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Brums [2.3K]
3 years ago
12

Which of the following is an example of an experimental scientific investigation

Chemistry
2 answers:
sleet_krkn [62]3 years ago
7 0

Scientific investigation is the process in which scientist solve the problems by using different systematic approach. It can be initiated in different ways.

Experimental Scientific investigation: The investigation in which scientist answer the question on the basis of experimental results. Experimental investigation contains both dependent and independent variables, and only one variable is tested at a time is possible. This process makes uses of collecting observation and identifying the process in physical word.

Noting the colour of a solution when heated is the experimental scientific investigation

Therefore, other given options are part of scientific observation whereas Noting the colour of a solution when heated explains the experimental scientific investigation.



Serggg [28]3 years ago
3 0

The correct answer is option D.

Noting the color of a solution when heated is an example of  an experimental scientific investigation.

Option A, noting the PH of a solution on a digital meter is an example of experimental data recording.

Similarly, option B recording the sound of glass breaking, is an example of experimental record keeping.

Option C, placing test tubes in an ice bath for 15 min is an example of experimental procedure.

But option D, noting the color of a solution when heated is an example of experimental scientific investigation. Because the scientist is make an acute observation of the color change whenever the solution is heated.

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How would you prepare 500 ml of the following solutions : Sodium succinate buffer (0.1 mol/dm3 pH 5.64)
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Answer:

8.10g of sodium succinate must be added and 247mL of 0.1M HCl adding enough water until make 500mL

Explanation:

<em>Succinic acid has a pKa₂ of 5.63</em>

To solve this question we must find the amount of sodium succinate and 0.1M HCl that we have to add using H-H equation:

pH = pKa + log [A-] / [HA]

5.64 = 5.63 + log [Na₂Succ.] / [HSucc⁺]

0.01 = log [Na₂Succ.] / [HSucc⁺]

1.0233 = [Na₂Succ.] / [HSucc⁺] <em>(1)</em>

As:

0.1M = [Na₂Succ.] + [HSucc⁺] <em>(2)</em>

Replacing (2) in (1):

1.0233 = 0.1M - [HSucc⁺] / [HSucc⁺]

1.0233[HSucc⁺] = 0.1M - [HSucc⁺]

2.033[HSucc⁺] = 0.1M

[HSucc⁺] = 0.0494M

[Na₂Succ] = 0.0506M

Both [Na₂Succ⁺] and [HSucc⁺] ions comes from the same sodium succinate we have to find the moles of sodium succinate in 500mL of 0.1M. Then, based on the reaction:

Na₂Succ + HCl → HSucc⁺ + Cl⁻

The moles of HCl added = Moles HSucc⁺ we need:

<em>Moles Na₂Succ:</em>

0.500L * (0.1mol/L) = 0.0500 moles

<em>Mass -Molar mass sodium succinate: 162.05g/mol-:</em>

0.0500mol * (162.05g/mol) = 8.10g of sodium succinate must be added

<em>Moles HCl:</em>

0.0494M * 0.500L = 0.0247 moles HCl * (1L / 0.1mol) = 0.247L =

And 247mL of 0.1M HCl adding enough water until make 500mL

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