Consider a reaction that connects the ends of a chain of carbons to form a ring. The closure of the ring results in the formation of a new chemical bond, with the bonding electrons going into a bonding sigma molecular orbital at a lower energy. Which of the following statements correctly describes the resulting enthalpy, entropy, and heat changes? Mark all that are correct.
A. The entropy of the surroundings decreases (AS surr < 0) because the closure of the ring results in fewer conformations for the system.
B. The system releases energy to the environment, because its internal energy decreases.
C. The entropy of the surroundings increases (AS surr > 0) because the surroundings absorb heat released by the reaction, and become more disordered.
D. The entropy of the system decreases (AS sys < 0) because the closure of the ring reduces the number of conformations the ring can adopt.
E. The enthalpy or internal energy of the system decreases (AH sys < 0) because the electrons that form the bond enter an orbital that is at a lower energy.

Answer:

26.982 g

i think so..........

Answer:

the answer is D . our galaxy

Answer:

our galaxy of course

Divide the number of molecules N by avogadro's constant.

Let n be the number of moles then,

[tex]n=\frac{N}{N_A}=\frac{3.3\cdot10^{24}}{6.022\cdot10^{23}}=5.48\mathrm{mol}[/tex]

Hope this helps.

Answer:

Follows are the solution to the given question:

Explanation:

Dry Soil weight = solid soil weight = [tex]284 \ grams[/tex]

solid soil volume =[tex]205 \ cc[/tex]

saturated mass soil = [tex]361 \ g[/tex]

The weight of the soil after drainage is =[tex]295 \ g[/tex]

Water weight for soil saturation = [tex](361-284) = 77 \ g[/tex]

Water volume required for soil saturation =[tex]\frac{77}{1} = 77 \ cc[/tex]

Sample volume of water: [tex]= \frac{\text{water density}}{\text{water density input}}[/tex]

[tex]= 361- 295 \\\\ = 66 \ cc[/tex]

Soil water retained volume = (draining field weight - dry soil weight)

                                             [tex]= 295 - 284 \\\\ = 11 \ cc.[/tex]

[tex]\text{POROSITY}= \frac{\text{Vehicle volume}}{\text{total volume Soil}}[/tex]

                    [tex]= \frac{77}{(205 + 77)} \\\\= \frac{77}{(282)} \\\\ = 27.30 \%[/tex]

(Its saturated water volume is equal to the volume of voids)

[tex]\text{YIELD SPECIFIC} = \frac{\text{Soil water volume}}{\text{Soil volume total}}[/tex]

                              [tex]= \frac{66}{(205+77)}\\\\= \frac{66}{(282)}\\\\=0.2340\\\\ = 0.23[/tex]

[tex]\text{Specific Retention}= \frac{\text{Volume of soil water}}{\text{Total soil volume}}[/tex]

                            [tex]= \frac{11}{282} \\\\= 0.0390 \\\\ = 0.04[/tex]

Answer:

Explanation:

From the information above:

The equation for the reactor volume of CSTR is given by the formula:

[tex]V = \dfrac{C_{Ao}V_o - C_A V_o }{-rA}[/tex]

[tex]V = \dfrac{C_{Ao}V_o (1 - 0.01)}{KC_A^2}[/tex]

The reaction rate of A is:

[tex]-r_A = kC_A^2[/tex] and k = 3 dm³/mol.h

replacing the values in the above equation to solve for v;

∴

[tex]V = \dfrac{(0.5 \ mol/dm^3) (10 \ dm^3/h) (1 - 0.01)}{ ( 3 \ dm^3/mol.h) (0.01 \times 0.5 \ mol/dm^3)^2}[/tex]

V = 66,000 dm³

Thus, the reactor volume of CSTR is 66000 dm³

Taking the differential mole balance equation for PFR is as follows:

[tex]\dfrac{d(C_AV_o)}{dV} = -r_A[/tex]

[tex]\dfrac{d(C_AV_o)}{dV} = Kc^2A[/tex]

Taking the integral between initial and final concentrations;

[tex]\dfrac{v_o}{k} \int ^{C_A}_{C_{Ao}} \dfrac{dC_A}{C^2A}= - \int ^V_o \ dV[/tex]

[tex]V = \dfrac{V_o}{K} \Big [\dfrac{1}{C_A}- \dfrac{1}{C_{Ao}} \Big ][/tex]

replace the values in the above equation;

[tex]V= \dfrac{(10 \ dm^3/h) }{(3 \ dm^3/mol.h)}\Big [ \dfrac{1}{0.01(0.5 \ mol/dm^3)} - \dfrac{1}{0.5 \ mol/dm^3} \Big ][/tex]

[tex]\mathbf{V = 66 0 \ dm^3}[/tex]

Thus, the reactor volume of PFR is 660 dm³

Answer:

150.17 g/mol

Explanation:

Answer:

The number of particles in the 41.8g sample of Bismuth is 12.044 × 10²³

Explanation:

In a 15.5 g sample of bismuth there are 4.47 x 10²² atoms are present.

