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Delvig [45]
3 years ago
11

The following equation shows the equilibrium in an aqueous solution of ammonia: NH3(aq)+H2O(l)⇌NH4+(aq)+OH−(aq)NH3(aq)+H2O(l)⇌NH

4+(aq)+OH−(aq) Which of the following represents a conjugate acid-base pair?a) NH3 and H2O b) NH4+ and OH− c) H2O and OH− d) NH3 and OH−
Chemistry
1 answer:
yarga [219]3 years ago
8 0

Answer:

c) H2O and OH−

Explanation:

Acids are the species which furnish hydrogen ions in the solution or is capable of forming bonds with electron pair species as they are electron deficient species.

When an acid donates a proton, it changes into a base which is known as its conjugate base.

Bases are the species which furnish hydroxide ions in the solution or is capable of forming bonds with electron deficient species as they are electron rich species. When a base accepts a proton, it changes into a acid which is known as its conjugate acid.

The acid and the base which is only differ by absence or presence of the proton are known as acid conjugate base pair.

Thus,  for the reaction,

NH_3+H_2O\rightleftharpoons Nh_4^++OH^-

The base is NH_3 and the conjugate acid of the base is NH_4^+.

Also, The base is OH^- and the conjugate acid of the base is H_2O.

<u>Correct option is :- c) H2O and OH−</u>

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Explanation:

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6 0
3 years ago
Read 2 more answers
PLEASE HELP ME ASAP! CHEMISTRY TUTOR<br><br> SEE ATTACHED
Masteriza [31]

Answer:

\large \boxed{\text{-827.4 kJ}}

Explanation:

We have three equations:

1. 2H₂S(g)            + O₂(g)   ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ

2. S(s, rhombic)  + O₂(g)   ⟶ SO₂(g);                                 ∆H = -296.8 kJ

3. PbO(s)             + H₂S(g) ⟶ PbS(s)               + SO₂(g);    ∆H =  -104.3 kJ

From these, we must devise the target equation:

4. 2PbS(s)            + 3O₂(g) ⟶2PbO(s)             + 2SO₂(g); ΔH = ?

The target equation has PbS(s) on the left, so you reverse Equation 3 and double it.

When you reverse an equation, you reverse the sign of its ΔH.

When you double an equation, you double its ΔH.

5. 2PbS(s) + 2H₂O(g) ⟶ 2PbO(s) + 2H₂S(g); ∆H = 208.6 kJ

Equation 5 has 2H₂O on the left. That is not in the target equation.

You need an equation with 2H₂O on the right, so you copy Equation 1.  

6. 2H₂S(g) + O₂(g) ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ  

Equation 6 has 2S(s, rhombic) on the right. That is not in the target equation.

You need an equation with 2S(s, rhombic) on the left, so you double Equation 2.  

7. 2S(s, rhombic)  + 2O₂(g) ⟶ 2SO₂(g); ∆H = -593.6 kJ

Now, you add equations 5, 6, and 7, cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you add their ΔH values.

You get the target equation 4:

5. 2PbS(s)  + <u>2H₂O(g</u>)  ⟶ 2PbO(s) + <u>2H₂S(g</u>);  ∆H =  208.6 kJ

6. <u>2H₂S(g)</u> + O₂(g)        ⟶ <u>2S(s</u>)     + <u>2H₂O(g)</u> ; ∆H = -442.4 kJ

<u>7</u><u>. </u><u>2S(s)</u><u>      + 2O₂(g)      ⟶ 2SO₂(g);                   ∆H = -593.6 kJ </u>

4 . 2PbS(s) + 3O₂(g)      ⟶ 2PbO(s) + 2SO₂(g); ΔH = -827.4 kJ

\Delta H \text{ for the reaction is $ \large \boxed{\textbf{-827.4 kJ}}$}

8 0
3 years ago
9. To become like a noble gas the element P will have a charge of
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Answer:

to become a noble gas element P will have 2 electrons in it's outer most energy level if it has one energy level

and eight in the last energy level if more than one

6 0
3 years ago
Which biogeochemical cycle has the least activity because the essential element is mostly stored in rock?
loris [4]
<span>The biogeochemical cycle that has the least activity because its essential element is mostly stored in rock is the phosphorus cycle. Phosphorus, unlike the other essential elements, is commonly found as a solid. This is why the atmosphere does not play a role in this cycle. Instead, phosphorus remains on land, and is mostly found in rocks and minerals.</span>
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3 years ago
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A single serving bag of snack chips contains 65.0 Cal. Assuming that all of the energy from eating these chips goes toward keepi
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Answer:

= 62.1 hours

Explanation:

Energy provide by the serving is 65 cal

= 65 cal  × 4.184 Kj = 271.96 kJ

271.96 KJ = 271960 J

Energy required for 1minute of energy

= 73 x 1

= 73 J/min

So, 271960 joules will be required for 271960 heart beat

Minutes = 271960 / 73

= 3593.94 minutes  

Time in hours = 3725.429 / 60

= 62.1 hours

5 0
4 years ago
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