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vivado [14]
3 years ago
5

i’m currently focusing on my algebra so i need a little help on a science study guide, i can’t post multiple pictures on the pag

es so please message me if you want to help

Chemistry
1 answer:
riadik2000 [5.3K]3 years ago
8 0

which question do you need help on

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173.1f is the answer I believe, please let me know if I'm wrong then I would try to make up for it
4 0
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Help! I need to finish work. Brainlist will be given to the most helpful!
Soloha48 [4]

Answer:

1. matches with elements.

2. matches with compounds.

3. matches with atoms

4. matches with weight

5. matches with gas

6. matches with carbon dioxide

7. matches with Mendeleev (there's an element named after him)

8. matches with IUPAC

Hope that helped :)

7 0
2 years ago
Do parts a, b and c
Leona [35]

Answer:- (a)The pH of the buffer solution is 3.90.

(b) the pH of the solution after addition of HCl would be 3.60.

(c) the pH of the buffer solution after addition of NaOH is 4.32.

Solution:- (a) It is a buffer solution so the pH could easily be calculated using Handerson equation:

pH=Pka+log(\frac{base}{acid})

pKa can be calculated from given Ka value as:

pKa=-logKa

pKa=-log(6.3*10^-^5)

pKa = 4.20

let's plug in the values in the Handerson equation:

pH=4.20+log(\frac{0.025}{0.05})

pH = 4.20 - 0.30

pH = 3.90

The pH of the buffer solution is 3.90.

(b) Let's say the acid is represented by HA and the base is represented by A^- .

Original mili moles of HA from part a = 0.05(100) = 5

original mili moles of A^- from part a = 0.025(100) = 2.5

mili moles of HCl that is H^+ added = 0.100(10.0) = 1

This HCl reacts with the base present in the buffer to make HA as:

A^-+H^+\rightarrow HA

Total mili moles of HA after addition of HCl = 5+1 = 6

mili moles of base after addition of HCl = 2.5-1 = 1.5

Let's plug in the values in the Handerson equation again. Here, we could use the mili moles to calculate the pH. The answer remains same even if we use the concentrations also as the final volume is same both for acid and base.

pH=4.20+log(\frac{1.5}{6})

pH = 4.20 - 0.60

pH = 3.60

So, the pH of the solution after addition of HCl would be 3.60.

(c) mili moles of NaOH or OH^- added to the original buffer = 0.05(15.0) = 0.75

This OH^- reacts with HA to form A^- as:

HA+OH^-\rightarrow H_2O+A^-

mili moles of HA after addition of NaOH = 5-0.75 = 4.25

mili moles of A^- after addition of NaOH = 2.5+0.75 = 3.25

Let's plug in the values again in Handerson equation:

pH=4.20+log(\frac{3.25}{4.25})

pH = 4.20 - 0.12

pH = 4.32

So, the pH of the buffer solution after addition of NaOH is 4.32.

7 0
3 years ago
Kon kon rajasthan ka h​
xenn [34]

Answer:

I am from Pakistan

Explanation:

Nice to meet you!

8 0
2 years ago
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