The term molar mass is defined as the ratio between the mass and the amount of substance of any sample of said compound. The molar mass is a property of a substance.

The molar mass of a compound can be calculated by adding the standard atomic masses in g/mol of the constituent atoms.

Given:

Mass of Bismuth  = 15.5g

Number of atoms  = ?

To find the number atoms, find the number of moles in this element first.

Number of moles  = mass/ molar mass

Number of moles  =  15.5g / 209    

= 0.074mole

1 mole of a substance contains 6.023 x 10²³ atoms

0.074 mole of Bismuth will contain 0.074 x 6.023 x 10²³ atoms

= 4.47 x 10²² atoms

Thus, 4.47 x 10²² atoms are found in a 15.5 g sample of bismuth.

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Given

Atomic symbol

Required

Atomic composition

Solution

Atomic number = number of protons = number of electrons

Mass Number (A) is the sum of protons and neutrons

Number of protons and Number of Neutrons  in nucleus

Number of electrons in the shell

From the picture:

protons = 3

neutrons = 4

electrons = 3

atomic number = protons = electrons = 3

mass number = protons + neutrons = 3+4 = 7

Answer:

C. It may act as an insulator or a conductor.

Explanation:

Group B are the metalloids, so they have properties of both metals and nonmetals.

The statement which is most likely true for an element present in section B is: C. It may act as an insulator or a conductor.

A periodic table can be defined as an organized tabular array of all the chemical elements, which is typically arranged in order of increasing atomic number or number of protons (in rows).

In this diagram, the chemical element that is present in section B of the periodic table is a metalloid.

A metalloid can be defined as a chemical element that are typically intermediate between those of semiconductors, non-metals and metals.

In this context, the statement which is most likely true for an element present in section B is that it may act as an insulator or a conductor because it is a metalloid.

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Answer:

Continuing on across the periodic table we see that fluorine is the next element after oxygen. ... Rather than forming seven bonds fluorine only forms a single bond for basically the same reasons that oxygen only forms two bonds. Hydrogen fluoride, HF, has one bond, but four centers of electron density around the fluorine.

Answer:

combustion reaction. hope this helps!

Answer:

water/d

Explanation:

Answer:

jayfeather friend me

Explanation:

Hence it is quite important to understand and learn the processes of the water cycle.

Step 1: Evaporation. The water cycle starts with evaporation. ...

Step 2: Condensation. ...

Step 3: Sublimation. ...

Step 4: Precipitation. ...

Step 5: Transpiration. ...

Step 6: Runoff. ...

Step 7: Infiltration.

Answer:

atoms

Explanation:

i hope that help you

Answer:

atoms

Explanation:

Answer:

Explanation:

A peptide is any class of organic compounds composed of different numbers of amino acids in which the amine of one is reacted with the carboxylic acid of the next to form an amine bond. The peptide of DTLH is composed of the following amino acids:

Asp-Thr-Leu-His

Their structures are first drawn out in the image attached below. This is followed by the isoelectric structure of the peptide DTLH

Answer:

ccc

Explanation:

Answer:

A

Explanation:

trust

Answer:

If coal and petroleum will get exhausted it will be very difficult for us to transport because most vehicles depends on petroleum, Transport on Earth will became complicated, and if coal will get exhausted we will lose an unique fossil fuel. Coal is used in various domestic and industrial purposes.

Answer:

The smaller numbers that result from measuring with larger units are easier to use .

That’s the one, good luck!

The smaller numbers that result from measuring with larger units are easier to use.

Light year is a unit of astronomical distance that measures the distance traveled by light in one year. The distance between stars can also be measured using parsecs.

The equivalent of this distance in various units is written as follows;

Using light year to represent the unit of distance is more concise because big measurement can be represented easily with small numbers using this large unit.

Thus, we can conclude that the smaller numbers that result from measuring with larger units are easier to use.

