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Gala2k [10]
3 years ago
15

Can i titrate a solution of unknown concentration with another solution of unknown concentration and still get a meaningful answ

er
Chemistry
1 answer:
olga2289 [7]3 years ago
6 0

Answer:

No, I don't think so at least

Explanation:

In chemistry, you do calculations to find the concentration of a solution with another solution of KNOWN concentration. Without concentrations of either solution, were would you get values from? Where would you start? :3

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a simple voltaic cell is made by immersing one zinc plate and one copper plate inside water diluted sulfuric acid solution.

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Doctors use the radioactive isotope chromium-51 to label red blood cells in the human body. Chromium-51 gives off relatively ___
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Alpha particles are relatively heave and can be stopped by a sheet of paper.
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Please help!!<br><br>Show the electronic configuration of magnesium atom​.
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Explanation:

Magnesium has atomic number 12. It will be distributed in K, L, M shell in the following way:

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18 An important environmental consideration is the appropriate disposal of cleaning solvents. An environmental waste treatment c
Katyanochek1 [597]

Answer:

a) Percentage by mass of carbon: 18.3%

   Percentage by mass of hydrogen: 0.77%

b)  Percentage by mass of chlorine: 80.37%

c) Molecular formula: C_{2} H Cl_{3}

Explanation:

Firstly, the mass of carbon must be determined by using a conversion factor:

0.872g CO _{2} *\frac{12g C}{44g CO_{2} } = 0.238g CO_{2}

The same process is used to calculate the amount of hydrogen:

0.089g H_{2}O*\frac{2g H}{18g H_{2}O }  = 0.010g H

The percentage by mass of carbon and hydrogen are calculated as follows:

%C\frac{0.238g}{1.3g} *100%= 18.3%

%H\frac{0.010g}{1.3g} *100%=0.77%

From the precipation data it is possible obtain the amount of chlorine present in the compound:

1.75 AgCl*\frac{35.45g Cl}{143.45g AgCl}= 0.43g AgCl

Let's calculate the percentage by mass of chlorine:

%Cl=\frac{0.43g}{0.535g} * 100%= 80.37%

Assuming that we have 100g of the compound, it is possible to determine the number of moles of each element in the compound:

18.3g C*\frac{1mol C}{12g C} = 1.52mol C

0.77g H*\frac{1mol H}{1g H} = 0.77mol H

80.37gCl*\frac{1molCl}{35.45g Cl} = 2.27mol Cl

Dividing each of the quantities above by the smallest (0.77mol), the  subscripts in a tentative formula would be

C=\frac{1.52}{0.77} = 1.97 ≈ 2

H = \frac{0.77}{0.77} = 1

Cl =\frac{2.27}{0.77}=2.94≈3

The empirical formula for the compound is:

C_{2} H Cl_{3}

The mass of this empirical formula is:

mass of C + mass of H + mass of Cl= 24g +1+ 106.35 =131.35g

This mass matches with the molar mass, which means that the supscript in the molecular formula are the same of the empirical one.

5 0
3 years ago
A gas occupies 800ml at a temperature of 27C. What is the volume at 132C?
pishuonlain [190]

V
1
​
/T
1
​
=V
2
​
/T
2
​

(900.0 mL) / (300.0 K) = (x) / (405.0 K); x = 1215 mL.

Change the 900 to 800, and the 300 to 27, then change the 405 to 132. And solve
3 0
3 years ago
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