It is possible to increase a reaction rate by adding very small amounts of a catalyst, and this is because...

If the aluminum block is initially at 25 ∘C∘C, what is the final temperature of the block after the evaporation of the alcohol?

Effectus [21]

Answer:

Final temperature of aluminum block = 12.1°C

<em>Note: The question is not complete. The complete question is given below:</em>

<em>If the aluminum block is initially at 25 ∘C, what is the final temperature of the block after the evaporation of the alcohol? Assume that the heat required for the vaporization of the alcohol comes only from the aluminum block and that the alcohol vaporizes at 25 ∘C. Heat of vaporization of the alcohol at 25 ∘C is 45.4 kJ/mol Suppose that 1.12 g of rubbing alcohol (C3H8O) evaporates from a 73.0 g aluminum block.</em>

Explanation:

Heat lost by aluminum block = heat required for vaporization of alcohol

Heat required to vaporize ethanol, H = mass of alcohol * heat of vaporization of alcohol

Mass of alcohol = 1.12 g; molar mass of rubbing alcohol = 60 g/mol

Heat of vaporization = (45.4 kJ/mol)/ 60 g/mol = 0.75666 kJ/g

H = 1.12 g × 0.7566 kJ/g

H = 0.8474 kJ = 847.4 J

Heat lost by aluminum block, Q = -(mass × specific heat capacity × temperature change)

Mass of aluminum block = 73.0 g

Specific heat capacity of aluminum = 0.900 J/g

Temperature change = (Final temperature, T - 25)

Q = -73.0 g × 0.900 J/g × (T - 25)

Q = -65.7 J × (T - 25°C)

Since, Heat lost by aluminum block = heat for vaporization of alcohol

-65.7 J × (T - 25°C) = 847.4 J

T - 25 = 847.4/-65.7

T - 25 = -12.9

T = -12.9 + 25

T = 12.1°C

5 0

3 years ago

Generally, which of the three families of elements (metals, nonmetals, or inert gases) has the least tendency to form ionic bond

Sergio [31]

Answer:

Inert gases

Explanation:

Inert elements have a stable electron configuration meaning their shells/orbitals are full with their requisite number of electrons. Therefore, gaining or losing an electron would take high ionization energy. Therefore they are less likely to be involved in chemical reaction unless a high amount of energy is used. An example of an inert gas is Helium.

6 0

3 years ago

Please help I just need those last 4! :)

mojhsa [17]

Answer:

Series:

3. Thermistor

Parallel:

1. ???

2. ???

3. ???

5 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

Assuming an efficiency of 30.80%, calculate the actual yield of magnesium nitrate formed from 147.4g of magnesium and excess cop

Reil [10]

Answer:

The answer to your question is 280 g of Mg(NO₃)₂

Explanation:

Data

Efficiency = 30.80 %

Mg(NO₃)₂ = ?

Magnesium = 147.4 g

Copper (II) nitrate = excess

Balanced Reaction

Mg + Cu(NO₃)₂ ⇒ Mg(NO₃)₂ + Cu

Reactants Elements Products

1 Mg 1

1 Cu 1

2 N 2

6 O 6

Process

1.- Calculate the theoretical yield

Molecular weight Mg = 24

Molecular weight Mg(NO₃)₂ = 24 + (14 x 2) + (16 x 6)

= 24 + 28 + 96

= 148 g

24 g of Mg -------------------- 148 g of Mg(NO₃)₂

147.4 g of Mg ------------------- x

x = (147.4 x 148) / 24

x = 908.96 g of Mg(NO₃)₂

2.- Calculate the Actual yield

yield percent = $\frac{actual yield}{theoretical yield}$

Solve for actual yield

Actual yield = Yield percent x Theoretical yield

Substitution

Actual yield = $\frac{30.8}{100}$ x 908.96

Actual yield = 279.95 ≈ 280g

4 0

3 years ago