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docker41 [41]
3 years ago
15

Aluminum chloride acts as a strong Lewis acid since

Chemistry
1 answer:
ryzh [129]3 years ago
8 0

Answer: Since it became stronger

Explanation: ok

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Solve the following equation:13x - 22 = 30​
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Answer:

52/13

Explanation:

13X=30+22

x=52/13

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Identify what is in that photo please!! (10 pts)
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Electromagnet

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2. The number of molecules in 1.0 mole of So2 is the same as the number of molecules in:
m_a_m_a [10]

Answer:

1.0 moles of N2

Explanation:

since

1.0 × avogadro's no# = same answer for SO2 and N2

avogadro's no#= 6.02× 10²³

6 0
3 years ago
Does phosphate have the ability to conduct electricity?
baherus [9]

Answer:

Red phosphorous can vary in colour from orange to purple, due to slight variations in its chemical structure. The third form, black phosphorous, is made under high pressure, looks like graphite and, like graphite, has the ability to conduct electricity.

Explanation:

4 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 −
irina1246 [14]

<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

3 0
4 years ago
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