1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Serga [27]
3 years ago
10

An aqueous solution of ammonia is found to be basic. This observation can be explained by the net ionic equationHNO3(aq) + H2O(l

) → NO3-(aq) + H3O+(aq).NH4+(aq) + H2O(l) → NH3(aq) + H3O+(aq).NO3-(aq) + H2O(l) → HNO3(aq) + OH-(aq).NH3(aq) + H2O(l) → NH4+(aq) + OH-(aq).
Chemistry
1 answer:
jeka57 [31]3 years ago
6 0

Answer:

NH_3_{(aq)}+H_2O_{(l)}\rightarrow NH_4^+_{(aq)}+OH^-_{(aq)}

Explanation:

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.

Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.  Only the species which are present in aqueous state dissociate. So, the net ionic equation of aqueous solution of ammonia is shown below as:-

NH_3_{(aq)}+H_2O_{(l)}\rightarrow NH_4^+_{(aq)}+OH^-_{(aq)}

You might be interested in
Manganese(IV) oxide is reduced to manganese(II) oxide by hydrogen gas. Include all phases in answer
liq [111]
<span><span>Mn<span>O<span>2<span>(s)</span></span></span>+<span>H<span>2<span>(g)</span></span></span>→Mn<span>O<span>(s)</span></span>+<span>H2</span><span>O<span>(g)</span></span></span></span>
4 0
3 years ago
What is the sum of kinetic and potential energy particles in matter
beks73 [17]
Mechanical I believe because this are things that move and don’t move so that would mean that it would be mechanical since mechanical means movement
5 0
3 years ago
A 50.0 mL solution of 0.129 M KOH is titrated with 0.258 M HCl. Calculate the pH of the solution after the addition of each of t
kobusy [5.1K]

Answer:

A- pH = 13.12

B- pH = 12.91

C- pH = 12.71

D- pH = 12.43

E- pH = 11.55

F- pH = 7

G- pH = 2.46

H- pH = 1.88

Explanation:

This is a titration of a strong base with a strong acid. The neutralization reaction is: KOH (aq) + HCl (aq) →  H₂O(l) + KCl(aq)

Our pH at the equivalence point is 7, because we have made a neutral salt.

To determine the volume at that point we state the formula for titration:

mmoles of base = mmoles of acid

Volume of base  . M of base = Volume of acid . M of acid

50mL . 0.129M = 0.258 M . Volume of acid

Volume of acid = (50mL . 0.129M) / 0.258 M →  25 mL (Point <u>F</u>)

When we add 25 mL of HCl, our pH will be 7.

A- At 0 mL of acid, we only have base.

KOH → K⁺ + OH⁻

[OH⁻] = 0.129 M

To make more easy the operations we will use, mmol.

mol . 1000 = mmoles → mmoles / mL = M

- log 0.129 = 0.889

14 - 0.889 = 13.12

B-  In this case we are adding, (7 mL . 0.258M) = 1.81 mmoles of H⁺

Initially we have  0.129 M . 50 mL = 6.45 mmoles of OH⁻

1.81 mmoles of H⁺ will neutralize, the 6.45 mmoles of OH⁻ so:

6.45 mmol - 1.81 = 4.64 mmoles of OH⁻

This mmoles of OH⁻ are not at 50 mL anymore, because our volume has changed. (Now, we have 50 mL of base + 7 mL of acid) = 57 mL of total volume.

[OH⁻] = 4.64 mmoles / 57 mL = 0.0815 M

- log 0.0815 M = 1.09 → pOH

pH = 14 - pOH → 14 - 1.09 = 12.91

C- In this case we add (12.5 mL . 0.258M) = 3.22 mmoles of H⁺

<em>Our initial mmoles of OH⁻ would not change through all the titration. </em>

Then 6.45 mmoles of OH⁻ are neutralized by 3.22 mmoles of H⁺.

6.45 mmoles of OH⁻ - 3.22 mmoles of H⁺ = 3.23 mmoles of OH⁻

Total volume is: 50 mL of base + 12.5 mL = 62.5 mL

[OH⁻] = 3.23 mmol / 62.5 mL = 0.0517 M

- log  0.0517 = 1.29 → pOH

14 - 1.11 = 12.71

D- We add (18 mL . 0.258M) = 4.64 mmoles of H⁺

6.45 mmoles of OH⁻ are neutralized by 4.64 mmoles of H⁺.

