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Studentka2010 [4]
3 years ago
12

What do you use to test ph in swimming pool

Chemistry
1 answer:
Alona [7]3 years ago
7 0

Answer:

to make sure the cholerie is not to high

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QUESTION 20
Assoli18 [71]

Answer:

49.56

Explanation:

5 0
2 years ago
Calculate the pH of a buffer solution that contains 0.25 M benzoic acid (C 6H 5CO 2H) and 0.15M sodium benzoate (C
gtnhenbr [62]

Answer:

pH=3.97

Explanation:

Hello,

In this case, for the calculation of the pH of the given buffer we need to use the Henderson-Hasselbach equation:

pH=pKa+log(\frac{[base]}{[acid]} )

Whereas the pKa for benzoic acid is 4.19, the concentration of the base is 0.15 M (sodium benzoate) and the concentration of the acid is 0.25 M (benzoic acid), therefore, the pH turns out:

pH=4.19+log(\frac{0.15M}{0.25M} )\\\\pH=3.97

Regards.

4 0
3 years ago
What is the molecular weight of one mole of H2CO3? g/mole
eimsori [14]
The answer is 62.026 g/mole
6 0
3 years ago
Read 2 more answers
HI! I REALLY NEED HELP, THANK YOU!!
worty [1.4K]

Answer:

Nuclear fusion is a reaction in which two or more atomic nuclei are combined to form one or more different atomic nuclei and subatomic particles. The difference in mass between the reactants and products is manifested as either the release or the absorption of energy. 

Explanation: Read this and you might be able to figure it out for yourself ☺️☺️☺️

6 0
3 years ago
A 0.1510 gram sample of a hydrocarbon produces 0.5008 gram CO2 and 0.1282 gram H2O in combustion analysis. Its
Over [174]
In a combustion of a hydrocarbon compound, 2 reactions are happening per element:

C + O₂ → CO₂
2 H + 1/2 O₂ → H₂O

Thus, we can determine the amount of C and H from the masses of CO₂ and H₂O produced, respectively.

1.) Compute for the amount of C in the compound. The data you need to know are the following:
Molar mass of C = 12 g/mol
Molar mass of CO₂ = 44 g/mol
Solution:
0.5008 g CO₂*(1 mol CO₂/ 44 g)*(1 mol C/1 mol CO₂) = 0.01138 mol C
0.01138 mol C*(12 g/mol) = 0.13658 g C

Compute for the amount of H in the compound. The data you need to know are the following:
Molar mass of H = 1 g/mol
Molar mass of H₂O = 18 g/mol
Solution:
0.1282 g H₂O*(1 mol H₂O/ 18 g)*(2 mol H/1 mol H₂O) = 0.014244 mol H
0.014244 mol H*(1 g/mol) = 0.014244 g H

The percent composition of pure hydrocarbon would be:
Percent composition = (Mass of C + Mass of H)/(Mass of sample) * 100
Percent composition = (0.13658 g + 0.014244 g)/(<span>0.1510 g) * 100
</span>Percent composition = 99.88%

2. The empirical formula is determined by finding the ratio of the elements. From #1, the amounts of moles is:

Amount of C = 0.01138 mol
Amount of H = 0.014244 mol

Divide the least number between the two to each of their individual amounts:
C = 0.01138/0.01138 = 1
H = 0.014244/0.01138 = 1.25

The ratio should be a whole number. So, you multiple 4 to each of the ratios:
C = 1*4 = 4
H = 1.25*4 = 5

Thus, the empirical formula of the hydrocarbon is C₄H₅.

3. The molar mass of the empirical formula is

Molar mass = 4(12 g/mol) + 5(1 g/mol) = 53 g/mol
Divide this from the given molecular weight of 106 g/mol
106 g/mol / 53 g/mol = 2
Thus, you need to multiply 2 to the subscripts of the empirical formula.

Molecular Formula = C₈H₁₀

4 0
3 years ago
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