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andrey2020 [161]

3 years ago

5

Helppp nowwww plsss

2 answers:

Keith_Richards [23]3 years ago

6 0

Option 3 is the correct answer

Gnom [1K]3 years ago

4 0

It’s humid it’s number 3

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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

Annette [7]

<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>

6 0

2 years ago

A solid powder is known to be a mixture of NaCl and Na2CO3, but the relative amounts of each compound in the sample are unknown.

AveGali [126]

Answer:

Concentration of sodium carbonate in the solution before the addition of HCl is 0.004881 mol/L.

Explanation:

$Na_2CO_3(aq) +2 HCl(aq)\rightarrow 2NaCl(aq) + H_2O(l)+CO_2(g)$

Molarity of HCl solution = 0.1174 M

Volume of HCl solution = 83.15 mL = 0.08315 L

Moles of HCl = n

$molarity=\frac{moles}{Volume (L)}$

$0.1174 M=\frac{n}{0.08315 L}$

$n=0.1174 M\times 0.08315 L=0.009762 mol$

According to reaction , 2 moles of HCl reacts with 1 mole of sodium carbonate.

Then 0.009762 mol of HCl will recat with:

$\frac{1}{2}\times 0.009762 mol=0.004881 mol$

Moles of Sodium carbonate = 0.004881 mol

Volume of the sodium carbonate containing solution taken = 1L

Concentration of sodium carbonate in the solution before the addition of HCl:

$[Na_2CO_3]=\frac{0.004881 mol}{1 L}=0.004881 mol/L$

4 0

3 years ago