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alexdok [17]
2 years ago
6

14. Which of the following alkanols on oxidation is most likely to yield the

Chemistry
2 answers:
kari74 [83]2 years ago
6 0

Answer:

A

Explanation:

I think that is the ans but am not sure

Mila [183]2 years ago
4 0

Answer:

2-methylpropan-1-ol.

Explanation:

{ \tt{CH _{3}CH(CH _{3})CH _{2}OH}}

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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
The mass of an unidentified metal sphere is 133 grams. Students determine the
Rainbow [258]

Given the data from the question, the identity of the unknown metal having a of mass 133 g is Cobalt

<h3>What is density? </h3>

The density of a substance is simply defined as the mass of the subtance per unit volume of the substance. Mathematically, it can be expressed as

Density = mass / volume

<h3>How to determine the density </h3>
  • Mass = 133 g
  • Volume of water = 25 mL
  • Volume of water + metal = 40 mL
  • Vol of metal = 40 – 25 = 15 mL
  • Density =?

Density = mass / volume

Density = 133 / 15

Density = 8.86 g/mL

Comparing the density of the unknown metal (i.e 8.86 g/mL) with those given in the chart in the question above, we can conclude that the unknown metal is Cobalt

Learn more about density:

brainly.com/question/952755

4 0
2 years ago
A quantity of water is heated from 25.0°C to 36.4°C by absorbing 325 J of heat energy. If the specific heat of water is 4.18 J /
Arlecino [84]

Answer:

6,8 g

Explanation:

c = 4.18 J/(g * °C) = 4180 J / (kg * °C)

t_{1} = 25 °C

t_{2} = 36,4 °C

Q = 325 J

The formula is: Q = c * m * (t_{2} - t_{1})

m = \frac{Q}{c * (t_{2} - t_{1} )}

Calculating:

m = 325 / 4180 * (36,4 - 25) ≈ 0,0068 kg = 6,8 g

6 0
3 years ago
In the rock cycle, which rock type may be weathered to become sediment, and eventually sedimentary rock?
iragen [17]

i think it is I. and III.

:)

6 0
2 years ago
Read 2 more answers
El agua oxigenada se descompone rápidamente en agua y oxígeno cuando se le agrega una peueña cantidad de dióxido de manganeso, q
katovenus [111]

Answer:

IDK

Explanation:

I dont speak spanish

6 0
3 years ago
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