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fredd [130]
3 years ago
9

Name an element that has the same number of electron energy levels as Calcium (Ca)?

Chemistry
1 answer:
Lerok [7]3 years ago
6 0

Answer:

aluminum?

Explanation:

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What is the heat of vaporization of ethanol, given that ethanol has a normal boiling point of 63.5°C and the vapor pressure of e
Serjik [45]

Explanation:

According to Clausius-Claperyon equation,

       ln (\frac{P_{2}}{P_{1}}) = \frac{-\text{heat of vaporization}}{R} \times [\frac{1}{T_{2}} - \frac{1}{T_{1}}]

The given data is as follows.

         T_{1} = 63.5^{o}C = (63.5 + 273) K

                         = 336.6 K

        T_{2} = 78^{o}C = (78 + 273) K

                         = 351 K

         P_{1} = 1 atm,             P_{2} = ?

Putting the given values into the above equation as follows.    

        ln (\frac{P_{2}}{P_{1}}) = \frac{-\text{heat of vaporization}}{R} \times [\frac{1}{T_{2}} - \frac{1}{T_{1}}]

       ln (\frac{1.75 atm}{1 atm}) = \frac{-\text{heat of vaporization}}{8.314 J/mol K} \times [\frac{1}{351 K} - \frac{1}{336.6 K}]

                      \Delta H = \frac{0.559}{1.466 \times 10^{-4}} J/mol

                                  = 0.38131 \times 10^{4} J/mol

                                  = 3813.1 J/mol

Thus, we can conclude that the heat of vaporization of ethanol is 3813.1 J/mol.

4 0
4 years ago
The "bleaching" of plant leaves is caused by: A.water pollution B.poor soil quality C.air pollution D.excess bacteria
devlian [24]

Answer: The "bleaching" of plant leaves is caused by air pollution.

8 0
3 years ago
The lipid portion of a typical bilayer is about 30 Å thick. a) Calculate the minimum number of residues in an α-helix required t
Mrrafil [7]

Answer:

a) The minimum number of residues in alfa helix to span 30 Å, is 20 residues

b) The minimum number of residues in a Beta sheet to span 30 Å, is 8.6 residues.

c) Should be better if you show an image to do this question.

d) Look answer below.

Explanation:

a) Due to every 5,4 Å there are 3,6 residues. So doing the calculus give us 20 residues.

b)  That’s it because every 7 Å there are 2 residues

c) First of all, is important to know that the residues of a protein define the secondary structure due to intermolecular and intramolecular interactions. Taking in account that the membrane bilayer has a lipid interior α-helical domains which are hydrophobic principally are embedded in membranes by hydrophobic interactions with the lipid interior of the bilayer favored by Van der Waals interactions and probably also by ionic interactions with the polar head groups of the phospholipids.

3 0
4 years ago
When a 0.235-g sample of benzoic acid is combusted in a bomb calorimeter, the temperature rises 1.643 ∘C . When a 0.275-g sample
andrew-mc [135]

Answer : The heat of combustion per mole of caffeine at constant volume is 76197.18 kJ/mole

Explanation :

First we have to calculate the specific heat calorimeter.

Formula used :

Q=m\times c\times \Delta T

where,

Q = heat of combustion of benzoic acid = 26.38 kJ/g = 26380 J/g

m = mass of benzoic acid = 0.235 g

c = specific heat of calorimeter = ?

\Delta T = change in temperature = 1.643^oC

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

26380J/g=0.235g\times c\times 1.643^oC

c=68323.38J/^oC

Thus, the specific heat of calorimeter is 68323.38J/^oC

Now we have to calculate the heat of combustion of caffeine.

Formula used :

Q=c\times \Delta T

where,

Q = heat of combustion of caffeine = ?

c = specific heat of calorimeter = 68323.38J/^oC

\Delta T = change in temperature = 1.584^oC

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

Q=68323.38J/^oC\times 1.584^oC

Q=108224.23J=108.2kJ

Now we have to calculate the moles of caffeine.

\text{Moles of caffeine}=\frac{\text{Mass of caffeine}}{\text{Molar mass of caffeine}}

Mass of caffeine = 0.275 g

Molar mass of caffeine = 194.19 g/mole

\text{Moles of caffeine}=\frac{0.275g}{194.19g/mole}=0.00142mol

Now we have to calculate the heat of combustion per mole of caffeine at constant volume.

\text{Heat of combustion per mole of caffeine}=\frac{108.2kJ}{0.00142mol}=76197.18kJ/mole

Therefore, the heat of combustion per mole of caffeine at constant volume is 76197.18 kJ/mole

4 0
3 years ago
How many moles are in 2.3 grams of phosphorus? :(
tatiyna

Answer:

there are 0.074 moles in 2.3 grams of phosphorus

4 0
4 years ago
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