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hammer [34]
3 years ago
6

What would be the effect of each of the following sources of error on the molarity of H2SO4 determined in this experiment? State

whether the concentration of H2SO4 obtained would be too high or too low. Explain your answer briefly.
a. The burette is contaminated with an acid solution.
[1 mark]

•

b. The burette contains a large air bubble in the tip, which disappears during the titration [1mark]

c. A small volume of the acid is spilled when you transfer it into the Erlenmeyer flask
[1 mark]
Chemistry
1 answer:
Dafna11 [192]3 years ago
7 0

Answer:

The burette is contaminated with an acid solution- The concentration is too high

The burette contains a large air bubble in the tip, which disappears during the titration- The concentration of the acid is too low.

A small volume of the acid is spilled when you transfer it into the Erlenmeyer flask- The concentration of the acid is too low.

Explanation:

The molality of a solution is the number of moles of solute per liter of solution. The concentration of an unknown solution is obtained by titration against a standard solution of acid or base whose concentration  is known. It is a volumetric method of analysis.

If the burette is contaminated with the acid, then the concentration of the base that reacts with the acid is less than the stated amount and consequently the concentration of the acid calculated is higher than it should be.

If the tip of the burette contains air bubbles, the volume of the acid reported is less than the volume of acid actually delivered during the titration hence the calculated concentration of the acid is too low.

If some volume of acid is spilled when you transfer the acid into the Erlenmeyer flask, the volume of the acid reacted is decreased and consequently the calculated concentration of the acid is too low.

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An electrolysis reaction is
Llana [10]

Answer:

the electrolysis reaction is a non- spontaneous reaction

Explanation:

Since electrons flow from it, the anode in an electrolytic cell is positive, while the cathode is negative when electrons flow into it. The device functions like a galvanic cell in that direction. In an electrolytic cell, an external voltage is applied and that is what causes a non spontaneous reaction

3 0
2 years ago
A compound is 80.0% carbon and 20.0% hydrogen by mass. assume you have a 100.-g sample of this compound. the molar mass of the c
ch4aika [34]
Basis of the calculation: 100g
 
For Carbon: 
 Mass of carbon = (100 g)(0.80) = 80 g
  Number of moles of carbon = (80 g)(1 mole / 12g) = 20/3

For Hydrogen:
  Mass of hydrogen = (100 g)(0.20) = 20 g
     Number of moles of hydrogen = (20 g)(1 mole / 1 g) = 20

Translating the answer to the formula of the substance,
     C20/3H20

Dividing the answer,
    CH3

The molar mass of the empirical formula is:
    12 + 3 = 15 g/mol

Since, the molar mass given for the molecular formula is 30.069 g/mol, the molecular equation is,
    C2H6

ANSWER: C2H6

 
4 0
3 years ago
If you were sitting near a gram of protactinium-234 and a gram of uranium-234, both solid materials, which would you consider mo
Novosadov [1.4K]

aswer: protactinium-234 is more dang erous than uranium

6 0
2 years ago
How many atoms are there in 8.88 g Si?
Mariana [72]

Answer:

\boxed {\boxed {\sf 1.90 \times 10^{23} \ atoms \ Si}}

Explanation:

We are asked to find how many atoms are in 8.88 grams of silicon.

<h3>1. Grams to Moles </h3>

First, we convert grams to moles. We use the molar mass or the mass of 1 mole of a substance. These values are found on the Periodic Table as they are equal to the atomic masses, but the units are grams per mole instead of atomic mass units.

Look up silicon's molar mass.

  • Si:  28.085 g/mol

We will convert using dimensional analysis. Set up a conversion factor with the molar mass.

\frac { 28.085 \ g \  Si}{1 \ mol \ Si}

We are converting 8.88 grams of silicon to moles, so we multiply by this value.

8.88 \ g \ Si *\frac { 28.085 \ g \  Si}{1 \ mol \ Si}

Flip the fraction so the units of grams of silicon cancel.

8.88 \ g \ Si *\frac{1 \ mol \ Si} { 28.085 \ g \  Si}

8.88  *\frac{1 \ mol \ Si} { 28.085 }

\frac {8.88} { 28.085 } \ mol \ Si

0.316183015845 \ mol \ Si

<h3>2. Moles to Atoms </h3>

Next, we convert moles to atoms. We use Avogadro's Number or 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this case, the particles are atoms of silicon.

Set up another conversion factor.

\frac {6.022 \times 10^{23} \ atoms \ Si}{1 \ mol \ Si}

Multiply by the number of moles we calculated.

0.316183015845\ mol \ Si *\frac {6.022 \times 10^{23} \ atoms \ Si}{1 \ mol \ Si}

The units of moles of silicon cancel.

0.316183015845 * \frac {6.022 \times 10^{23} \ atoms \ Si}{1}

0.316183015845 * {{6.022 \times 10^{23} \ atoms \ Si}

1.90405412 \times 10^{23} \ atoms \ Si

<h3>3. Significant Figures</h3>

The original measurement of 8.88 grams has 3 significant figures, so our answer must have the same.

For the number we calculated, that is the hundredth place. The 4 in the thousandth place tells us to leave the 0 in the hundredth place.

1.90 \times 10^{23} \ atoms \ Si

<u>8.88 grams of silicon contains 1.90 ×10²³ atoms of silicon.</u>

6 0
2 years ago
The molecular weight of a gas is ________ g/mol if 3.5 g of the gas occupies 2.1 l at stp
bija089 [108]
<span>Pre-1982 definition of STP: 37 g/mol Post-1982 definition of STP: 38 g/mol This problem is somewhat ambiguous because the definition of STP changed in 1982. Prior to 1982, the definition was 273.15 K at a pressure of 1 atmosphere (101325 Pascals). Since 1982, the definition is 273.15 K at a pressure of exactly 100000 Pascals). Because of those 2 different definitions, the volume of 1 mole of gas is either 22.414 Liters (pre 1982 definition), or 22.71098 liters (post 1982 definition). And finally, there's entirely too many text books out there that still use the 35 year obsolete definition. So let's solve this problem using both definitions and you need to pick the correct answer for the text book you're using. First, determine how many moles of gas you have. Just simply divide the volume you have by the molar volume. Pre-1982: 2.1 / 22.414 = 0.093691443 moles Post-1982: 2.1 / 22.71098 = 0.092466287 moles Now determine the molar mass. Simply divide the mass by the moles. So Pre-1982: 3.5 g / 0.093691443 moles = 37.35666667 g/mol Post-1982: 3.5 g / 0.092466287 moles = 37.85163333 g/mol Finally, round to 2 significant figures. So Pre-1982: 37 g/mol Post-1982: 38 g/mol</span>
5 0
3 years ago
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