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MA_775_DIABLO [31]
3 years ago
11

A student dropped a pea size amount of K2CO3 into a solution of HCl(aq). He observed the formation of gas bubbles and collected

the gas into another test tube. The student performed a splint test and observed that the splint was extingished when he placed the splint into the test tube of the gas. What can be said about the results of this students experiment?
a. The student completed the experiment correctly and there were no errors in the experiment.
b. The experiment was performed incorrectly. K2CO3 doesn't react with HCl. Therefore, the student picked up the wrong compound when conducting the experiment.
c. The student performed the splint test incorrectly. He should of observed the splint flare up when the splint was placed in the test tube.
d. The student performed the splint test incorrectly. He should of observed a popping sound when the splint was placed in the test tube.
Chemistry
1 answer:
Wittaler [7]3 years ago
3 0

Answer:

The student completed the experiment correctly and there were no errors in the experiment.

Explanation:

When a pea size amount of K2CO3 is dropped into a solution of HCl, the following reaction occurs;

K2CO3(s) + 2HCl(aq) ----> 2KCl(aq) + CO2(g) + H2O(l)

The gas CO2 does not support burning hence, when the student performed a splint test and observed that the splint was extinguished when he placed the splint into the test tube of the gas.

Hence, the experiment was properly conducted and the student completed the experiment correctly and there were no errors in the experiment.

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Artist 52 [7]

Answer:

Molar absorptivity or molar extinction co-effecient = 2120.14 cm⁻¹M⁻¹

Explanation:

First convert Concentration from ppm inM or mol/l

⇒ Molar mass of KMnO₄ = 158.03 g

⇒ 4.48 ppm = 4.48 mg/l = 4.48 x 10⁻³ g/l

⇒ Molarity = \frac{4.48 X10^{-3} }{158.03X 1(lit)} = 2.83 x 10⁻⁵ molar

Absorbance (A) = - log(T)     ( T = % transmittance)

                          = - log(0.859)

                          = 0.06

According to Lambert Beer's law

     

                 ε = \frac{A}{C X l}

      or,      ε = \frac{0.06}{2.83 X 10^{-5}X1 cm }

      or,      ε = 2120.14 cm⁻¹M⁻¹

Where

    ε = Molar absorptivity

    A = absorbance

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     l = length  

6 0
3 years ago
Answer questions 2 and 3
nlexa [21]

Answer:

2:a-heterogenous

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c-heterogenous

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Explanation:

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4 0
4 years ago
In a labratory activity the density of a sample of vanadium is determined to be 6.9g/cm^3 at room temperature. what is the perce
Snowcat [4.5K]

Answer:

c : 13%

Explanation:

Data Give:

Experimental density of vanadium = 6.9 g/cm³

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Solution:

Formula used to calculate % error

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The reported accepted density value for vanadium = 6.11 g/cm³

Put value in the above equation

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                  % error = [ 0.129] x 100

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Round to the 2 significant figure

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