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8_murik_8 [283]
1 year ago
8

How many bonding electrons are in the Lewis structure of PCl₃?

Chemistry
1 answer:
Blababa [14]1 year ago
3 0

Answer:

B.) 6

Explanation:

Attached below is the Lewis structure of PCl₃ . Since phosphorus (P) has 5 valence electrons and chlorine (Cl) has 7 valence electrons, there should be 26 valence electrons (5 + 7(3) = 26) in the Lewis structure.

Bonding electrons are the electrons present in the chemical bonds between two atoms.

There are 2 electrons shared in every single bond. Within PCl₃, there are 3 single bonds. As such, there are 6 bonding electrons in the Lewis structure of PCl₃.

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How many carbon dioxide molecules must be added to rubp to make a single molecule of glucose?
Pani-rosa [81]

Answer:

6 carbon dioxide molecules

Explanation:

The Calvin cycle generates the necessary reactions for the fixation of carbon in a solid structure for the formation of glucose and, in turn, regenerates the molecules for the continuation of the cycle.

The Calvin cycle is also known as the dark phase of photosynthesis or also called the carbon fixation phase. It is known as the dark phase because it is not light dependent as is the first phase or light phase .

This second stage of photosynthesis fixes the carbon of the absorbed carbon dioxide and generates the precise number of biochemical elements and processes necessary to produce sugar and recycle the remaining material for continuous production.

The Calvin cycle uses the energy produced in the light phase of photosynthesis to fix the carbon dioxide (CO2) carbon in a solid structure such as glucose, in order to generate energy.

The glucose molecule composed of a six-carbon main structure will be further processed in glycolysis for the preparatory phase of the Krebs cycle, both part of the cellular respiration.

The Calvin cycle produces in six turns a six-carbon glucose molecule and regenerates three RuBP that will be catalyzed again by the RuBisCo enzyme with CO2 molecules for the restart of the Calvin cycle.

The Calvin cycle requires six molecules of CO2, 18 ATP and 12 NADPH produced in the light phase of photosynthesis to produce a glucose molecule and regenerate three RuBP molecules.

7 0
2 years ago
A mutation can be defined as a change in an organism's
larisa86 [58]

Answer:

Genotype

Explanation:

A mutation is the change is DNA, aka the genotype

4 0
2 years ago
Molecules in the gas phase move faster than the same molecules love in the liquid true or false
vredina [299]

true gas molicules bounce off one another much faster liquid just slides

5 0
2 years ago
Read 2 more answers
An herbicide contains only C, H, Cl, and N. The complete combustion of a 200.0 mg sample of the herbicide in excess oxygen produ
MAVERICK [17]

Answer:

%C = 56,1%

%H = 5,5%

%Cl = 27,6%

%N = 10,8%

Explanation:

The moles of CO₂ are the same than moles of C in the herbicide.

Moles of H₂O are ¹/₂ of moles of H in the herbicide.

Moles of CO₂ are obtained using:

n = PV/RT

Where, in STP: P is 1 atm; V is 0,2092L; R is 0,082atmL/molK; T is 273 K

moles of CO₂ are: 9,345x10⁻³ mol≡ mol of C×12,01g/mol = <em>0,1122 g of C ≡ 112,2mg of C</em>

In the same way, moles of H₂O are 5,450x10⁻³mol×2 =0,1090 mol of H×1,01g/mol = <em>0,0110 g of H ≡ 11,0mg of H</em>

As you have 55,14 mg of Cl, the mg of N are:

200,0mg - 112,2 mg of C - 11,0 mg of H - 55,14 mg of Cl = 21,66 mg of N

Thus, precent composition of the herbicide is:

%C = \frac{112,2 mgC}{200,0mg}×100 = 56,1%C

%H = \frac{11,0 mgH}{200,0mg}×100 = 5,5%H

%Cl = \frac{55,14 mgCl}{200,0mg}×100 = 27,6%Cl

%N = \frac{21,66 mgN}{200,0mg}×100 = 10,8%N

I hope it helps!

3 0
3 years ago
When a hydrogen atom makes the transition from the second excited state to the ground state (at -13.6 eV) the energy of the phot
viktelen [127]

Answer : The energy of the photon emitted is, -12.1 eV

Explanation :

First we have to calculate the 'n^{th}' orbit of hydrogen atom.

Formula used :

E_n=-13.6\times \frac{Z^2}{n^2}ev

where,

E_n = energy of n^{th} orbit

n = number of orbit

Z = atomic number  of hydrogen atom = 1

Energy of n = 1 in an hydrogen atom:

E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6eV

Energy of n = 2 in an hydrogen atom:

E_3=-13.6\times \frac{1^2}{3^2}eV=-1.51eV

Energy change transition from n = 1 to n = 3 occurs.

Let energy change be E.

E=E_-E_3=(-13.6eV)-(-1.51eV)=-12.1eV

The negative sign indicates that energy of the photon emitted.

Thus, the energy of the photon emitted is, -12.1 eV

3 0
2 years ago
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