Chemistry College

What are the possible values of 1 and m for
n=4 ?​

Answers

Answer 1

Answer:

If n = 4, then the possible values of 1 and m depend on the equation or expression being used. Without more information, it is impossible to determine what the possible values of 1 and m might be. Can you please provide more context or information about the problem you are trying to solve?

Related Questions

A 75.0- mL
volume of 0.200 M
NH3
( Kb=1.8×10−5
) is titrated with 0.500 M
HNO3
. Calculate the pH
after the addition of 17.0 mL
of HNO3
.

Answers

Answer:

ok, here is your answer

Explanation:

i am going to solve this problem by using the ICE table method which is an easy method to determine the pH of a weak base with the given data of the problem.Given:Initial volume of NH3 solution (Vi) = 75.0 mLInitial concentration of NH3 solution (Ci) = 0.200 MInitial moles of NH3 solution (Ni) = Ci x Vi = 0.200 M x 75.0 mL = 0.0150 molesKb = 1.8 x 10^-5Moles of HNO3 added (n) = 0.500 M x 17.0 mL = 0.00850 molesVolume of NH3 solution after the addition of HNO3 (Vf) = 75.0 mL + 17.0 mL = 92.0 mLConcentration of NH3 solution after the addition of HNO3 (Cf) = Ni / Vf = 0.0150 moles / 92.0 mL = 0.163 MTo find the pH after the addition of 17.0 mL of HNO3, we need to use the ICE table method.ICE table method:Initial: NH3 + H2O ⇌ NH4+ + OH-Change: -x 0 +x +xEquilibrium: 0.0150 - x 0 x xKb = [NH4+][OH-] / [NH3]1.8 x 10^-5 = x^2 / 0.163Solving for x, x = 0.00171 M[OH-] = 0.00171 M[OH-] = Kw / [H3O+] = 1.0 x 10^-14 / [H3O+][H3O+] = 5.85 x 10^-12pH = -log[H3O+]pH = -log(5.85 x 10^-12)pH = 11.23Therefore, the pH after the addition of 17.0 mL of HNO3 is approximately 11.23.

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Why is it easy to describe things we can see?

Answers

Answer:

when you're trying to describe what you see, focus on describing the formal elements of a visual: line, colour, composition, texture, depth, symmetry, size etc. before elaborating on the value of what you want to see.

a student has a 1 L solution of 2 M HCL and wants to increase the HCL concentration to 3 M

Answers

The student needs to add approximately 83.3 mL of 12 M HCl solution to the existing 1 L of 2 M HCl solution to increase the concentration to 3 M. It is important to handle concentrated acids with caution and follow proper safety procedures.

To increase the concentration of a 1 L solution of 2 M HCl to 3 M, the student needs to calculate the volume of concentrated HCl needed and add it to the existing solution. Here's how the calculation can be done:

Given:

Initial concentration of HCl solution = 2 M

Final concentration desired = 3 M

Initial volume of HCl solution = 1 L

Step 1: Calculate the moles of HCl in the initial solution.

Moles of HCl = Initial concentration × Initial volume = 2 M × 1 L = 2 moles

Step 2: Calculate the moles of HCl needed for the desired concentration.

Moles of HCl needed = Final concentration × Final volume = 3 M × 1 L = 3 moles

Step 3: Calculate the moles of HCl to be added.

Moles of HCl to be added = Moles needed - Moles present = 3 moles - 2 moles = 1 mole

Step 4: Convert the moles of HCl to the required volume of concentrated HCl.

To calculate the volume, we need to know the concentration of the concentrated HCl solution. Assuming it is 12 M, we can use the following formula:

Volume of concentrated HCl = Moles of HCl to be added / Concentration of concentrated HCl

Volume of concentrated HCl = 1 mole / 12 M = 0.0833 L or 83.3 mL

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10. When dissolved in water, most Group 1 metal salts can be described as
strong electrolytes.
strong acids.
weak electrolytes.
A
B
C
D
non-electrolytes.
(1)

Answers

When dissolved in water, most Group 1 metal salts can be described as strong electrolytes.

When Group 1 metal salts are dissolved in water, they can be described as strong electrolytes. This is because Group 1 metals, such as lithium (Li), sodium (Na), potassium (K), and so on, readily lose their outermost valence electron to form positive ions (cations). These cations then dissociate completely in water, separating from the anions to which they were originally bonded.

The dissociation of Group 1 metal salts in water results in the formation of positively charged metal ions and negatively charged non-metal ions (anions). These ions are free to move and conduct electric current, making the solution a good conductor of electricity. The complete dissociation of Group 1 metal salts in water and the presence of freely moving ions make them strong electrolytes.

Strong electrolytes are substances that ionize completely or almost completely in solution, producing a high concentration of ions. This is in contrast to weak electrolytes, which only partially ionize and produce a lower concentration of ions.

In summary, when Group 1 metal salts are dissolved in water, they form strong electrolytes due to their ability to dissociate completely into ions, leading to a high concentration of freely moving ions in the solution, thus enabling efficient electrical conductivity.

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