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sweet [91]
4 years ago
10

What is the oxidation number for Al in Al₂O₃?

Chemistry
1 answer:
sesenic [268]4 years ago
4 0
Oxygen in oxides has oxidation number -2.

<span>Al⁺³₂O⁻²₃

+3*2 -2*3=0</span>
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C2H6 is a chemical compound
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Read 2 more answers
How many grams of glucose are needed to prepare 400. g of a 2.00% (m/m) glucose solution g?
Dominik [7]
The grams   of glucose  are  needed  to  prepare  400g  of  a 2.00%(m/m)  glucose  solution  g  is  calculated  as  below

=% m/m =mass  of the solute/mass  of  the  solution  x100

let mass of   solute  be represented  by  y
mass  of solution = 400 g
 % (m/m) = 2% = 2/100

 grams  of  glucose  is  therefore =2/100 =  y/400
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y= 8.0 grams



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3 years ago
____________ is the process used to separate the solvent from a solution
bixtya [17]
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2 years ago
The study of the composition of matter is?
NemiM [27]

Chemistry. More specifically, analytical chemistry.

6 0
3 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
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