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KiRa [710]
3 years ago
6

Calculate the hfg and sfg of steam at 120oC from the Clapeyron equation, and compare them to the tabulated values.

Chemistry
1 answer:
den301095 [7]3 years ago
5 0

Answer:

The tabulated values at 120 C of <u><em>hfg = 2202.1</em></u><u> kJ/Kg </u>and <u><em>sfg </em></u><u>= 5.6 </u>

<u>kJ/Kg.K</u>

Explanation:

According to Clapeyron Equation :

h_{fg}=TV_{fg}\left (\frac{\partial P}{\partial T}  \right )_{sat}

=T(V_{g}-V_{f})_{120^{0}}\left ( \frac{\Delta P}{\125^{0}-115^{0}} \right )

Taking P2 at temperature 125 and P1 at 115 degree Celsius .

T = 120 + 273.15 = 393.15 K

P2 = 232.23 kPa and P1 = 169.18kPa

(these are experimental values given in the Data)

= 393.15(0.88133 -0.001060\ m^{3}/Kg)\frac{232.23-169.18 kPa}{10}

h_{fg} = 2206.8 kJ/Kg

S_{fg}=\frac{h_{fg}}{T}

\frac{2206.8}{393.15}

= 5.6131 kJ/Kg.K

The tabulated values at 120 C of <em>hfg = 2202.1</em> kJ/Kg and <em>sfg </em>= 5.6131

kJ/Kg.K

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What is the pOH of 0.5 M KOH?
jenyasd209 [6]

Answer:

pOH = 0.3

Explanation:

As KOH is a strong base, the molar concentration of OH⁻ is equal to the molar concentration of the solution. That means that in this case:

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With that information in mind we can<u> calculate the pOH </u>by using the following formula:

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5 0
3 years ago
Chemistry!! Please help asap!! Will mark brainiliest!!
Whitepunk [10]
<h2>Answer:</h2><h3>Part 1. </h3>

Option B is correct option.

The half-reaction 2MnO2 + H2O + 2e- Mn2O3 is missing  OH- ions.

Explanation:

Full equation:

                                          2MnO2 + H2O + 2e-  → Mn2O3 + 2OH-

<h3>Part 2:</h3>

The option B which is Mg is stronger reducing agent than Ag is correct option.

Explanation:

Equation:

                               Mg(s) + 2Ag+ (aq) → Mg2+ (aq) + 2Ag(s)

According to equation Mg converts to Mg+2 which means it gives to electron to reduce Ag. So it act as an reducing agent.

<h3>Part 3:</h3>

The correct option is B. Which is  5, 1, 8, 5, 1, 4.

Explanation:

Full equation :

                  5 Fe²⁺ (aq) + MnO₄⁻ (aq) + 8 H⁺ (aq) --> 5 Fe³⁺ (aq) + Mn²⁺ (aq) + 4 H₂O (l)


<h3 />

5 0
3 years ago
Hello
nordsb [41]

Answer:

The ratio of the number of atoms of gold and silver in the ornament is 197 : 1080.

Explanation:

Let the mass of silver ornament be x.

Mass of gold polished on an ornament = 1% of x= 0.1 x

Moles of silver =\frac{x}{108 g/mol}

Number of atoms = Moles\times N_A

Where : N_A = Avogadro number

Moles of gold =\frac{0.1x}{197 g/mol}

Silver atoms =\frac{x}{108 g/mol}\times N_A

Gold atoms =\frac{0.1x}{197 g/mol}\times N_A

The ratio of the number of atoms of gold and silver in the ornament:

=\frac{\frac{0.1x}{197 g/mol}\times N_A}{\frac{x}{108 g/mol}\times N_A}

= 197 : 1080

3 0
3 years ago
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