to make the mixture = x L

Let the volume of 10% alcohol used to make the mixture = y L

Total volume of the mixture = x + y = 50 L .................. (1)

For 50% alcohol:

C₁ = 50% , V₁ = x L

For 10% alcohol :

C₂ = 10% , V₂ = y L

For the resultant alcohol solution:

C₃ = 35% , V₃ = 50 L

Using  

C₁V₁ + C₂V₂ = C₃V₃

50×x + 10×y = 35×50

So,  

5x + y = 175 .................. (2)

Solving 1 and 2 we get,

<u>x = 31.25 L </u>

<u>y = 18.75 L</u>

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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
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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

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When the forest burns we will lose a huge amount of oxygen and animals will lose their home.

Explanation:

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Explain the process of initiation and elongation.
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Initiation <span>occurs when mRNA, tRNA, and an amino acid meet up inside the ribosome. </span>Elongation occurs when amino acids are continually added to the line, forming a long chain bound together by peptide bonds.
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3 years ago
Atoms of metallic elements tend to...
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The answer is: lose electrons and form positive ions.

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For example, magnesium has atomic number 12, which means it has 12 protons and 12 electrons. It lost two electrons to form magnesium cation (Mg²⁺) with stable electron configuration like closest noble gas neon (Ne) with 10 electrons.  

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3 0
3 years ago
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Margaret [11]
As mentioned above, phosphoric acid has 3 pKa values, and after 3 ionization it gives 3 types of ions at different pKa values:

H₃PO₄(aq) + H₂O(l) ⇌ H₃O⁺(aq) + H₂PO₄⁻ (aq)         pKₐ₁ 
<span>

</span>H₂PO₄⁻(aq) + H₂O(l) ⇌ H₃O⁺(aq) + HPO₄²⁻ (aq)       pKₐ₂


HPO₄²⁻(aq) + H₂O(l) ⇌ H₃O⁺(aq) + PO₄³⁻ (aq)          pKₐ₃ 


At the highest pKa value (12.4) of phosphoric acid, the last OH group will lose its hydrogen. On the picture I attached, it is shown required protonated form of phosphoric acid before reaction whose pKa value is 12.4.


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