CHEMISTRY COLLEGE

PLEASE HELP FAST!!If metal X is lower than metal Y on the activity series, then:
A. X will react in water, but only if the temperature is low enough
B. Y will form oxides of X, but only indirectly
C. X will replace ions of Y in a solution
D. Y will replace ions of X in a solution

Answers

Answer 1
Answer:

Answer:

D. Y will replace ions of X in a solution

Explanation:

If metal X is lower than metal Y on the activity series, then Y will replace ions of X in a solution.

This is the crux of single displacement reactions.

  • In a single displacement reaction, a metallic ion in solution is replaced by a metal higher in the activity series than the metal in solution.
  • On the activity series, metals higher are more reactive and will displace the lower and less reactive ones.
  • Reactivity increases up the group.

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Answers

Answer:

Mhmmm I Agree

Explanation:

Its Because I Strongly Agree

An imaginary line dividing the earth's surface into two hemisphere the northern and southern hemisphere, it is locatedat 0⁰, which of the following imaginary lines is being described? a.equator b.latitude c.longitude d.prime meridian​

Answers

Answer:

A. Equator

Explanation:

The equator is located in the centre of the Earth, dividing the northern and southern hemispheres.

Answer:

Equator because equator divides the earth into Northern and Southern hemisphere.

Which molecule does not exhibit hydrogen bonding?a. HFb. CH3NH2c. CH2F2d. HOCH2CH2OH

Answers

Answer:

(c) CH₂F₂

Explanation:

Hydrogen bonds are weak intermolecular forces. They are the strongest kind of intermolecular forces, although they are weaker than the covalent bonds.

Hydrogen bonds arise from molecules which contain a hydrogen atom which is bonded to one of the most electronegative elements such as N, O or F.

(a) HF,  → has H-F bond

(b) CH₃NH₂,   → has N-H bond

(c) CH₂F₂,  → has no H-F bond ( F- C- F)

(d) HOCH₂CH₂OH, → has O-H bond

Therefore, only CH₂F₂ does not exhibit hydrogen bonding.

In the following structure, carbons (I),(2),(3) and (4) are classified respectively as

Answers

Answer:

Carbon (i) : quaternary carbon

Carbon (ii) : secondary carbon

Carbon (iii) : tertiary carbon

Carbon (iv) : secondary carbon

Explanation:

Carbons can be classified into 4 categories:

(1) Primary carbon(1^o): These are the atoms where the carbon atom is attached to one other carbon atom.

(2) Secondary carbon(2^o): These are the atoms where the carbon atom is attached to two other carbon atoms.

(3) Tertiary carbon(3^o): These are the atoms where the carbon atom is attached to three other carbon atoms.

(4) Quaternary carbon(4^o): These are the atoms where the carbon atom is attached to four other carbon atoms.

In the given structure:

Carbon (i) is attached to 4 further carbon atoms and hence, it is a quaternary carbon.

Carbon (ii) is attached to 2 further carbon atoms and hence, it is a secondary carbon.

Carbon (iii) is attached to 3 further carbon atoms and hence, it is a tertiary carbon.

Carbon (iv) is attached to 2 further carbon atoms and hence, it is a secondary carbon.

Calculate the pH of a solution prepared by adding 20.0 mL of 0.100 M HCl to 80.0 mL of a buffer that is comprised of 0.25 M C2H5NH2 and 0.25 M C2H5NH3Cl. Kb of C2H5NH2 = 9.5 x 10-4.

Answers

Answer: The pH of resulting solution is 10.893

Explanation:

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}* 1000}{\text{Volume of solution (in mL)}}

  • For ethylamine:

Molarity of ethylamine solution = 0.25 M

Volume of solution = 80 mL

Putting values in above equation, we get:

0.25M=\frac{\text{Moles of ethylamine}* 1000}{80mL}\n\n\text{Moles of ethylamine}=(0.25* 80)/(1000)=0.02mol

  • For HCl:

Molarity of HCl = 0.100 M

Volume of solution = 20.0 mL

Putting values in above equation, we get:

0.100M=\frac{\text{Moles of HCl}* 1000}{20.0mL}\n\n\text{Moles of HCl}=(0.100* 20)/(1000)=0.002mol

  • For C_2H_5NH_3Cl:

Molarity of C_2H_5NH_3Cl solution = 0.25 M

Volume of solution = 80 mL

Putting values in above equation, we get:

0.25M=\frac{\text{Moles of }C_2H_5NH_3Cl* 1000}{80mL}\n\n\text{Moles of }=(0.25* 80)/(1000)=0.02mol

The chemical reaction for ethylamine and HCl follows the equation:

