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vagabundo [1.1K]
3 years ago
5

A 100.0 ml sample of 0.18 m hclo4 is titrated with 0.27 m lioh. determine the ph of the solution before the addition of any lioh

.
Chemistry
1 answer:
snow_lady [41]3 years ago
3 0
Thank you for posting your question here at brainly. Below is the solution:

 <span>moles HClO4 = 0.100 L x 0.18 M = 0.018 
moles LiOH = 0.030 L x 0.27 = 0.0081 
moles H+ in excess = 0.018 - 0.0081 = 0.0099 
total volume = 0.130 L 
[H+] = 0.0099/ 0.130= 0.0762 M 
pH = 1.12</span>
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Which of the following lists describes characteristics of a base? (3 points)
mel-nik [20]

Answer:

slippery, high pH, and caustic (last option)

Explanation:

when we say base we should think soap. soap is slippery. Bases give OH- ion. when OH- is combined with H+ ion it will create water which raises the pH. Since base can dissolve fats, ex: using dish liquid to cut grease on pots and pans etc.. they are caustic. Biologically they can disrupt the cell memebrane making it caustic to cell tissue.

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2 years ago
Which set of elements have the most similar chemical properties?
EleoNora [17]

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Would a strong acid have a large or a small ka? explain.
Zolol [24]
Very small pka. can also be - pka. 
7 0
2 years ago
In a plant, 1500kg of nitrogen oxide is consumed per day to produce 1,500 kg of nitrogen per day. What is the prevent yield?
slavikrds [6]

Answer:

65.21 percent

Explanation:

We are to find the percentage yield

We have this equation,

2no+o2 -->2no2

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= Mno2/46

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We cross multiply from here

1500x46 = 30xMnO2

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The percentage yield would be

1500/2300 *100

= 0.6521 x 100

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This answers the question

5 0
2 years ago
How would you prepare 2.00 l of a 0.25 m acetate buffer at ph= 4.50 from concentrated acetic acid (17.4 m) and 1.00 m naoh? the
zvonat [6]

The reaction of acetic acid with sodium hydroxide is:

CH_3COOH + NaOH    CH_3COONa + H_2O

The ratio of A-/HA is calculated as follows:

According to Henderson Hasslebach equation:


[A-]/[HA] = 10^p^H^-^p^K^_a

= 10^4^.^7^6^-^4^.^5^ = ^2^.^1^8

The total concentration of HA and A- = 2.0 L * 0.25 M = 0.5 mol.

[ A- ]+ [ HA ]= 0.5

[ A^- ] = 0.5 – [ HA]

[ HA] = 0.156 mol and [ A- ]= 0.343 mol

Total of 0.5 moles of acetic acid is required:

0.5 mol HA * 60.0 g

HA = 30.0 g acetic acid

Conversion of 0.343 moles of the acetic acid to acetate can be performed by adding NaOH

0.343 mol HA * (1 mol NaOH/1 mol HA) * (1000 mL/1 mol NaOH)

= 343 mL  

Thus, 343 mL of 1M NaOH is required

3 0
2 years ago
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