All categories

3 years ago

Calculate [H+] for each of the following solutions, and indicate whether the solution is acidic, basic, or neutral.

Chemistry

1 answer:

vesna_86 [32]3 years ago

3 0

Answer:

Part A: 10.7.

Part B: 2) the solution is basic.

Part C: 5.9.

Part D: 1) the solution is acidic.

Part E: 8.1.

Part F: 2) the solution is basic.

Explanation:

Part A [OH−]= 4.5×10−4 M . Express your answer using two significant figures.

∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.

[OH⁻] = 4.5×10⁻⁴ M.

∴ [H₃O⁺] = 10⁻¹⁴/[OH⁻] = 10⁻¹⁴/(4.5×10⁻⁴ M) = 2.22 × 10⁻¹¹ M.

∵ pH = - log[H₃O⁺]

∴ pH = - log(2.22 × 10⁻¹¹) = 10.65 ≅ 10.7.

Part B 1) the solution is acidic 2) the solution is basic 3) the solution is neutral

The solution is basic, because the pH is higher than 7.

Part C [OH⁻]= 7.7×10⁻⁹ M . Express your answer using two significant figures.

∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.

[OH⁻] = 7.7×10⁻⁹ M.

∴ [H₃O⁺] = 10⁻¹⁴/[OH⁻] = 10⁻¹⁴/(7.7×10⁻⁹ M) = 1.29 × 10⁻⁶ M.

∵ pH = - log[H₃O⁺]

∴ pH = - log(1.29 × 10⁻⁶) = 5.88 ≅ 5.9.

Part D 1) the solution is acidic 2) the solution is basic 3) the solution is neutral

The solution is acidic, because the pH is lower than 7.

Part E A solution in which [OH−] is 140 times greater than [H+]. Express your answer using two significant figures.

∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.

∵ [OH⁻] = 140[H₃O⁺]

∴ [H₃O⁺](140[H₃O⁺]) = 10⁻¹⁴.

∴ 140 [H₃O⁺]² = 10⁻¹⁴.

[H₃O⁺]² = 10⁻¹⁴/ 140 = 7.14 x 10⁻¹⁷.

∴ [H₃O⁺] = √(7.14 x 10⁻¹⁷) = 8.45 x 10⁻⁹ M.

∵ pH = - log[H₃O⁺]

∴ pH = - log(8.45 x 10⁻⁹) = 8.07 ≅ 8.1.

Part F 1) the solution is acidic 2) the solution is basic 3) the solution is neutral

The solution is basic, because the pH is higher than 7.

You might be interested in

Below is a list of substances. Some are mixtures and some are not. Select all of the substances which are homogeneous.

Papessa [141]

B the atmosphere

D. gasoline

C. a carbonated soft drink (without bubbles)

6 0

3 years ago

A syringe initially holds a sample of gas with a volume of 285 mL at 355 K and 1.88 atm. To what temperature must the gas in the

Ivahew [28]

Answer: The temperature to which the gas in the syringe must be heated is 720.5 K

Explanation:

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.

The equation follows:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1,V_1\text{ and }T_1$ are the initial pressure, volume and temperature of the gas

$P_2,V_2\text{ and }T_2$ are the final pressure, volume and temperature of the gas

We are given:

$P_1=1.88atm\\V_1=285mL\\T_1=355K\\P_2=2.50atm\\V_2=435mL\\T_2=?K$

Putting values in above equation, we get:

$\frac{1.88atm\times 285mL}{355K}=\frac{2.50atm\times 435mL}{T_2}\\\\T_2=\frac{2.50\times 435\times 355}{1.88\times 285}=720.5K$

Hence, the temperature to which the gas in the syringe must be heated is 720.5 K

8 0

3 years ago

What gives the gem amethyst its purplish color?

Alina [70]

iron Presence of trace elements, irradiation and iron impurities give the gem amethyst its purplish color!

7 0

3 years ago

Photosynthesis was another biological phenomenon that occupied the attention of the chemists of the late 18th century. The demon

balu736 [363]

Answer:

In the 1770s, the English clergyman Joseph Priestley (who is credited with the discovery of O2) established the production of oxygen by vegetables recognizing that the process was, apparently, the inverse of animal respiration, which consumed such chemical element.

Explanation:

In 1772, Joseph Priestley in his Recherches sur diversces especes d'air differentiated the air of animal respiration from that emitted by vegetables in the presence of light. Of the latter, which he called "dephlogistic air", he highlighted his purifying property of the environment indicating that: plants far from affecting the air in the same way as animal respiration, produce the opposite effects, and tend to preserve the sweet and healthy atmosphere , when it becomes harmful as a result of the life and breathing of the animals or their death and their rot.

In 1780, Jean Ingeshousz in his Experiences sur les vegetaux completed and reaffirmed the observations of Joseph Priestley. At the same time, he could deny Charles Bonnet's hypothesis, by demonstrating that the air expelled from the leaves comes from inside, and that the stimulating factor of the gaseous emission was not the heat produced by the sun, but the intensity of the light .

It was, finally, Jean Senebier that between 1782 and 1784, found that the "fixed air" dissolved in the water favors the vegetation. From these observations, he hypothesized that "fixed air" (carbon dioxide) is absorbed by the plants, which take it from the atmosphere with the humidity it has and in which it is mixed. Once this gas has been captured, both from the atmosphere and from the ground, it is decomposed in the presence of light by the leaves, releasing the "vital air" (oxygen) and leaving the carbon in the plant.

Thus, at the end of the century the participation of the atmosphere in plant dynamics was already seated, although the how and why of this participation were still unknown and no theory had been formulated to explain the nutritional process as a whole.

3 0

3 years ago

All of the following reactions can be described as displacement reactions except:____________.

Lady_Fox [76]

Answer:

Explanation:

The reaction that is not a displacement reaction from all the options is $C_6H_6_{(l)} + Cl_{2(g)} --> C_6H_5Cl_{(l)} + HCl_{(g)}$

In a displacement reaction, a part of one of the reactants is replaced by another reactant. In single displacement reactions, one of the reactants completely displaces and replaces part of another reactant. In double displacement reaction, cations and anions in the reactants switch partners to form products.

Options a, c, d, and e involves the displacement of a part of one of the reactants by another reactant while option b does not.

Correct option = b.

8 0

3 years ago