All categories

2 years ago

Match the followings.

Chemistry

1 answer:

insens350 [35]2 years ago

5 0

Answer:

Explanation:

Here, we are to copy and complete the paragraph with the correct options as given below in the question.

Before that, let us have an understanding of what a conversion factor is all about.

A conversion factor refers to a formula that is being used to change over an estimation in a given arrangement of units to its comparable estimation in another arrangement of units. Subsequently, the formula comprises both a number and its unit.

So;

$\text{You can check that you have written the correct conversion factors for an }$

$\text{equality by looking at the numbers and units in each; the information on }$

$\text{one side of the equity appears in the of a conversion factor while the }$

$\text{information on the other side of the equity appears in the denominator.}$

$\text{To generate a second conversion factor, we "turn over" the first }$

$\text{conversion factor so the numbers and units in the numerator and }$

$\text{denominator switch places.}$

You might be interested in

Ka/KbMIXED PRACTICE108.Calculate the [H3O+(aq)], the pH, and the % reaction for a 0.50 mol/L HCN solution. ([H3O+(aq)] = 1.8 x 1

NeTakaya

Answer:

a) [H₃O⁺] = 1.8x10⁻⁵ M

b) pH = 4.75

c) % rxn = 3.5x10⁻³ %

Explanation:

a) The dissociation reaction of HCN is:

HCN(aq) + H₂O(l) ⇄ H₃O⁺(aq) + CN⁻(aq)

0.5 M - x x x

The dissociation constant from the above reactions is given by:

$Ka = \frac{[H_{3}O^{+}][CN^{-}]}{[HCN]} = 6.17 \cdot 10^{-10}$

$6.17 \cdot 10^{-10} = \frac{x*x}{(0.5 - x)}$

$6.17 \cdot 10^{-10}*(0.5 - x) - x^{2} = 0$

By solving the above quadratic equation we have:

x = 1.75x10⁻⁵ M = 1.8x10⁻⁵ M = [H₃O⁺] = [CN⁻]

Hence, the [H₃O⁺] is 1.8x10⁻⁵ M.

b) The pH is equal to:

$pH = -log[H_{3}O^{+}] = -log(1.75 \cdot 10^{-5} M) = 4.75$

Then, the pH of the HCN solution is 4.75.

c) The % reaction is the % ionization:

$\% = \frac{x}{[HCN]} \times 100$

$\% = \frac{1.75 \cdot 10^{-5} M}{0.5 M} \times 100$

$\% = 3.5 \cdot 10^{-3} \%$

Therefore, the % reaction or % ionization is 3.5x10⁻³ %.

I hope it helps you!

6 0

2 years ago

Lindsey is trying to gain credibility for her studies. She completed her experiment and discussed her finding with colleagues. H

noname [10]

Answer:

Option B:Publishing scientific journals

Explanation:

We are told that Lindsey is trying to gain credibility for her studies.

Since she completed her experiment and discussed her finding with colleagues, the most logical next step would be to publish scientific journals. This is because the other options given are not steps that should be taken because she has completed the research and therefore has no need to speak at a conference next nor even create new charts which they must have done during the research. No need for her to make sure the topic is popular.

Option B is correct

8 0

3 years ago

How many electrons does H have with a -1 charge

pentagon [3]

Answer:The electron has a negative charge and the proton has a positive charge, and these charges work against each other to make the electromagnetic force that holds the entire atom together.

Explanation:

This is the answer

5 0

3 years ago

Only_________are involved in forming chemical bonds betwen two atoms

olchik [2.2K]

Only valence electrons are involved in forming chemical bonds betwen two atoms.

Hope this helped you- have a good day bro cya)

4 0

3 years ago

The acid-dissociation constants of HC3H5O3 and CH3NH3+ are given in the table below. Which of the following mixtures is a buffer

sergey [27]

Answer:

A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH

Explanation:

The pH of a buffer solution is calculated using following relation

$pH=pKa+log(\frac{salt}{acid} )$

Thus the pH of buffer solution will be near to the pKa of the acid used in making the buffer solution.

The pKa value of HC₃H₅O₃ acid is more closer to required pH = 4 than CH₃NH₃⁺ acid.

pKa = -log [Ka]

For HC₃H₅O₃

pKa = 3.1

For CH₃NH₃⁺

pKa = 10.64

pKb = 14-10.64 = 3.36 [Thus the pKb of this acid is also near to required pH value)

A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH

Half of the acid will get neutralized by the given base and thus will result in equal concentration of both the weak acid and the salt making the pH just equal to the pKa value.

8 0

3 years ago