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Post UTME Online Tutorials (studypaedia) by ThanosDr: 1:20am On Apr 18, 2020
Due to the current pandemic halting traditional method of Education all over the world.

We are making available amazing online learning resource for putme candidates to ensure optimum performance in their examination.

Once this pandemic is curtailed there will be a mad rush in all areas of our lives, therefore, you shouldn't be caught unawares as it might end up being disastrous to all the gains you made during your UTME.

Your mantra should be:
Stay at home, stay safe and learn something relevant every day.
Re: Post UTME Online Tutorials (studypaedia) by ThanosDr: 11:57am On Apr 18, 2020
Biology: Answered

SOLUTION

Cactus are mostly found in arid (dry/hot) regions. They possess swollen stem and needle-like leaves which help to conserve water hereby preventing drying out.

Which makes D the answer to this question.

Re: Post UTME Online Tutorials (studypaedia) by ThanosDr: 12:02pm On Apr 18, 2020
English: Answered

Solution:
To mean "concerning" we can only have three forms
1. Regarding √
2. As regards √ (trick: note the "s" at the end of AS and REGARDS)
3. With regard to √ (trick: WITH has "t" and should be followed with "TO" having "t" too)

Only one of the three will be among the options as any of them is correct.

Re: Post UTME Online Tutorials (studypaedia) by ThanosDr: 12:03pm On Apr 18, 2020
Chemistry: Answered

SOLUTION

Dissolution of Na2CO3 in water is basic ...

Na2CO3 + 2H2O -----> 2NaOH + H2CO3.

Since the concentration is 3 and the base is a strong base, it dissociates completely in water.

2NaOH ----> 2Na+ + 2OH–

The concentration of sodium ion and hydroxyl ions is the same.

That is, [Na+] = [OH–].

n(Na+) = CV.

n(Na+) = 3 × 0.07 = 0.21mol.

n(Na+) = 2 × 0.21 = 0.42mol


For the second compound,

NaHCO3 + H2O ------> NaOH + H2CO3

NaOH ----> Na+ + OH–.

n(Na+) = CV.

n(Na+) = 1 × 0.03 = 0.03mol.

Total volume = (70 + 30)cm³ = 100cm³ = 0.1dm³.

Total number of moles = 0.42 + 0.03 = 0.45mol.

Concentration = n/V.

C = 0.45/0.1

C = 4.5moldm–³.

Re: Post UTME Online Tutorials (studypaedia) by ThanosDr: 7:08pm On Apr 18, 2020
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Re: Post UTME Online Tutorials (studypaedia) by ThanosDr: 7:08pm On Apr 18, 2020
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Re: Post UTME Online Tutorials (studypaedia) by ThanosDr: 7:09pm On Apr 18, 2020
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Re: Post UTME Online Tutorials (studypaedia) by ThanosDr: 7:11pm On Apr 18, 2020
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