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gizmo_the_mogwai [7]
3 years ago
7

Which of the following is not a chemical change?

Chemistry
1 answer:
MaRussiya [10]3 years ago
3 0

Answer:

May be boiling is the answer because all other options are a proper chemical changes.

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Which of the following compounds is always part of an aqueous solution
harkovskaia [24]
The answer is water.
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20 mL of an approximately 10% aqueous solution of ethylamine, CH3CH2NH2, is titrated with 0.3000 M aqueous HCl. Which indicator
Vinvika [58]

20 mL of an approximately 10% aqueous solution of ethylamine, CH3CH2NH2, is titrated with 0.3000 M aqueous HCl. Which indicator would be most suitable for this titration? The pKa of CH3CH2NH3+ is 10.75.

Answer:

Bromocresol green, color change from pH = 4.0 to  5.6

Explanation:

The equation for the reaction is as follows:

C_2H_5NH_{2(aq)     +     H^+_{(aq)      ⇄        C_2H_5NH_{3(aq)}^+

Given that concentration of C_2H_5NH_{2(aq) = 10%

i.e 10 g of C_2H_5NH_{2(aq) in 100 ml solution

molar mass = 45.08 g/mol

number of moles = \frac{10}{45.08}

= 0.222 mol

Molarity of C_2H_5NH_{2(aq) = 0.222 × \frac{1000}{100}mL

= 2.22 M

However, number of moles of C_2H_5NH_{2(aq) in 20 mL can be determined as:

number of moles of C_2H_5NH_{2(aq) = 20 mL × 2.22 M

= 44*10^{-3} mole

Concentration of C_2H_5NH_{2(aq) = \frac{44*10^{-3}*1000}{20}

= 2.22 M

Similarly, The pKa Value of C_2H_5NH_{2(aq)  is given as 10.75

pKb value will be: 14 - pKa

= 14 - 10.75

= 3.25

Finally, the pH value at equivalence point is:

pH= \frac{1}{2}pKa - \frac{1}{2}pKb-\frac{1}{2}log[C]

pH = \frac{14}{2}-\frac{3.25}{2}-\frac{1}{2}log [2.22]

pH = 5.21

∴

The indicator that is best fit for the given titration is Bromocresol Green Color change from pH between 4.0 to 5.6.

6 0
4 years ago
HELPPPP MEEE PLEASEEE!!!!!!
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Therefore, 40 +30=70
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Which group has 4 valence electrons
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Group 14: Carbon family has valence electrons. 
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23. The reaction between solid sodium and iron(ill) oxide 15
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A. Limiting reactant

Explanation:

Mark me as a brainliest happy po ako sa pag sagot sana tama ako kung mali paki. Comment lang dito kasi solve ako ulit

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