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poizon [28]
3 years ago
9

Suppose the concentration of the standard stock solution is 50.04 ppm and that you delivered 4.60 mL of this standard stock solu

tion from your buret into your 25.00mL volumetric flask. What is the concentration of the standard dilution?
Chemistry
1 answer:
12345 [234]3 years ago
4 0

Answer:

9.21 ppm

Explanation:

Considering

concentration_{working\ solution}\times Volume_{working\ solution}=concentration_{stock\ solution}\times Volume_{stock\ solution}

Given  that:

concentration_{working\ solution}=?

Volume_{working\ solution}=25.00mL

Volume_{stock\ solution}=4.60mL

concentration_{stock\ solution}=50.04 ppm

So,  

concentration_{working\ solution}\times 25.00mL=50.04 ppm\times 4.60mL

concentration_{working\ solution}=\frac{50.04\times 4.60}{25.00}\ ppm=9.21\ ppm

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All of the following institutions awards certificates except: a.) Community college b.) liberal arts college c. Junior college d
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a solution with a transmittance of 0.44 is analyzed in a spectrophotometer with 6% stray light. calculate the absorbance reporte
IgorLugansk [536]

The absorbance reported by the defective instrument was 0.3933.

Absorbance A = - log₁₀ T

Tm = transmittance measured by spectrophotometer

Tm = 0.44

Absorbance reported in this equipment = -log₁₀ (0.44) = 0.35654

True absorbance can be calculated by true transmittance, Tm = T+S(α-T)

S = fraction of stray light = 6%= 6/100 = 0.06

α= 1, ideal case

T = true transmittance of the sample

Tm = T+S(α-T)

now, T= Tm-S/ 1-S = 0.44-0.06/ 1-0.06 = 0.404233

therefore, actual reading measured is A = -log₁₀ T = -log₁₀ (0.404233)

i.e; 0.3933

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3 0
1 year ago
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)
stiv31 [10]

¹/3 C3H8(g) + ⁵/3 O2(g)

Explanation:

The coefficient before every molecule is representative of the number of moles. We can represent it in ration form so as to calculate the question;

C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(l) means;

For every 1 mole of C₃H₈(g) and 5 moles of  O₂(g) produces  3 moles of  CO₂(g) and 4 moles of H₂O(l).

Therefore to produce 1.00 mole of CO₂(g);

We represent it in ratio;

C₃H₈(g) : CO₂(g)

1  :     3

What about ;

? (x)   :  1

We cross multiply;

3x = 1 * 1

X = 1/3

We evaluate the same for O₂;

O₂(g) : CO₂(g)

5 :     3

What about

? (x) :     1

3x = 5 * 1

x = 5/3

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7 0
3 years ago
If the theoretical yield of a reaction is 0.110 g and the actual yield is 0.104 g , what is the percent yield?
Westkost [7]
Theoretical Yield is an Ideal yield with 100 % conversion of reactant to product. It is in fact a paper work.

While,

Actual Yield is the yield which is obtained experimentally. It is always less than theoretical yield because it is not possible to have 100% conversion of reactants into products. Even some amount of product is lost while handling it during the process.

Percentage Yield is Calculated as,

                    %age Yield  =  Actual Yield / Theoretical Yield × 100

Data Given:
                   Actual Yield  =  0.104 g

                   Theoretical Yield  =  0.110 g

Putting Values,

                    %age Yield  =  0.104 g / 0.110 g × 100

                    %age Yield  =  94.54 %
3 0
3 years ago
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