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Naily [24]
2 years ago
13

Review the list of common titration errors. Determine whether each error would cause the calculation for moles of analyte to be

too high, too low, or have no effect. Representation of a buret clamped to a ring stand. A funnel is in the top of the buret and a beaker is positioned underneath the buret. A solution is being poured from a bottle into the buret via the funnel. adding titrant past the color change of the analyte solution recording the molarity of titrant as 0.1 M rather than its actual value of 0.01 M spilling some analyte out of the flask during the titration starting the titration with air bubbles in the buret filling the buret above the 0.0 mL volume mark
Chemistry
1 answer:
kakasveta [241]2 years ago
3 0

Answer and Explanation:

<em>A funnel is in the top of the buret and a beaker is positioned underneath the buret:</em> This is correct and is necessary to fill the buret, but the funnel and the beaker has to be removed before the titration starts. The calculation for moles of analyte does not affect.

<em>A solution is being poured from a bottle into the buret via the funnel:</em> Using a funnel helps to fill the burette but it must be removed to filling the buret at 0.0 mL. In this case, the calculation for moles of analyte do not affect.

<em>Adding titrant past the color change of the analyte solution</em>: In this case, an excess of titrant is added, thus the calculation for moles of anality will be higher than it should be.

<em>Recording the molarity of titrant as 0.1 M rather than its actual value of 0.01 M</em>: In this case, the titrant is considered more concentrated than it is hence, the calculation for moles of anality will be higher than it should be.

<em>Spilling some analyte out of the flask during the titration</em>: The excess of titrant spilled out of the flask higher up the volume of titrant measured. Therefore, the calculation for moles of anality will be higher than it should be.

<em>Starting the titration with air bubbles in the buret</em>: The air inside the burette occupies measured volume, thus the volume of titrant measured will be higher than the real volume spilled in the flask. Hence the calculation for moles of anality will be higher than it should be.

<em>Filling the buret above the 0.0 mL volume mark</em>: Some volume of titrant will be spilled inside the flask but will no be measured since the buret measures the titrant below the 0.0mL mark, thus the calculation for moles of anality will be lower than it should be.

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The answer:
<span>The equation of its dissolution in water is: AgNO3 → Ag + (aq)  +  NO3- (aq)

and     </span>AgNO3 →   Ag + (aq)  +  NO3- (aq)
          1 mol          1mol                1mol 
             ?  --------   0.854mo
so for finding the value, it is sufficients to complute 1 x 0.854 mol =0.854 mol
so,  0.854 mol is required for the reaction to form 0.854 mol of Ag

5 0
2 years ago
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What is the empirical formula? A compound is used to treat iron deficiency in people. It contains 36.76% iron, 21.11% sulfur, an
Mandarinka [93]

Answer: The empirical formula of the compound is Fe_1S_1O_4

Explanation:

Empirical formula is defined formula which is simplest integer ratio of number of atoms of different elements present in the compound.

Percentage of iron in a compound = 36.76 %

Percentage of sulfur in a compound = 21.11 %

Percentage of oxygen in a compound = 42.13 %

Consider in 100 g of the compound:

Mass of iron in 100 g of compound = 36.76 g

Mass of iron in 100 g of compound = 21.11 g

Mass of iron in 100 g of compound = 42.13 g

Now calculate the number of moles each element:

Moles of iron=\frac{36.76 g}{55.84 g/mol}=0.658 mol

Moles of sulfur=\frac{21.11 g}{32.06 g/mol}=0.658 mol

Moles of oxygen=\frac{42.13 g}{16 g/mol}=2.633 mol

Divide the moles of each element by the smallest number of moles to calculated the ratio of the elements to each other

For Iron element = \frac{0.658 mol}{0.658 mol}=1

For sulfur element = \frac{0.658 mol}{0.658 mol}=1

For oxygen element = \frac{2.633 mol}{0.658 mol}=4.001\approx 4

So, the empirical formula of the compound is Fe_1S_1O_4

4 0
1 year ago
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A mineral sample is analyzed for its cobalt and calcium content. A sample is dissolved, and then the cobalt and calcium are prec
bekas [8.4K]

Answer:

9.88

Explanation:

As higher is the Ksp, more soluble is the compound. So, Co(OH)₂ is the less soluble hydroxide.

The maximum concentration of it must be 1x10⁻⁶ M, and the reaction is:

Co(OH)₂(s) ⇄ Co⁺²(aq) + 2OH⁻(aq)

So, [Co⁺²] = 1x10⁻⁶M

Ksp =  [Co⁺²] *[OH⁻]²

[OH⁻]² = 5.9x10⁻¹⁵/1x10⁻⁶

[OH⁻] = √(5.9x10⁻⁹)

[OH⁻] = 7.6811x10⁻⁵

pOH = -log[OH⁻]

pOH = -log(7.6811x10⁻⁵)

pOH = 4.11

Knowing that pH + pOH = 14

pH = 14 - 4.11

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7 0
2 years ago
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A metal oxide with the formula mo contains 15.44% oxygen. in the box below, type the symbol for the element represented by m.
lana66690 [7]

<u>Answer:</u> The element represented by M is Strontium.

<u>Explanation:</u>

Let us consider the molar mass of metal be 'x'.

The molar mass of MO will be = Molar mass of oxygen + Molar mass of metal = (16 + x)g/mol

It is given in the question that 15.44% of oxygen is present in metal oxide. So, the equation becomes:

\frac{15.44}{100}\times (x+16)=16g/mol\\\\(x+16)=\frac{16g/mol\times 100}{15.44}\\\\x=(103.626-16)g/mol\\\\x=87.62g/mol

The metal atom having molar mass as 87.62/mol is Strontium.

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CO + 2H2 = CH3OH

The Keq is the ration of the amount of the product and the reactant. We use the ICE table for this. We do as follows:

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I         .42           .42                    0
C     -0.13      -2(0.13)            0.13
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E =    .29           0.16               0.13

Therefore, 

Keq = [CH3OH] / [CO2] [H2]^2 = 0.13 / 0.29 (0.16^2)
Keq = 17.51
4 0
2 years ago
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