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Gwar [14]
3 years ago
10

In the spectrophotometric analysis of benzene, the calibration curve has a slope of 195 AU/M and an intercept of 0.079 AU. If an

unknown has an absorbance of 0.517 AU, what is the concentration of benzene in the sample?
a. 2.09 x10^-4 M
b. 2.25 x 10^-3 M
c. 2.65 x 10^-3 M
d. 0.438 M
Chemistry
1 answer:
zloy xaker [14]3 years ago
5 0

Answer:

b. 2.25 x 10^-3 M

Explanation:

Please remember that according to Beer´s, law the absorbance is given by

A = Acl where A = absorbance, A is the molar absorptivity, c is concentration (M) and l is the path length of the cubette of used.

Now it follows from this equation that the absorbance is linearly related to concentration and what we are being told is that experimentally the linear equation obtained during calibration was:

A = 195 c + 0.079

We have all the information required to solve this question to solve for c:

0.517 = 195 c + 0.079 ⇒ c = (0.517 - 0.079) AU/ 195 AU/M = 2.25 x 10⁻³ M

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allochka39001 [22]

Explanation:

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2. (2 Pts) Consider the reaction N2(g) + 3H2(g) =; 2NH3(g). The production of ammonia is an

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4. (3 Pts) ff 1ven th e o £ 11 owmg d t a a £ or th ere action: A(g) + 2B(s) =; AB2(g)

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300 1.5x104

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900 3.4 X 10-3

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4 0
3 years ago
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patriot [66]

Answer:

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3 0
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Calcium oxalate (CaC₂O₄) is obtained by the reaction of 5 g of potassium oxalate (K₂C₂O₄).

We can calculate the moles of CaC₂O₄ obtained considering the following relationships.

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5 g K_2C_2O_4 \times \frac{1molK_2C_2O_4}{184.24gK_2C_2O_4}  \times \frac{1molCaC_2O_4}{2molK_2C_2O_4} = 0.03molCaC_2O_4

5 g of potassium oxalate react to produce 0.03 moles of calcium oxalate.

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7 0
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mixas84 [53]
1.00*10^3
You’d need to lower the exponent because rounding to 3 sig figs changes the 9’s to - 1000. Keep the 0’s.
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4 0
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