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Lady bird [3.3K]
3 years ago
15

Suppose two scientists perform the same experiment separately, but their results are very different. What could they do to find

out why this happened?
Chemistry
1 answer:
Lynna [10]3 years ago
5 0
They should identify the confounding variable.

Some condition that is not examined by the scientist might alter the experiment result. That condition is called confounding variable. If the method of the experiment same but result is very different, there should be unidentified confounding variable. It could be air humidity, temperature, ventilation, light, time of the year or anything that might not be seen by naked eye.
Try to redo the experiment with controlling variable as much as possible.
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Three. Also known as a triplet code.

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Why some elements are radioactive (unstable). When the atoms of an element have extra neutrons or protons it creates extra energy in the nucleus and causes the atom to become unbalanced or unstable. Whether radioactive elements can become stable and if so, how. The unstable nucleus of radioactive atoms emit radiation.

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The equilibrium concentrations of the reactants and products are [HA]=0.280 M, [H+]=4.00×10−4 M, and [A−]=4.00×10−4 M. Calculate
Tems11 [23]

Answer:

6.24

Explanation:

The following data were obtained from the question:

Concentration of HA, [HA] = 0.280 M,

Concentration of H+, [H+] = 4×10¯⁴ M

Concentration of A-, [A−] = 4×10¯⁴ M

pKa =.?

Next, we shall write the balanced equation for the reaction. This is given below:

HA <===> H+ + A-

Next, we shall determine the equilibrium constant Ka for the reaction. This can be obtained as follow:

Equilibrium constant for a reaction is simply the ratio of concentration of the product raised to their coefficient to the concentration of the reactant raised to their coefficient.

The equilibrium constant for the above equation is given below:

Ka = [H+] [A−] /[HA]

Concentration of HA, [HA] = 0.280 M,

Concentration of H+, [H+] = 4×10¯⁴ M

Concentration of A-, [A−] = 4×10¯⁴ M

Equilibrium constant (Ka) =

Ka = (4×10¯⁴ × 4×10¯⁴) / 0.280

Ka = 1.6×10¯⁷/ 0.280

Ka = 5.71×10¯⁷

Therefore, the equilibrium constant for the reaction is 5.71×10¯⁷

Finally, we shall determine the pka for the reaction as follow:

Equilibrium constant, Ka = 5.71×10¯⁷

pKa =?

pKa = – Log Ka

pKa = – Log 5.71×10¯⁷

pKa = 6.24

Therefore, the pka for the reaction is 6.24.

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3 years ago
PLSS HELP! The gas carbon dioxide is a pure substance. Which of the following is true about carbon dioxide? (4 points)
Nadusha1986 [10]

Answer:

The answer is option A..

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