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Yuliya22 [10]
3 years ago
5

What is the concentration of the base (KOH) in the following titration, given the data below?

Chemistry
2 answers:
omeli [17]3 years ago
8 0

HBr and KOH, molarity is equal to normality. Substituting the known values,

          (0.75 M) x (22.6 mL - 0 mL) = Nb x (37.5 mL - 0.5 mL) 

                                         Nb = 0.46 N

                                          Mb = 0.46 M

Anettt [7]3 years ago
4 0
To calculate the concentration of the base based on the titration, the concept used is the equal of number of equivalence of the acid used to that of the base. From this,
                                 Na x Va = Nb x Vb
For HBr and KOH, molarity is equal to normality. Substituting the known values,
          (0.75 M) x (22.6 mL - 0 mL) = Nb x (37.5 mL - 0.5 mL) 
                                         Nb = 0.46 N
                                          Mb = 0.46 M
Thus, the concentration of the base is approximately 0.46 M. 
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louder sounds

Explanation:

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3 years ago
The complete combustion of a sample of propane produced 2.641 grams of carbon dioxide and 1.442 grams of water as the only produ
ohaa [14]
A general equation for a combustion reaction would be expressed as follows:

CxHy + (x+y/2)O2 = xCO2 + y/2H2O

Propane would obviously would only have carbon and hydrogen in its structure. Assuming a complete combustion, all of the carbon atoms would go to carbon dioxide and all of the hydrogen atoms to water. To determine the empirical, we determine the number of carbon and hydrogen atoms present.

moles C = 2.461 g CO2 ( 1 mol / 44.01 g ) ( 1 mol C / 1 mol CO2 ) = 0.06 mol C

moles H = 1.442 g H2O ( 1 mol / 18.02 g ) ( 2 mol H / 1 mol H ) = 0.16 mol H

Then, we divide the smallest amount to the each mole of the atoms. We do as follows:

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Then we multiply a number in order to obtain a whole number ratio between the atoms.

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8 0
3 years ago
I need help, plz help me with this problem
Svetlanka [38]

Answer:

It's b

Explanation:

I had the same exact question

5 0
2 years ago
Acetic acid has a Kb of 2.93 °C/m and a normal boiling point of 118.1 °C. What would be the boiling point of a solution made by
alexdok [17]

Boiling point elevation is given as:

ΔTb=iKbm

Where,

ΔTb=elevation in the boiling point

that is given by expression:

ΔTb=Tb (solution) - Tb (pure solvent)

Here Tb (pure solvent)=118.1 °C

i for CaCO3= 2

Kb=2.93 °C/m

m=Molality of CaCO₃:

Molality of CaCO₃=Number of moles of CaCO₃/ Mass of solvent (Kg)

=(Given Mass of CaCO3/Molar mass of CaCO₃)/ Mass of solvent (Kg)

=(100.0÷100 g/mol)/0.4

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So now putting value of m, i and Kb in the boiling point elevation equation we get:

ΔTb=iKbm

=2×2.93×2.5

=14.65 °C

boiling point of a solution can be calculated:

ΔTb=Tb (solution) - Tb (pure solvent)

14.65=Tb (solution)-118.1

Tb (solution)=118.1+14.65

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3 0
3 years ago
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Svetradugi [14.3K]

Answer:

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4 0
3 years ago
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