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Fed [463]
3 years ago
15

As you move down group one of the periodic table, what happens to the atomic radii? They increase, because of the higher number

of occupied energy levels. They increase, because of the stronger attraction between electrons and the nucleus. They decrease, because of the weaker effective nuclear charge. They decrease, because of the lower number of protons.
Chemistry
2 answers:
Dennis_Churaev [7]3 years ago
6 0

The correct answer is the atomic radius increase because of the higher number of occupied energy levels.  

Atomic radius increases when going down a group. The valence electrons hold the higher levels because of the enhancing quantum number (n). As a consequence, the valence electrons get further away from the nucleus with the increase in n. The electron shielding inhibits these outer electrons from getting fascinated towards the nucleus. Therefore, they are held, and the resulting atomic radius is greater.  


masya89 [10]3 years ago
5 0

Answer:

they increase, because of the higher number of occupied energy levels

Explanation:

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What is the pH when 10.0 mL of 0.20 M potassium hydroxide is added to 30.0 mL of 0.10 M cinnamic acid, HC9H7O2 (Ka = 3.6 × 10–5)
astra-53 [7]

Answer:-

Solution:- As is clear from the given Ka value, Cinnamic acid is a weak acid. let's calculate the moles of acid and KOH added to it from their given molarities and mL.

For KOH,  10.0mL(\frac{1L}{1000mL})(\frac{0.20mol}{1L})

= 0.002 mol

For Cinnamic acid,  30.0mL(\frac{1L}{1000mL})(\frac{0.10mol}{1L})

= 0.003 mol

Acid and base react as:

HC_9H_7O_2(aq)+KOH(aq)\rightleftharpoons KC_9H_7O_2(aq)+H_2O(l)

The reaction takes place in 1:1 mol ratio. Since the moles of acid are in excess, the acid is still remaining when all the kOH is used.

0.002 moles of KOH react with 0.002 moles of Cinnamic acid to form 0.002 moles of potassium cinnamate. Excess moles of Cinnamic acid = 0.003 - 0.002 = 0.001

As the solution have weak acid and it's salt(or we could say conjugate base), it is a buffer solution and the pH of the buffer solution could easily be calculated using Handerson equation:

pH=pKa+log(\frac{base}{acid})

pKa could be caluted from given Ka value using the formula:

pKa = - log Ka

pKa=-log3.6*10^-^5

pKa = 4.44

let's plug in the values in Handerson equation and calculate the pH:

pH=4.44+log(\frac{0.002}{0.001})

pH = 4.44+0.30

pH = 4.74

So, the first choice is correct, pH is 4.74.

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