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Cloud [144]
2 years ago
6

A 5.00-mL sample of blood was treated with trichloroacetic acid to precipitate proteins. After centrifugation, the resulting sol

ution was brought to a pH of 3 and was extracted with two 5-mL portions of methyl isobutyl ketone containing the organic lead complexing agent APCD. The extract was aspirated directly into an air-acetylene flame yielding an absorbance of 0.454 at 283.3 nm. Five-milliliter aliquots of standard solutions containing 0.240 and 0.475 ppm Pb were treated in the same way and yielded absorbances of 0.412 and 0.642. Calculate the concentration Pb (ppm) in the sample assuming that Beer’s law is followed.
Chemistry
1 answer:
Anarel [89]2 years ago
3 0

Explanation:

According to the Beer-lambert' s law,

        Absrobance = absrobptitvity × lenght × concentration

So, if we will plot a graph between absrobance and concentration then we will obtain a straight line

Hence, formula to calculate slope of the graph is as follows.

   The slope of graph = \frac{\text{difference in absrobance at two points}}{\text{difference in concentrations at those points}}

The given data is as follows.

     A_{1} = 0.412 (first point)

     A_{2} = 0.642 (second point)

     C_{1} = 0.240 ppm (first point)

     C_{2} = 0.475 ppm (second point)

Hence, putting these values into the above formula we will calculate the value of slope as follows.

           Slope = \frac{\text{difference in absrobance at two points}}{\text{difference in concentrations at those points}}

                     = \frac{0.642 - 0.412}{0.475 - 0.240}

                      = \frac{0.23}{0.235}

                     = 0.978

It is known that for the graph, line equation can be written as follows.

                       y = mx + c  ........... (1)

where,      C = intercept

                 m = slope

Hence, calculate the value of intercept as follows.

           0.412 = 0.24 \times 0.978 + c

               c = 0.178

As it is given that the absorbance values 0.454 (the y-axis value). Therefore, putting this value into equation (1) we get the following.

                     y = mx + c

                  0.454 = x \times 0.978 + 0.178

                   0.276 = 0.978x

                         x = 0.282 ppm

Thus, we can conclude that the concentration of Pb in the given sample is 0.282 ppm.

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(45 pts) What is the theoretical yield (in g) of iron(III) carbonate that can be produced from 1.72 g of iron(III) nitrate and a
mote1985 [20]

Answer:

1.04g of iron III carbonate

Explanation:

First, we must put down the equation of reaction because it must guide our work.

2Fe(NO3)3(aq) + 3Na2CO3(aq)→Fe2(CO3)3(s) + 6NaNO3(aq)

From the question, we can see that sodium carbonate is in excess while sodium nitrate is the limiting reactant.

Number of moles of iron III nitrate= mass of iron III nitrate reacted/ molar mass of iron III nitrate

Mass of iron III nitrate reacted= 1.72g

Molar mass of iron III nitrate= 241.88 g∙mol–1

Number of moles of iron III nitrate= 1.72g/241.88 g∙mol–1= 7.11×10^-3 moles

From the equation of the reaction;

2 moles of iron III nitrate yields 1 mole of iron III carbonate

7.11×10^-3 moles moles of iron III nitrate yields 7.11×10^-3 × 1/ 2= 3.56×10^-3 moles of iron III carbonate

Theoretical mass yield of iron III carbonate = number of moles of iron III carbonate × molar mass

Theoretical mass yield of iron III carbonate = 3.56×10^-3 moles ×291.73 g∙mol–1 = 1.04g of iron III carbonate

8 0
2 years ago
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A student is given the question: "what is the mass of a gold bar that is 7.379 × 10–4 m3 in volume? the density of gold is 19.3
IRINA_888 [86]

<em>The mass of a gold bar  = 1.424 x 10⁴ gram</em>

<em></em>

<h3><em>Further Explanation</em></h3>

Density is a quantity derived from the mass and volume

density is the ratio of mass per unit volume

With the same mass, the volume of objects that have a high density will be smaller than type

The unit of density can be expressed in g / cm3 or kg / m3

Density formula:

\large{\boxed{{\bold{\rho~=~\frac{m}{V}}}}

ρ = density

m = mass

v = volume

A common example is the water density of 1 gr / cm3

The mass itself is often equated with weight, even though it is different. Mass is the amount of matter in the matter while weight is related to the gravitational force

<em>Known variable</em>

volume gold bar 7.379 × 10⁻⁴ m³

a density of 19.3 g/cm³

<em>Asked</em>

the mass of a gold bar

<em>Answer</em>

volume is known to be 7.379 × 10⁻⁴ m³, then we change it first to units of cm³ adjusting to units of density

7.379 × 10⁻⁴ m³ = 7.379 × 10² cm³

then:

mass = volume x density

mass = 7.379 × 10² cm³ x 19.3 g/cm³

mass = 1.424 x 10⁴ gram

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2 years ago
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. Calculate the mass of O2 produced if 3.450 g potassium chlorate is completely decomposed by heating in presence of a catalyst
Vesnalui [34]
 <span>2 KClO3(s) → 3 O2(g) + 2 KCl(s) 

</span><span>Note: MnO2 (Manganese Dioxide) is not part of the reaction. A catalyst lowers the activation energy and increases both forward and reverse reactions at equal rates. 
</span>
molar mass of KClO3 = 122.5
Moles of KClO3 =  3.45 / 122.55 = 0.028

Moles of O2 produce = \frac{3}{2} \times 0.028

= 0.042 moles

molar mass of O2 = 32

so, mass of O2 = 32 x 0.042  = 1.35 g



5 0
2 years ago
The standard heats of combustion (δh∘) per mole of 1,3-butadiene, c4h6(g); butane, c4h10(g); and h2(g) are −2540.2,−2877.6, and
ryzh [129]

solution:

Hydration is the addition of water; hydrogenation is the addition of hydrogen.  

desire rxn: _C4H6(g) + 2 H2(g)-----> C4H10(g)___dHhy = ??  

knowns:  

__________C4H6 + 11/2 O2 --------> 4CO2 + 3H2O______dHox = -2540.2 kJ/mole  

__________4CO2 + 5H2O -----------> C4H10 + 13/2 O2___-dHox = 2877.6 kJ/mole  

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Basic mathematics is a prerequisite to chemistry – I just try to help you with the methodology of solving the problem


5 0
2 years ago
How many grams of sulfur must be burned to give 100.0 g of So2
andriy [413]

Answer:

50 g of S are needed

Explanation:

To star this, we begin from the reaction:

S(s) + O₂ (g) →  SO₂ (g)

If we burn 1 mol of sulfur with 1 mol of oxygen, we can produce 1 mol of sulfur dioxide. In conclussion, ratio is 1:1.

According to stoichiometry, we can determine the moles of sulfur dioxide produced.

100 g. 1mol / 64.06g = 1.56 moles

This 1.56 moles were orginated by the same amount of S, according to stoichiometry.

Let's convert the moles to mass

1.56 mol . 32.06g / mol = 50 g

4 0
2 years ago
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