                  C_2H_5NH_2+HCl\rightarrow C_2H_5NH_3Cl

Initial:           0.02          0.002         0.02

Final:            0.018          -                0.022

Volume of solution = 20.0 + 80.0 = 100 mL = 0.100 L    (Conversion factor:  1 L = 1000 mL)

To calculate the pOH of basic buffer, we use the equation given by Henderson Hasselbalch:

pOH=pK_b+\log(([salt])/([base]))

pOH=pK_b+\log(([C_2H_5NH_3Cl])/([C_2H_5NH_2]))

We are given:

pK_b = negative logarithm of base dissociation constant of ethylamine = -\log(9.5* 10^(-4))=3.02

[C_2H_5NH_3Cl]=(0.022)/(0.100)

[C_2H_5NH_2]=(0.018)/(0.100)

pOH = ?

Putting values in above equation, we get:

pOH=3.02+\log((0.022/0.100)/(0.018/0.100))\n\npOH=3.107

To calculate pH of the solution, we use the equation:

pH+pOH=14\npH=14-3.107=10.893

Hence, the pH of the solution is 10.893

The pH of the solution is 10.9

Data;

  • Volume of buffer = 80mL
  • Volume of HCL = 20.0mL
  • conc. of C2H5NH2 = 0.25M
  • conc. of C2H5NH3Cl = 0.25
  • Kb of C2H5NH2 = 9.5*10^-4

pH of a Solution

The pH of buffer can be calculated by using Henderson-Hasselbalch's equation

pOH = _pKb+ log ([salt])/([base])

The initial moles of salt present is calculated as

0.25 * 80*10^-^3 = 0.02mmoles

The initial moles of base present is calculated as

0.25*80*10^-^3 = 20mmoles

On adding HCl the following reaction will occurs

C_2H_5NH_2 + HCl \to C_2H_5NH_3Cl

This will lead to formation of extra moles of salt that is  equal to moles of acid added and eventually lead to decrease in number of moles of base by equal measure.

Moles of HCl added is

moles of HCL= 0.1 * 20 * 10^-^3 = 2mmoles

Adding the value

Moles of salt present = 20 + 2 = 22mmoles

Subtracting the value

Moles of base left = 20-2 = 18mmoles

Now using Henderson-Hasselbalch's equation we can calculate the pOH of solution

pKb = -logKb = -log (9.5*10^-^4) = 3.02

The pOH of the base can be calculated as

pOH = 3.02 + log ((22)/(18)) = 3.107

Using the above, we can solve for the pH of the solution.

pH = 14 - pOH = 10.893

The pH of the solution is 10.9

Learn more on pH of a solution using Henderson-Hasselbalch equation here;

brainly.com/question/13557815

The value of delta for the [C_rF_6]^3- complex is 182 kJ/mol. Calculate the expected wavelength of the absorption corresponding to promotion of an electron from the lower-energy to the higher-energy of orbital set in this complex. (Remember to divide by Avogadro's number.) Should the complex absorb in the visible range?

Answers

Answer:  Yes the absorb in the visible range.

Explanation:

The relationship between wavelength and energy of the wave follows the equation:

E=(Nhc)/(\lambda)

where,

E = energy of the wave  = 182 kJ/mol  = 182000 J/mol

N = avogadro's number =  6.023* 10^(23)

h = plank constant = 6.6* 10^(-34)Js^(-1)

c = speed of light = 3* 10^8m/s

\lambda = wavelength of the wave = ?

Putting all the values:

182000=(6.023* 10^(23)* 6.6* 10^(-34)* 3* 10^8m/s)/(\lambda)

\lambda=0.65* 10^(-6)m=650nm    (1nm=10^(-9)m)

The wavelength range for visible rays is 400 nm to 750 nm, thus the complex absorb in the visible range.

Final answer:

The expected wavelength of the absorption in the [CrF6]3- complex cannot be calculated without knowing the exact value of ΔE. Therefore, it is unclear if the complex will absorb in the visible range.

Explanation:

In this complex, the value of delta (Δ) is 182 kJ/mol. To calculate the expected wavelength of the absorption corresponding to the promotion of an electron from the lower-energy orbital to the higher-energy orbital, we can use the equation: Wavelength = (hc)/(ΔE), where h is Planck's constant and c is the speed of light. By substituting the given values and dividing by Avogadro's number, we can calculate the expected wavelength.

However, the information given in the question does not provide the exact value for ΔE. Without this information, it is not possible to calculate the expected wavelength accurately. Therefore, we cannot determine if the complex will absorb in the visible range.

Learn more about Calculating expected wavelength of absorption in a complex here:

brainly.com/question/38332205

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