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4 years ago

15

Which of the following atoms most easily loses an electron Ba Cl Te C P

1 answer:

solong [7]4 years ago

7 0

Ba! It is a metal and wants to lose 2 electrons ASAP

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Consider the electrolysis of molten barium chloride (bacl2). (a) write the half-reactions. include the states of each species..

IrinaK [193]

<span>Molten barium chloride is separetes:</span><span>
BaCl</span>₂(l) → Ba(l) + Cl₂(g), <span>
but first ionic bonds in this salt are separeted because of heat:
BaCl</span>₂(l) → Ba²⁺(l) + 2Cl⁻(l).

Reaction of reduction at cathode(-): Ba²⁺(l) + 2e⁻ → Ba(l).

Reaction of oxidation at anode(+): 2Cl⁻(l) → Cl₂(g) + 2e⁻.

The anode is positive and the cathode is negative.

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3 years ago

In a Lewis structure, why must pairs of unshared of lone pairs if dots be placed around the outside of some of the atoms

Maksim231197 [3]

Answer:

In a Lewis symbol, the symbol for the element is used to represent the atom and its core electrons. Dots placed around the atom are used to indicate the valence electrons. ... In 1916, Gilbert Newton Lewis, an American chemist, suggested that molecules were formed when atoms shared pairs of outer electrons.

Explanation:

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3 years ago

Question 9 Water molecules require a lot of energy to separate them<br> because...

ira [324]

Answer:

they are highly adhesive

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3 years ago

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sineoko [7]

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does that answer it?

8 0

3 years ago

Phosphoric acid is a triprotic acid ( K a1 = 6.9 × 10 − 3 Ka1=6.9×10−3, K a2 = 6.2 × 10 − 8 Ka2=6.2×10−8, and K a3 = 4.8 × 10 −

PilotLPTM [1.2K]

<u>Answer:</u> To calculate the pH of the buffer composed of $H_2PO_4^-\text{ and }HPO_4^{-2}$ , we use the $K_a2$

<u>Explanation:</u>

Phosphoric acid is a triprotic acid and it will undergo three dissociation reaction each having their respective dissociation constants.

The chemical equation for the first dissociation reaction follows:

$H_3PO_4\rightleftharpoons H_2PO_4^-+H^+;K_a1=6.9\times 10^{-3}$

The chemical equation for the second dissociation reaction follows:

$H_2PO_4^-\rightleftharpoons HPO_4^{2-}+H^+;K_a2=6.2\times 10^{-8}$

The chemical equation for the third dissociation reaction follows:

$HPO_4^{2-}\rightleftharpoons PO_4^{3-}+H^+;K_a3=4.8\times 10^{-13}$

To form a buffer composed of $H_2PO_4^-\text{ and }HPO_4^{-2}$ , we use the $K_a$ of second dissociation process

To calculate the $pK_a$ , we use the equation:

$pK_a=-\log (K_a)\\\\pK_a=-\log(6.2\times 10^{-8})\\\\pK_a=7.21$

To calculate the pH of buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a2+\log(\frac{[\text{conjugate base}]}{[\text{weak acid}]})$

$pH=pK_a2+\log(\frac{[HPO_4^{2-}]}{[H_2PO_4^-]})$

We are given:

$pK_a2$ = negative logarithm of second acid dissociation constant of phosphoric acid = 7.21

$[HPO_4^{2-}]$ = concentration of conjugate base

$[H_2PO_4^{-}]$ = concentration of weak acid

Hence, to calculate the pH of the buffer composed of $H_2PO_4^-\text{ and }HPO_4^{-2}$ , we use the $K_a2$

8 0

3 years ago