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Natalija [7]
3 years ago
13

Calculate the [H3O+] of solutions a and b; calculate the [OH-] solutions c and d.

Chemistry
1 answer:
patriot [66]3 years ago
3 0

Answer:

I got the answers but it won't let me post it correctly on here....

Explanation:

9.) 10-2.76 =0.0174 [H30+]= 1.74*10-3 M

10.)10-3.65=0.00224  [H3O+] =2.24*10-2 M

11.)10-3.65=0.00224 [OH-]= 2.224*10-4M

12.)10-6.87=0.00000135  [OH-]= 1.35*10-7M

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What is the conjugate acid in the following equation:
yanalaym [24]

Answer:

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Explanation:

An acid is a proton donor; a base is a proton acceptor.

Thus, NO₂⁻ is the base, because it accepts a proton from the water.

H₂O is the acid, because it donates a proton to the nitrite ion.

The conjugate base is what's left after the acid has given up its proton.

The conjugate acid is what's formed when the base has accepted a proton.

NO₂⁻/HNO₂ make one conjugate acid/base pair, and H₂O/OH⁻ are the other conjugate acid/base pair.

NO₂⁻ + H₂O ⇌ HNO₂ + OH⁻

base     acid     conj.      conj.

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5 0
3 years ago
How many molecules are in 1kg of water
Mila [183]

Answer:

334.2× 10²³ molecules

Explanation:

Given data:

Mass of water = 1 Kg ( 1000 g )

Number of molecules = ?

Solution:

Number of moles of water:

Number of moles = mass/ molar mass

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Number of moles = 55.5 mol

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55.5 mol×6.022× 10²³ molecules

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8 0
3 years ago
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Explanation:

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ludmilkaskok [199]

Answer:

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