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polet [3.4K]
3 years ago
10

One of your classmates is having trouble understanding ions. He explains the formation of a cation like this:

Chemistry
1 answer:
Anettt [7]3 years ago
3 0
Electrons carry a negative charge adding 2 electrons together would give a positive charge because 2 negatives make a positive.
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Enough of a monoprotic acid is dissolved in water to produce a 0.0121 M solution. The pH of the resulting solution is 6.49. Calc
Hitman42 [59]
Calculate the H positive from the pH equation: pH equals -log (H positive). This would be 10 to the -6.49. Let's call the acid HA. To calculate Ka in this equation, Ka equals H positive times A- over HA. HA is going to be the 0 0121. So, Ka=(10^-6.49)^2/0.0121. This equals 1.05*10^-13/0.0121. Ka then equals 8.65*10^-12.
4 0
3 years ago
Kc= [3.1*10^-2]^2. (mol/L^2) divide by [8.5*10^-1] [3.1*10^-3]^3 (mol/L)^4​
jok3333 [9.3K]

Answer:

Kc = 12.58

Explanation:

Kc = [0.229]^2*[0.687]^6/[0.221]^4*[0.5685]^3

Kc = (0.052441)(0.10513)/(0.002385)(0.18373)

Kc = 0.0005513/0.000438

Kc = 12.58

Hope that helps!!

5 0
3 years ago
An analytical chemist is titrating of a solution of propionic acid with a solution of 224.9 ml of a 0.6100M solution of propioni
Svetllana [295]

<u>Answer:</u> The pH of acid solution is 4.58

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}    .....(1)

  • <u>For KOH:</u>

Molarity of KOH solution = 1.1000 M

Volume of solution = 41.04 mL

Putting values in equation 1, we get:

1.1000M=\frac{\text{Moles of KOH}\times 1000}{41.04}\\\\\text{Moles of KOH}=\frac{1.1000\times 41.04}{1000}=0.04514mol

  • <u>For propanoic acid:</u>

Molarity of propanoic acid solution = 0.6100 M

Volume of solution = 224.9 mL

Putting values in equation 1, we get:

0.6100M=\frac{\text{Moles of propanoic acid}\times 1000}{224.9}\\\\\text{Moles of propanoic acid}=\frac{0.6100\times 224.9}{1000}=0.1372mol

The chemical reaction for propanoic acid and KOH follows the equation:

                 C_2H_5COOH+KOH\rightarrow C_2H_5COOK+H_2O

<u>Initial:</u>          0.1372         0.04514  

<u>Final:</u>           0.09206          -                0.04514

Total volume of solution = [224.9 + 41.04] mL = 265.94 mL = 0.26594 L     (Conversion factor:  1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[\text{salt}]}{[acid]})

pH=pK_a+\log(\frac{[C_2H_5COOK]}{[C_2H_5COOH]})

We are given:  

pK_a = negative logarithm of acid dissociation constant of propanoic acid = 4.89

[C_2H_5COOK]=\frac{0.04514}{0.26594}

[C_2H_5COOH]=\frac{0.09206}{0.26594}

pH = ?  

Putting values in above equation, we get:

pH=4.89+\log(\frac{(0.04514/0.26594)}{(0.09206/0.26594)})\\\\pH=4.58

Hence, the pH of acid solution is 4.58

7 0
3 years ago
How many electrons does nitrogen (N) need to gain to have a stable outer electron shell
earnstyle [38]
It need "3 electrons" to have a stable electronic configuration. 

'cause it has 5 electrons in it's outer shell & every atom needs 8 electrons. So, it requires 3 more!

Hope this helps!
7 0
2 years ago
Read 2 more answers
Which subscripts would properly complete the formula unit Al_N_?
MakcuM [25]
I believe that it would be Al1N1.
4 0
3 years ago
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