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2 years ago

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what is the pH of a solution prepared from solid, neutral 2-nitrophenol providing a fromal concentration of 0.0353M, given that

the Ka of 2nitrophenol is 5.98 x 10^-8

1 answer:

REY [17]2 years ago

7 0

Answer:

pH = 4.34

Explanation:

pH= -1/2(logKa) -1/2(log C)

= -1/2( log 5.98*10^-8) -1/2(log 0.0353)

=-1/2(-7.22)-1/2(-1.45)

=3.61+0.725= 4.34

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A bottle containing 415 g of cleaning solution is used to clean hospital equipment. If the cleaning solution has a specific grav

neonofarm [45]

Answer: 483 mL of the cleaning solution are used to clean hospital equipment

Explanation:

The question requires us to calculate the volume, in mL, of solution is used to clean hospital equipment, given that 415g of this solution are used and the specific gravity of the solution is 0.860.

Measurements > Density

Specific gravity is defined as the ratio between the density of a given substance to the density of a reference material, such as water:

$Specific\text{ gravity = }\frac{density\text{ of substance}}{density\text{ of reference substance}}$

The density of a substance is defined as the ratio between the mass and the volume of this substance:

$density=\frac{mass}{volume}$

Considering the reference substance as water and its density as 1.00 g/mL, we can determine the density of the substance which specific gravity is 0.860:

$0.860=\frac{density\text{ of substance}}{1.00g/mL}\rightarrow density\text{ of substance = 0.860g/mL}$

Thus, taking water as the reference substance, we can say that the density of the cleaning solution is 0.860 g/mL.

Now that we know the density of the cleaning solution (0.860 g/mL) and the mass of solution that is used to clean hospital equipment (415g), we can calculate the volume of solution that is used to clean the equipment:

$\begin{gathered} density=\frac{mass}{volume}\rightarrow volume=\frac{mass}{density} \\ \\ volume=\frac{415g}{0.860g/mL}=483mL \end{gathered}$

Therefore, 483 mL of the cleaning solution are used to clean hospital equipment.

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10 months ago

Water vapor enters a compressor at 35 kpa and 160°c and leaves at 300 kpa with the same specific entropy as at the inlet. What a

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<h3><em><u>solution</u></em><em><u>:</u></em></h3>

<em><u>The initial entropy is obtained from the initial pressure and temperature with data from A-6 using interpolation:</u></em>

<em><u>s</u></em><em><u>=</u></em><em><u> </u></em><em><u>8</u></em><em><u>.</u></em><em><u>26</u></em><em><u>5</u></em><em><u>2</u></em><em><u> </u></em><em><u>kJ</u></em><em><u>/</u></em><em><u>kgK</u></em>

<em><u>The final temperature is determined from the entropy and the final pressure with data from A-6 using interpolation:</u></em>

<em><u>T₂ = T₁+</u></em><em><u> </u></em><em><u>T₂ - </u></em><em><u>T₁</u></em><em><u>/</u></em><em><u> </u></em><em><u>8</u></em><em><u>₂</u></em><em><u> </u></em><em><u>-</u></em><em><u> </u></em><em><u>8</u></em><em><u>₁</u></em><em><u> </u></em><em><u>(</u></em><em><u> </u></em><em><u>s</u></em><em><u> </u></em><em><u>-</u></em><em><u> </u></em><em><u>s</u></em><em><u>₁</u></em><em><u>)</u></em>

<em><u>= </u></em><em><u>(</u></em><em><u>400 +</u></em><em><u> </u></em><em><u>500 - 400</u></em><em><u>/</u></em><em><u>8.3271</u></em><em><u> </u></em><em><u>-</u></em><em><u> </u></em><em><u>8.0347</u></em><em><u> </u></em><em><u>(8.2652 - 8</u></em><em><u>)</u></em><em><u>)</u></em>

<em><u>= 478.83°C</u></em>

<em><u>The final enthalpy is determined in the same way:</u></em>

<em><u>h₂= h₁</u></em><em><u> </u></em><em><u>+</u></em><em><u> </u></em><em><u>h₂</u></em><em><u> </u></em><em><u>-</u></em><em><u> </u></em><em><u>h₁</u></em><em><u>/</u></em><em><u>s</u></em><em><u>₂</u></em><em><u> </u></em><em><u>-</u></em><em><u> </u></em><em><u>s</u></em><em><u>₁</u></em><em><u> </u></em><em><u>( s - s₁)</u></em>

<em><u>= (</u></em><em><u> </u></em><em><u>3275.5</u></em><em><u>+</u></em><em><u> </u></em><em><u>3486.6 </u></em><em><u>-</u></em><em><u> </u></em><em><u>3275.5</u></em><em><u>/</u></em><em><u> </u></em><em><u>8.3271</u></em><em><u> </u></em><em><u>-</u></em><em><u> </u></em><em><u>8.0347</u></em><em><u>)</u></em><em><u> </u></em><em><u>(8.265</u></em><em><u>)</u></em>

<em><u>=</u></em><em><u> </u></em><em><u>3441.91 </u></em><em><u>kJ</u></em><em><u>/</u></em><em><u>kg</u></em>

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1 year ago