In what direction does an applied force move an object?

Scoring Scheme: 3-3-2-1 Use one of your experimentally determined values of k, the activation energy you determined, and the Arr

Lena [83]

Answer:

K = 2.7x10⁻⁵ at 25ºC

Explanation:

A way to write Arrhenius equation is:

ln K = - Ea/R × (1/T) + lnA

If you graph ln K as Y and 1/T as X (Absolute temperature in K), the equation you will obtain is:

Y = -13815X +35.817

R² = 0.9927

(Taking the last k point as 0.0386) (ln 0.0386), <em>0.1386 has no sense</em>)

Your slope is -13815

-13815K = - Ea/R

-13815K×8.314J/molK = 114858J/mol = Ea

And your intercept =

lnA = 35.817

A = 3.59x10¹⁵

Now, you want to know rate constant at 25ºC = 298.15K. Replacing in the equation (Where Y is ln (activation energy) and X is 1/T):

Y = -13815X +35.817

Y = -13815(1/298.15K) +35.817

Y = -10.5187

lnK = -10.5187

8 0

3 years ago

Which of these solutions has the highest freezing point?

ANEK [815]

Answer: 1.0 M molecular sucrose (C₁₂H₂₂O₁₁).

Explanation:

1) The depression of the freezing point of a solvent when you add a solute is a colligative property.

2) Colligative properties are those physical properties of solutions that depends on the number of solute particles dissolved into the solution.

3) The relation between the number of solute particles and the depresson of the freezing point is proportional: the greater the number of solute particles the greater the freezing point depression.

4) You need to find the solution with the highest freezing point, this is the solution in which the freezing point decreased the least.

5) Then, that is the solution with least number of solute particles.

6) Since all the given solutions have the same molarity (1.0 M), you only have to deal with the possible ionization of the different solutes.

7) NaCl, CaBr₂, AlBr₃, and KCl are ionic compounds, so each unit of them will ionize into two, three, four, and two ions, respectively, while sucrose, being a covalent compound does not dissociate.

Then, 1.0 M solution of sucrose will have less solute particles than the others, and will exhitibit the lowest freezing point depression, meaning that it will have the highest freezing point of the given solutions.

5 0

2 years ago

Read 2 more answers

If there are no changes in the oxidation state of the reactants or products of a particular reaction, that reaction is not a red

dmitriy555 [2]

Yes, that is true. in order for it to be a redox reaction, both oxidation and reduction must be occurring.

3 0

3 years ago

in which of the following is the symbol for the ion and the number of electrons it contains given correctly? *

gregori [183]

Answer:Number of electrons that are present in an atom is determined by the electronic configuration of that atom.

If an ion is carrying a positive charge, it means that the atom has lost electrons and if an ion is carrying a negative charge, it means that the atom has gained electrons.

For the given options:

Option A: The atomic number of hydrogen atom is 1 and the electronic configuration for ion will be:

Thus, this atom does not have any electrons.

Option B: The atomic number of bromine atom is 35 and the electronic configuration for ion will be:

Thus, this atom has 36 electrons.

Option C: The atomic number of aluminium atom is 13 and the electronic configuration for ion will be:

Explanation:

3 0

3 years ago

Question 12 of 25

NISA [10]

Answer:

B. Equal amounts of all gases have the same volume at the same

conditions

Explanation:

Amedo Avogadro found the relationship between volume of a gas and the number of molecules contained in the volume.

The law states that "equal volumes of all gases at the same temperature and pressure contains equal number of molecules or moles".

The law describes the behavior of gases when involve in chemical reactions. It enables one to change over at will in any statement about gases from volumes to molecules and vice versa.

So, the right option is B which implies that equal amounts of all gases have the same volume at the same conditions.

8 0

2 years ago