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3 years ago

Is the indicator generally added to the titrant or the analyte in a titration?

2 answers:

5 0

In a titration process, the unknown or the analyte with a known volume is placed in a flask and the titrant whose concentration is known is placed in the burette. The indicator in the titration process is generally added to the flask with the analyte.

dsp733 years ago

5 0

Answer:

The correct answer is "the analyte".

Explanation:

Titration is a method of quantitative chemical analysis used to determine the unknown concentration of a reagent from a reagent with a known concentration.

A reagent called a "titrator" of known volume and concentration is used to react with an analyte solution of unknown concentration. The endpoint is the point at which the titration ends, and is determined by the use of an indicator, which is added to the analyte.

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What two quantities must be known to calculate the density of a sample of matter? A) mass and volume B) length and mass C) size

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The gas particles move faster.

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3 years ago

1.15 g of a metallic element needs 300 cm3 of oxygen for complete reaction, at 298 K and 1 atm

sashaice [31]

1) Calculate the number of moles of O2 (g) in 300 cm^3 of gas at 298 k and 1 atm

Ideal gas equation: pV = nRT => n = pV / RT

R = 0.0821 atm*liter/K*mol

V = 300 cm^3 = 0.300 liter

T = 298 K

p = 1 atm

=> n = 1 atm * 0.300 liter / [ (0.0821 atm*liter /K*mol) * 298K] = 0.01226 mol

2) The reaction of a metal with O2(g) to form an ionic compound (with O2- ions) is of the type

X (+) + O2 (g) ---> X2O or

2 X(2+) + O2(g) ----> X2O2 = 2XO or

4X(3+) + 3O2(g) ---> 2X2O3

In the first case, 1 mol of metal react with 1 mol of O2(g); in the second case, 2 moles of metal react with 1 mol of O2(g); in the third, 4 moles of X react with 3 moles of O2(g)

So, lets probe those 3 cases.

3) Case 1: 1 mol of metal X / 1 mol O2(g) = x moles / 0.01226 mol

=> x = 0.01226 moles of metal X

Now you can calculate the atomic mass of the hypotethical metal:

1.15 grams / 0.01226 mol = 93.8 g / mol

That does not correspond to any of the metal with valence 1+

So, now probe the case 2.

4) Case 2:

2moles X metal / 1 mol O2(g) = x / 0.01226 mol

=> x = 2 * 0.01226 = 0.02452 mol

And the atomic mass of the metal is: 1.15 g / 0.02452 mol = 46.9 g/mol

That is similar to the atomic mass of titanium which is 47.9 g / mol and whose valece is 2+.

4) Case 3

4 mol meta X / 3 mol O2 = x / 0.01226 => x = 0.01226 * 4 / 3 = 0.01635

atomic mass = 1.15 g / 0.01635 mol = 70.33 g/mol

That does not correspond to any metal.

Conclusion: the identity of the metallic element could be titanium.

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3 years ago

If the atomic number for oxygen is 8, how

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8 protons are in the nucleus

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3 years ago

What is the percent composition by mass of hydrogen in nh4hco3

Simora [160]

<u>Answer:</u> The percent composition by mass of hydrogen in given compound is 6.33 %

<u>Explanation:</u>

We are given:

A chemical compound having chemical formula of $NH_4HCO_3$

It is made up by the combination of 1 nitrogen atom, 5 hydrogen atoms, 1 carbon atom and 3 oxygen atoms

To calculate the percentage composition by mass of hydrogen in the compound, we use the equation:

$\%\text{ composition of Hydrogen}=\frac{\text{Mass of hydrogen}}{\text{Mass of compound}}\times 100$

Mass of compound = $[(1\times 14)+(5\times 1)+(1\times 12)+(3\times 16)]=79g/mol$

Mass of hydrogen = $(5\times 1)=5g/mol$

Putting values in above equation, we get:

$\%\text{ composition of Hydrogen}=\frac{5g/mol}{79g/mol}\times 100=6.33\%$

Hence, the percent composition by mass of hydrogen in given compound is 6.33 %

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3 years ago