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Sunny_sXe [5.5K]
2 years ago
11

Calculate the dipole moment for hf (bond length 0.917 å), assuming that the bond is completely ionic.

Chemistry
1 answer:
Gwar [14]2 years ago
7 0
The dipole moment  =  [the charge magnitude at one of the ends (you can choose either one)] [the distance between charges]
It is usually calculated in SI units of Debye which represents the positive and negative charges separated by 0.2082 angestron

Thus,
dipole moment = 0.917/0.2082 = 4.4044 debye.
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Adding a few drops of a strong acid to water will lower the pH appreciably. However, adding the same number of drops to a buffer
timurjin [86]

Answer:

Buffers are resistant to high pH changes.

Explanation:

This perfectly explains the reason why we use buffers. Buffers are substances which consist of a weak acid and its conjugate base. Buffers are resistant to significant pH changes upon addition of strong acids or bases. To illustrate this, let's say we have a buffer consisting of 0.1 mol of HF, a weak acid, and 0.1 mol of NaF (fluoride is a conjugate base of HF).

  • Let's say we add some strong acid, in a general form, this acid would be represented as H_3O^+. In this case, conjugate base will react and neutralize it to produce some amount of HF: H_3O^+ (aq) + F^- (aq)\rightarrow HF (aq) + H_2O (l).
  • Similarly, if we add some strong base OH^-, the acidic component will react with it to produce some amount of conjugate base: HF (aq) + OH^- (aq)\rightarrow F^- (aq) + H_2O (l). The ratio of HF to NaF in this case is held around the same value for addition of small amounts of strong acids/bases, so pH is kept almost constant, while in neutral water, pH would drastically increase or decrease.
5 0
2 years ago
34.62 mL of 0.1510 M NaOH was needed to neutralize 50.0 mL of an H2SO4 solution. What is the concentration of the original sulfu
Igoryamba

0.1045M is the concentration of the original sulfuric acid solution

Explanation:

Titration is done to know the volume or concentration of unknown electrolyte.

Data given:

volume of acid = 50 ml

molarity of acid =?

volume of base NaOH = 34.62 ml

molarity of the base = 0.1510

For titration the formula used is,

Macid x Vacid = Mbase  x V base

Putting the values in above equation:

M acid x 50 = 34.62 x 0.1510

Macid = \frac{5.22762}{50}

           = 0.1045 M is the molarity of sulphuric acid solution used in neutralization of 0.1510 M base solution.

5 0
2 years ago
The volume of a gas at 7.00°c is 49.0 ml. if the volume increases to 74.0 ml and the pressure is constant, what will the tempera
marshall27 [118]
Using charles law
v1/t1=v2/t2
v1=49ml
v2=74
t1=7+273=280k
t2=?
49/280=74/t2
0.175=74/t2  cross multiply
0.175t2=74
t2=74/0.175
t2=422k or 149celcius
8 0
2 years ago
A poisoned pill contains 0.00048 moles of KCN. How many molecules are in this sample?
mario62 [17]

Answer:

2.89 \times  {10}^{20}  \:  \: molecules

Explanation:

The number of molecules of KCN can be found by using the formula

<h3>N = n × L</h3>

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

N = 0.00048 × 6.02 × 10²³

We have the final answer as

2.89 \times  {10}^{20}  \:  \:  \: molecules

Hope this helps you

7 0
1 year ago
1. A student sees tiny bubbles clinging to the inside of an unopened plastic bottle full of carbonated soft drinks. The student
Lina20 [59]

Answer:

1) The bubbles will grow, and more may appear.

2)Can A will make a louder and stronger fizz than can B.

Explanation:

When you squeeze the sides of the bottle you increase the pressure pushing on the bubble, making it compress into a smaller space. This decrease in volume causes the bubble to increase in density. When the bubble increases in density, the bubble will grow and more bubbles will appear. Therefore, Changing the pressure (by squeezing the bottle) changes the volume of the bubbles. The number of bubbles doesn't change, just their size increases.

Carbonated drinks tend to lose their fizz at higher temperatures because the loss of carbon dioxide in liquids is increased as temperature is raised. This can be explained by the fact that when carbonated liquids are exposed to high temperatures, the solubility of gases in them is decreased. Hence the solubility of CO2 gas in can A at 32°C is less than the solubility of CO2 in can B at 8°C. Thus can A will tend to make a louder fizz more than can B.

3 0
2 years ago
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