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2 years ago

The structure of methyl acrylate is shown here. indicate the type of hybridization for each numbered atom.

Chemistry

2 answers:

Vsevolod [243]2 years ago

6 0

To find the hybridization of each atom, we need to count the number of bonds between each atom. 3 bonds is sp2, 4 bonds is sp3. Therefore, atoms 1, 2, 3, and 4 are sp2 and atom 5 is sp3.

12345 [234]2 years ago

3 0

The structure of methyl acrylate is

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For the following pair, indicate which element has the lower first ionization energy: Match the words in the left column to the

emmainna [20.7K]

Answer:

For 1: The correct answer is Ge.

For 2: The correct answer is Te.

For 3: The correct answer is Ba.

For 4: The correct answer is Ag.

Explanation:

Ionization energy is defined as the energy required to remove an electron from the outermost shell of an isolated gaseous atom. It is represented as $E_i$

$X(g)\rightarrow X^+(g)+1e^-;E_i$

Ionization energy increases as we move from left to right in a period. This happens because the atomic radius of an element decreases moving across a period, which increases the effective attraction between the negatively charged electrons and positively-charged nucleus. Hence, the removal of electron from the outermost shell becomes difficult and requires more energy.

Ionization energy decreases on moving from top to bottom in a group. This happens because the number of shells increases as we move down the group. The electrons get added in the new shell. So, the shielding of outermost electrons from the inner ones is more which decreases the attraction between the electrons and the nucleus. Hence, the removal of electron from the outermost shell becomes easy and requires less energy.

For the given options:

Option 1:

Chlorine is the 17th element of the periodic table belonging to Period 3 and Group 17.

Germanium is the 32nd element of the periodic table belonging to Period 4 and Group 14.

Hence, germanium will have smaller first ionization energy.

Option 2:

Tellurium is the 52nd element of the periodic table belonging to Period 5 and Group 16.

Selenium is the 34th element of the periodic table belonging to Period 4 and Group 16.

Hence, tellurium will have smaller first ionization energy.

Option 3:

Barium is the 56th element of the periodic table belonging to Period 6 and Group 2.

Titanium is the 22nd element of the periodic table belonging to Period 4 and Group 4.

Hence, barium will have smaller first ionization energy.

Option 4:

Copper is the 29th element of the periodic table belonging to Period 4 and Group 11.

Silver is the 47th element of the periodic table belonging to Period 5 and Group 11.

Hence, silver will have smaller first ionization energy.

5 0

2 years ago

Compute 4.659×104−2.14×104. Round the answer appropriately.

pentagon [3]

Answer: 25,200.

Explanation:

1) Given: 4.659 × 10⁴ - 2.14 × 10⁴

2) You have to deal with significant figures.

Since, the powers are the same (10⁴) you can directly subtract the decimal numbers. But first analyze the significant figures and the number of decimal digits.

3) The number 4.659 × 10⁴ has four significant figures (4, 6, 5, and 9), while the number 2.14 × 10⁴ has three significan figures (2, 1, and 4).

4) When you add or subtract numbers with diferent amount of decimal digits, the result must show the same number of decimal digits as the term with the least number of decimal digits.

5) Before subtracting, you must round all the terms to the least number of decimal digits. So, since 2.14 has two decimal digits and 4.659 has three decimal digits, you shall round 4.659 to 4.66.

6) Now you subtract 4.66 - 2.14 = 2.52

7) Multiply by the power of 10: 2.52 × 10⁴ = 25,200. And that is the answer.

5 0

2 years ago

Read 2 more answers

Interpret the following Arterial Blood Gases 1. pH 7.33 PaCO2 60 HCO3 34 A. Normal ABG values B. Respiratory acidosis without co

kicyunya [14]

Answer: 1) C; 2)D; 3)B; 4)B; 5) A

Explanation:Interpreting the following Arterial Blood gases, we have

1. pH 7.33 PaCO2 60 HCO3 34----Respiratory acidosis with partial compensation----C

2. pH 7.48 PaCO2 42 HCO3 30------. Metabolic alkalosis without compensation----D

3. pH 7.38 PaCO2 38 HCO3 24 ----- Normal---B

4. pH 7.21 PaCO2 60 HCO3 24------ Respiratory acidosis without compensation-----B

5. pH 7.48 PaCO2 28 HCO3 20 ----Respiratory alkalosis with partial compensation

The Arterial blood gas interpretation from analysis shows the pH and the partial pressures of oxygen and carbon dioxide in the arterial blood of an individual which can detect how well the lungs are functioning thereby making a physician make a diagnosis, estimate the severity of a condition and profer treatment.

4 0

2 years ago

What elements of paint can help forensic scientists compare samples to each other?

saul85 [17]

It would have to be paints consists of pigments,solvents, and binders. Once the [paint has been applied and has dried, the pigments are still able to determine the matched samples.

4 0

2 years ago

A precipitate forms when mixing solutions of sodium fluoride (NaF) and lead II nitrate (Pb(NO3)2). Complete and balance the net

Margarita [4]

Answer:

Pb^2+(aq) + 2F-(aq) → PbF2(s)

Explanation:

Step 1: Data given

sodium fluoride = NaF

lead(II)nitrate Pb(NO3)2

Step 2: The unbalanced equation

NaF(aq) + Pb(NO3)2(aq) → PbF2(s) + NaNO3(aq)

Step 3: Balancing the equation

NaF(aq) + Pb(NO3)2(aq) → PbF2(s) + NaNO3(aq)

On the left side we have 2x NO3 (in Pb(NO3)2), on the right side we have 1x NO3 (in NaNO3). To balance the amount of NO3 we hvae to multiply NaNO3 on the right side by 2.

NaF(aq) + Pb(NO3)2(aq) → PbF2(s) + 2NaNO3(aq)

On the left side we have 1x Na (in NaF), on the right side we have 2x Na (in 2NaNO3). To balance the amount of Na we have to multiply NaF on the left side by 2. Now the equation is balanced.

2NaF(aq) + Pb(NO3)2(aq) → PbF2(s) + 2NaNO3(aq)

Step 4: Calculate net ionic equation

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side, look like this:

2Na+(aq) + 2F-(aq) + Pb^2+(aq) + 2NO3-(aq) → PbF2(s) + 2Na+(aq) + NO3-(aq)

Pb^2+(aq) + 2F-(aq) → PbF2(s)

4 0

2 years ago