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Daniel [21]
2 years ago
14

PLEASE HELP WORTH 100 POINTS !!!!!

Chemistry
2 answers:
Rasek [7]2 years ago
5 0

Answer: Imidogen

Explanation: I just looked it up.

8_murik_8 [283]2 years ago
4 0

Answer:

Imidogen

Explanation:

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1. Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium
lilavasa [31]

Answer:

This question is incomplete, here's the complete question:

1. Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium ion present in 20.0 mL of each buffer.

Buffer A

Mass of sodium acetate used: 0.3730 g

Actual ph of the buffer 5.27

volume of the buffer used in buffer capacity titration 20.0 mL

Concentration of standardized NaOH 0.100M

moles of Naoh needed to change the ph by 1 unit for the buffer 0.00095mol

the buffer capacity 0.0475 M

Buffer B

Mass of sodium acetate used 1.12 g

Actual pH of the buffer 5.34

Volume of the buffer used in buffer capacity titration 20.0 mL

Concentration if standardized NaOH 0.100 M

moles of Naoh needed to change the ph by 1 unit 0.0019 mol

the buffer capacity 0.095 M

2.) A change of pH by 1 unit means a change in hydronium ion concentration by a factor of 10. Calculate the number of moles of NaOH that would theoretically be needed to decrease the moles of hydronium you calculated in #1 by a factor of 10 for each buffer. Are there any differences between your experimental results and the theoretical calculation?

3.) which buffer had a higher buffer capacity? Why?

Explanation:

Formula,

moles = grams/molar mass

molarity = moles/L of solution

1. Buffer A

molarity of NaC2H3O2 = 0.3731 g/82.03 g/mol x 0.02 L = 0.23 M

molarity of HC2H3O2 = 0. 1 M

Initial pH

pH = pKa + log(base/acid)

= 4.74 + log(0.23/0.1)

= 5.10

pH = -log[H3O+]

[H3O+] = 7.91 x 10^-6 M

In 20 ml buffer,

moles of H3O+ = 7.91 x 10^-6 M x 0.02 L

= 1.58 x 10^-7 mol

Buffer B

molarity of NaC2H3O2 = 1.12 g/82.03 g/mol x 0.02 L = 0.68 M

molarity of HC2H3O2 = 0.3 M

Initial pH

pH = pKa + log(base/acid)

= 4.74 + log(0.68/0.3)

= 5.10

pH = -log[H3O+]

[H3O+] = 7.91 x 10^-6 M

In 20 ml buffer,

moles of H3O+ = 7.91 x 10^-6 M x 0.02 L

= 1.58 x 10^-7 mol

2. let x moles of NaOH is added,

Buffer A,

pH = 5.10

[H3O+] = 7.91 x 10^-6 M

new pH = 4.10

new [H3O+] = 7.91 x 10^-5 M

moles of NaOH to be added = (7.91 x 10^-5 - 7.91 x 10^-6) x 0.02 L

= 1.42 x 10^-6 mol

3. Buffer B with greater concentration of NaC2H3O2 and HC2H3O2 has higher buffer capacity as it resists pH change to a wider range due to addition of acid or base to the system as compared to low concentration of Buffer A

5 0
3 years ago
Aluminum metal and manganese (IV) oxide react to produce aluminum oxide and manganese metal. What mass of manganese (IV) oxide r
gayaneshka [121]

Answer: 1.77 kg of manganese (IV) oxide reacts to produce 1.12kg of manganese metal.

Explanation:

The balanced chemical equation is:

4Al+3MnO_2\rightarrow 2Al_2O_3+3Mn

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

moles of manganese = \frac{1.12\times 1000g}{55g/mol}=20.4mol

According to stoichiometry :

3 moles of Mn is produced by = 3 moles of MnO_2

Thus 20.4 moles of Mn is produced by =\frac{3}{3}\times 20.4=20.4moles  of MnO_2

Mass of MnO_2=moles\times {\text {Molar mass}}=20.4moles\times 86.9g/mol=1773g=1.773kg    (1kg=1000g)

Thus 1.77 kg of manganese (IV) oxide reacts to produce 1.12kg of manganese metal.

6 0
3 years ago
Why were foreigners allowed to hold office for the Mongols
Slav-nsk [51]
They shots sides stoned found six end
6 0
3 years ago
HELP ME PLS I NEED HELP I HAVE 5MINS
Leviafan [203]

Answer:

K

Explanation:

Since the blood moves from the body to the right atrium to the right ventricle

3 0
2 years ago
Which law states that the entropy of the universe is always increasing?
vova2212 [387]

Second law of thermodynamics states entropy of the universe is always increasing. entropy always increases in a spontaneous process .

option B) second law of thermodynamics

for example if we keep a hot cup of coffee the heat is lost to surrounding and entropy( randomness increases) since it is a natural process entropy of universe increases.

5 0
2 years ago
Read 2 more answers
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