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sukhopar [10]
3 years ago
11

Determine the vapor pressure (atm) of rubbing alcohol (isopropanol) at 20.0 °C. The normal boiling point of isopropanol is 82.3

°C and the heat of vaporization (ΔHvap) is 39.9 kJ/mol.
Chemistry
1 answer:
jeyben [28]3 years ago
8 0
Here we apply the Clausius-Clapeyron equation:
ln(P₁/P₂) = ΔH/R x (1/T₂ - 1/T₁)

The normal vapor pressure is 4.24 kPa (P₁)
The boiling point at this pressure is 293 K (P₂)
The heat of vaporization is 39.9 kJ/mol (ΔH)
We need to find the vapor pressure (P₂) at the given temperature 355.3 K (T₂)

ln(4.24/P₂) = 39.9/0.008314 x (1/355.3 - 1/293)
P₂ = 101.2 kPa
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What will happen to the final pressure of a cylinder when the volume is doubled.
OLEGan [10]

If temperature were to double the pressure would likewise double. Increased temperature would increase the energy of the molecules and the number of collisions would also increase causing the increase in pressure.

7 0
3 years ago
What is the specific heat capacity of an unknown metal if 75.00 g of the metal absorbs 418.6J of heat and the temperature rises
EleoNora [17]

Answer:

The specific heat capacity of the unknown metal is 0.223 \frac{J}{g*C}

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

There is a direct proportional relationship between heat and temperature. The constant of proportionality depends on the substance that constitutes the body as on its mass, and is the product of the specific heat by the mass of the body. So, the equation that allows calculating heat exchanges is:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case, you know:

  • Q= 418.6 J
  • c= ?  
  • m= 75 g
  • ΔT= 25 C

Replacing:

418.6 J= c* 75 g* 25 C

Solving:

c=\frac{418.6 J}{75 g*25 C}

c= 0.223 \frac{J}{g*C}

<u><em>The specific heat capacity of the unknown metal is 0.223 </em></u>\frac{J}{g*C}<u><em></em></u>

<u><em> </em></u>

<u><em></em></u>

3 0
2 years ago
An amino acid A, isolated from the acid-catalyzed hydrolysis of a peptide antibiotic, gave a positive ninhydrin test and had a s
Ede4ka [16]
The answer could be 23
7 0
3 years ago
3. 2.15 x 3.1 x 100 =
jeka94

Answer:

666.5

Explanation:

Multiply 2.15 and 3.1 to get 6.665.

6.665×100≈666.5

Multiply 6.665 and 100 to get 666.5.

666.5

3 0
3 years ago
Read 2 more answers
Calculate the unit cell edge length for an 85 wt% fe-15 wt% v alloy. All of the vanadium is in solid solution, and, at room temp
Lady bird [3.3K]

Answer is 0.289nm.

Explanation: The wt % of Fe and wt % of V is given for a Fe-V alloy.

wt % of Fe in Fe-V alloy = 85%

wt % of V in Fe-V alloy = 15%

We need to calculate edge length of the unit cell having bcc structure.

Using density formula,

\rho_{ave}=\frac{Z\times M_{ave}}{a^3\times N_A}

For calculating edge length,

a=(\frac{Z\times M_{ave}}{\rho_{ave}\times N_A})^{1/3}

For calculating M_{ave}, we use the formula

M_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{V}}{M_V}}

Similarly for calculating (\rho)_{ave}, we use the formula

\rho_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{\rho_{Fe}}+\frac{(wt\%)_{V}}{\rho_V}}

From the periodic table, masses of the two elements can be written

M_{Fe}= 55.85g/mol

M_{V}=50.941g/mol

Specific density of both the elements are

(\rho)_{Fe}=7.874g/cm^3\\(\rho)_{V}=6.10g/cm^3

Putting  M_{ave} and \rho_{ave} formula's in edge length formula, we get

a=\left [\frac{Z\left (\frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{Fe}}{M_{Fe}}}  \right )}{N_A\left (\frac{100}{\frac{(wt\%)_V}{\rho_V}+\frac{(wt\%)_V}{\rho_V}}  \right )}  \right ]^{1/3}

a=\left [\frac{2atoms/\text{unit cell}\left (\frac{100}{\frac{85\%}{55.85g/mol}+\frac{15\%}{50.941g/mol}}  \right )}{(6.023\times10^{23}atoms/mol)\left (\frac{100}{\frac{85\%}{7.874g/cm^3}+\frac{15\%}{6.10g/cm^3}}  \right )}  \right ]^{1/3}

By calculating, we get

a=2.89\times10^{-8}cm=0.289nm

7 0
2 years ago
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