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3 years ago

Which of the following elements has the highest first ionization energy: Na, Mg, S or Cl?

Chemistry

1 answer:

storchak [24]3 years ago

3 0

Answer:

Cl has the highest ionization energy because all these elements lie in the same period and ionization energy increases from left to right across a period as Cl is the at the last in this period so it has the highest ionization energy

Explanation:

I hope it will help you:)

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leva [86]

Explanation: energy levels are three(counting number of cyecles) outer electrons are 3 also

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3 years ago

Hess’s law

Delvig [45]

From the statement of Hess' law, the enthalpy of the reaction A---> C is +90 kJ

Hess' law of constant heat summation states that for a multistep reaction, the standard enthalpy of reaction is always constant and is independent of the pathway or intermediate routes taken.

From Hess' law, the enthalpy change for the reaction A ----> C is calculated as follows:

A---> C = A ---> B + B ---> C

ΔH of A---> C = 30 kJ + 60 kJ

ΔH = 90 kJ

Therefore, the enthalpy of the reaction A---> C is +90 kJ

The above reaction A---> C can be shown in the enthalpy diagram below:

A -------------------> C (ΔH = +90 kJ)

\ /

\ / (ΔH = +60 kJ)

(ΔH = +30 J) \ /

> B

Learn more about enthalpy and Hess law at: brainly.com/question/9328637

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2 years ago

Glycolic acid, which is a monoprotic acid and a constituent in sugar cane, has a pKa of 3.9. A 25.0 mL solution of glycolic acid

Phoenix [80]

Answer:

pH = 8.0

Explanation:

First, we have to calculate the moles of NaOH.

$35.8 \times 10^{-3}L.\frac{0.020mol}{L} =7.2\times 10^{-4}mol$

Let's consider the balanced equation.

C₂H₄O₃ + NaOH ⇒ C₂H₃O₃Na + H₂O

The molar ratio C₂H₄O₃: NaOH: C₂H₃O₃Na is 1: 1: 1. So, when 7.2 × 10⁻⁴ moles of NaOH react completely with 7.2 × 10⁻⁴ moles of C₂H₄O₃ they form 7.2 × 10⁻⁴ moles of C₂H₃O₃Na.

The concentration of C₂H₃O₃Na is:

$\frac{7.2\times 10^{-4}mol}{60.8 \times 10^{-3}L} =0.012M$

C₂H₃O₃Na dissociates according to the following equation:

C₂H₃O₃Na(aq) ⇒ C₂H₃O₃⁻(aq) + Na⁺(aq)

C₂H₃O₃⁻ comes from a weak acid so it undergoes basic hydrolisis.

C₂H₃O₃⁻ + H₂O ⇄ C₂H₄O₃ + OH⁻

If we know that pKa for C₂H₄O₃ is 3.9, we can calculate pKb for C₂H₃O₃⁻ using the following expression:

pKa + pKb = 14

pKb = 14 -3.9 = 10.1

10.1 = -log Kb

Kb = 7.9 × 10⁻¹¹

We can calculate [OH⁻] using the following expression:

[OH⁻] = √(Kb.Cb) <em>where Cb is the initial concentration of the base</em>

[OH⁻] = √(7.9 × 10⁻¹¹ × 0.012M) = 9.7 × 10⁻⁷ M

Now, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log (9.7 × 10⁻⁷) = 6.0

pH + pOH = 14

pH = 14 - pOH = 14 - 6.0 = 8.0

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3 years ago