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Sphinxa [80]
3 years ago
8

Solution A is yellow when alizarin yellow is added and blue when thymol

Chemistry
1 answer:
olga2289 [7]3 years ago
4 0

Answer:

D. 6.3 x 10-5 mol/L NaOH

Explanation:

Alizarin yellow is an indicator that is yellow when pH < 10.1. In the same way, thymol blue is blue when pH > 9

That means the pH of the solution is between 9 - 10.1

Any acid as HCl could have a pH of these.

The solution of 3.2x10⁻⁴M NaOH has a pH of:

pOH = -log[OH-] = 3.49

pH = 14-pOH = 10.51. The pH of the solution is not 10.5

Now, the solution of 6.3x10⁻⁵M NaOH has a pH:

pOH = -log[OH-] = 4.2

pH = 14-pOH = 9.8

The pH of the solution could be 9.8. Right option is:

<h3>D. 6.3 x 10-5 mol/L NaOH</h3>
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3 years ago
Convert the density of water, 1.00 g/mL to lb/ft3. Provide only the numerical value in your answer with the correct number of si
PtichkaEL [24]

Answer:

The density of water 1.00 g/ml=62.43 lbs/ft^3.

Explanation:

Density of water = 1.00 g/mL

1 lb = 453.592 g

1 g=\frac{1}{453.592} lbs

ft^3=28316.8 mL

1 mL=\frac{1}{28316.8} ft^3

Density of the water in lb/ft^3:

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=62.43 lbs/ft^3

The density of water 1.00 g/ml=62.43 lbs/ft^3.

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4 years ago
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olga2289 [7]
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3 years ago
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Calculate the standard entropy of vaporization of ethanol at its boiling point, 352 K. The standard molar enthalpy of vaporizati
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Answer : The correct option is, (b) +115 J/mol.K

Explanation :

Formula used :

\Delta S=\frac{\Delta H_{vap}}{T_b}

where,

\Delta S = change in entropy

\Delta H_{vap} = change in enthalpy of vaporization = 40.5 kJ/mol

T_b = boiling point temperature = 352 K

Now put all the given values in the above formula, we get:

\Delta S=\frac{\Delta H_{vap}}{T_b}

\Delta S=\frac{40.5kJ/mol}{352K}

\Delta S=115J/mol.K

Therefore, the standard entropy of vaporization of ethanol at its boiling point is +115 J/mol.K

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