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Andrej [43]
2 years ago
8

Gaseous butano (CH,(CH),CH) will react with gaseous oxygen (0) to produce gaseous carbon dioxide (CO2) and goseous woter (11,0).

suppose
20.9 g of butane is mixed with 54, 9 of oxypen. Calculate the minimum mass of butone that could be left over by the chemical reaction. Be sure your answer
has the correct number of significant digits.
Chemistry
1 answer:
slega [8]2 years ago
5 0

Answer:

Firstly, We have to convert it in the Miles formula...

No. of moles = Mass given/Molar Mass

So, the final answer be come<em> </em>

<h3><em><u> </u></em><em><u>Ans</u></em><em><u> </u></em><em><u>-</u></em><em><u> </u></em><em><u>5</u></em><em><u>0</u></em><em><u>.</u></em><em><u>8</u></em><em><u> </u></em><em><u>gm</u></em><em><u> </u></em><em><u>and</u></em><em><u> </u></em><em><u>there</u></em><em><u> </u></em><em><u>same</u></em><em><u> </u></em><em><u>%</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>5</u></em><em><u>0</u></em><em><u>.</u></em><em><u>8</u></em><em><u>%</u></em><em><u> </u></em><em><u>butane</u></em><em><u> </u></em><em><u>in</u></em><em><u> </u></em><em><u>react</u></em><em><u>ion</u></em><em><u> </u></em></h3>
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A 25.0 ml sample of 0.150 m benzoic acid is titrated with a 0.150 m naoh solution. what is the ph at the equivalence point? the
Ad libitum [116K]

Solution:

At the equivalence point, moles NaOH = moles benzoic acid  

HA + NaOH ==> NaA + H2O where HA is benzoic acid  

At the equivalence point, all the benzoic acid ==> sodium benzoate  

A^- + H2O ==> HA + OH- (again, A^- is the benzoate anion and HA is the weak acid benzoic acid)  

Kb for benzoate = 1x10^-14/4.5x10^-4 = 2.22x10^-11  

Kb = 2.22x10^-11 = [HA][OH-][A^-] = (x)(x)/0.150  

x^2 = 3.33x10^-12  

x = 1.8x10^-6 = [OH-]  

pOH = -log [OH-] = 5.74  

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5 0
3 years ago
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Remember to balance the equations first
Semenov [28]

mol CO₂ = 9.6

mol N₂ = 4.8

mol O₂ = 0.8

mol H₂O = 8

<h3>Further explanation</h3>

Reaction

4C₃H₅O₉N₃ → 12CO₂ + 6N₂ + O₂ + 10H₂O

mol CO₂

\tt \dfrac{12}{4}\times 3.2=9.6

mol N₂

\tt \dfrac{6}{4}\times 3.2=4.8

mol O₂

\tt \dfrac{1}{4}\times 3.2=0.8

mol H₂O

\tt \dfrac{10}{4}\times 3.2=8

8 0
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Why was the international system of units adopted?
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To make everything easier to convert so everything universal

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The balanced equation below shows the products that are formed when pentane (C₅H₁₂) is combusted. <img src="https://tex.z-dn.net
masha68 [24]

Answer:

Option c → 8:1

Explanation:

This is the reaction:

C₅H₁₂ + 8O₂ → 10CO₂ + 6H₂O

1 mol of pentane needs 8 moles of oxygen to be combusted and this combustion produces 10 mol of carbon dioxide and 6 moles of water.

To determine the ratio, look the stoichiometry.

For every 8 moles of oxygen, I need 1 mole of pentane gas.

7 0
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