6.45 mmoles of OH⁻ - 4.64 mmoles of H⁺ = 1.81 mmoles of OH⁻

Total volume is: 50 mL of base + 18 mL = 68 mL

[OH⁻] = 1.81 mmol / 68 mL = 0.0265 M

- log  0.0265 = 1.57 → pOH

14 - 1.57 = 12.43

E- We add (24 mL . 0.258M) = 6.19 mmoles of H⁺

6.45 mmoles of OH⁻ are neutralized by 6.19 mmoles of H⁺.

6.45 mmoles of OH⁻ - 6.19 mmoles of H⁺ = 0.26 mmoles of OH⁻

Total volume is: 50 mL of base + 24 mL = 74 mL

[OH⁻] = 0.26 mmol / 74 mL = 3.51×10⁻³ M

- log  3.51×10⁻³  = 2.45 → pOH

14 - 2.45 = 11.55

F- This the equivalence point.

mmoles of OH⁻ = mmoles of H⁺

We add (25 mL . 0.258M) = 6.45 mmoles of H⁺

All the OH⁻ are neutralized.

OH⁻  +  H⁺  ⇄   H₂O              Kw

[OH⁻] = √1×10⁻¹⁴   →  1×10⁻⁷  →  pOH = 7

pH → 14 - 7 = 7

G- In this case we have an excess of H⁻

We add (26 mL . 0.258M ) = 6.71 mmoles of H⁺

We neutralized all the OH⁻ but some H⁺ remain after the equilibrium

6.71 mmoles of H⁺ - 6.45 mmoles of OH⁻ = 0.26 mmoles of H⁺

[H⁺] = 0.26 mmol / Total volume

Total volume is: 50 mL + 26 mL → 76 mL

[H⁺] = 0.26 mmol / 76 mL → 3.42×10⁻³ M

- log 3.42×10⁻³ = 2.46 → pH

H- Now we add (29 mL . 0.258M) = 7.48 mmoles of H⁺

We neutralized all the OH⁻ but some H⁺ remain after the equilibrium

7.48 mmoles of H⁺ - 6.45 mmoles of OH⁻ = 1.03 mmoles of protons

Total volume is 50 mL + 29 mL = 79 mL

[H⁺] = 1.03 mmol / 79 mL → 0.0130 M

- log 0.0130 = 1.88 → pH

After equivalence point, pH will be totally acid, because we always have an excess of protons. Before the equivalence point, pH is basic, because we still have OH⁻ and these hydroxides, will be neutralized through the titration, as we add acid.

5 0
3 years ago
Select the correct answer.
zalisa [80]

Well, this is somewhat difficult, because an electron already creates an electric field.  However, I know that when an electron moves it then creates a magnetic field.  So, I'm going to safely assume that when an electron moves, it creates an electric, a magnetic, and a gravitational field.  

I hope that helps!

8 0
3 years ago
Read 2 more answers
A chemical reaction that is expected to form 325.0 gof product only forms 123.8 g of product. What is the percent yield of this
Tasya [4]

Answer:

\boxed {\boxed {\sf A. \ 38.1 \%}}

Explanation:

Percent yield is the ratio of the amount actually produced to how much could theoretically be produced. It is found using this formula:

\% \ yield = \frac{actual \ yield}{theoretical \ yield} *100

For this reaction, the theoretical or expected yield is 325.0 grams. The actual yield is 123.8 grams.

\% \ yield = \frac{ 123.8 \ g }{325.0 \ g }*100

Divide.

\% \ yield = 0.380923077 *100

\% \ yield = 38.0923077

Round to the nearest hundredth. The 9 in the hundredth place tells us to round the 0 to a 1 .

\% \ yield \approx 38.1

The percent yield is about <u>38.1%</u>

5 0
2 years ago
Read 2 more answers
Other questions:
  • A decrease in velocity over time <br> A velocity <br> B time<br> C acceleration<br> D deceleration
    14·2 answers
  • It takes Serina 0.25 hours to drive to school. Her route is 16 meters long. What is
    5·1 answer
  • Volume in units of what?
    9·2 answers
  • Pyridine is a weak base that is used in the manufacture of pesticides and plastic resins. It is also a component of cigarette sm
    7·1 answer
  • Suggest 2 reasons why the noble gases become less ideal in their behaviour down the group from helium to xenon.
    5·1 answer
  • PLEASE HELP MEEE!!!!!!!!!
    14·1 answer
  • 2. What are some common properties of metals, nonmetals, and semimetals?
    15·2 answers
  • Does protecting animals so they won’t go extinct help with climate change?
    6·2 answers
  • How many moles of xenon trioxide are present in 1. 69 grams of this compound.
    10·1 answer
  • 3. Beaker A (Heat Meter) tells us how much heat energy left the hot
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!