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When hydrogen is passed over hot ferric oxide (FeO) hydrogen reacts with oxygen present in the compound and forms water (H2O) and pure Iron

Answer:

m = 79.1 grams

Explanation:

Given that,

Density, d = 2.26 g/mL

Volume occupies, V = 35 mL

We need to find the mass of the substance. We know that the density of an object is equal to the mass per unit volume. So,

d = m/V

[tex]m=d\times V\\\\m=2.26\ g/mL\times 35\ mL\\\\m=79.1\ g[/tex]

So, the mass of the substance is 79.1 grams.

Answer:bromine

Explanation:

The question is incomplete, the complete question is;

Aluminum metal and bromine liquid (red) react violently to make aluminum bromide (white powder). One way to represent this equilibrium is:

Al(s) + 3/2 Br2(l)AlBr3(s)

We could also write this reaction three other ways, listed below. The equilibrium constants for all of the reactions are related. Write the equilibrium constant for each new reaction in terms of K, the equilibrium constant for the reaction above.

1) 2 AlBr3(s) 2 Al(s) + 3 Br2(l)

2) 2 Al(s) + 3 Br2(l) 2 AlBr3(s)

3) AlBr3(s) Al(s) + 3/2 Br2(l)

Answer:

See explanation

Explanation:  

We have that; Al(s) + 3/2 Br2(l)AlBr3(s)

So;

Al(s) + 3/2 Br₂(l) = AlBr₃(s)

K = [  AlBr₃] / [ Al] [  Br₂]³/²

K² =  [  AlBr₃]² / [  Al ] ² [ Br₂]³

Now;

1) 2 AlBr₃ = 2 Al(s) + 3 Br₂(l) =

K₁ =  [  Al ] ² [ Br₂]³ /  [  AlBr₃]²

K₁ =  ( 1 / K² ) = K⁻²

For the second reaction;

2 ) 2 Al(s) + 3 Br₂(l) = 2 AlBr₃(s)

K₂ = [ AlBr₃ ]² / [  Al ]² [  Br₂ ]³

K₂ = K²

For the third reaction;

3 )

AlBr₃(s) =   Al(s) + 3/2 Br₂(l)

K₃  = [ Al ] [ Br₂ ] ³/² / [ AlBr₃ ]

=  ( 1 / K ) = K⁻¹

Answer:

option C : cellular respiration

Explanation:

cellular respiration is taking place in this diagram .

as in respiration food get oxidised to release energy and it releases CO2 and water

Answer:

6.576 g NaCl

Explanation:

2Na + Cl2 --> 2 NaCl

Step 1: Find out moles of Na and Cl to determine the limiting reactant:

moles Na = 12/23 = 0.52 moles Na

moles Cl2 = 16/71=0.225 moles Cl2

ratio Na:Cl is 2.3:1 so Cl is the limiting reactant.

Step 2: How many moles of NaCl are formed:

moles NaCl = moles Cl2 x (1 mol NaCl/2 moles Cl2) = 0.225/2 = 0.1125 moles NaCl

Step 3:

mass NaCl = moles NaCl x MM NaCl = 0.1125 x 58.45 = 6.576 g NaCl

Answer:

3

Explanation:

nitrogen atom in NH3 has 3 shared electrons.

Answer: We start by doing a 1/10 serial dilution, using 100 µL of 1M solution into 900 µL of water, until we get a 1.0x10^-5M (0.00001M) solution. Then use 1 mL of this 1.0x10^-5M solution into 9 mL of water.

Explanation:

To answer this question, we must use the law of conservation of mass, which states: In every chemical reaction mass is conserved, this means the total mass of the reactants is equal to the total mass of the products. The law implies that mass can neither be created nor destroyed, but it can be transformed. For example, in chemical reactions, the mass of the chemical components before the reaction is equal to the mass of the components after the reaction. Therefore, during any chemical reaction and low-energy thermodynamic processes in an isolated system, the total mass of the reactants or starting materials must be equal to the mass of the products. This law is quite accurate for low-energy processes, such as chemical reactions.

So, if the solution to be used is 1M , and a 10 mL of a 1.0x10^-6M solution is needed, we use the following equation:

Initial concentration x initial volume = final concentration x final volume.

The initial concentration is 1M, the final concentration is 1.0x10^-6M and the final volume required is 10 mL.

1M x initial volume = 1.0x10^-6M x 10 mL

initial volume=  1.0x10^-5 mL= 0.01 µL of 1M solution.

Since the final volume is 10 mL, we have to add the difference in volume with water, which is 10 mL - 1.0x10^-5 mL= 9.99999 mL.

However, since 0.01 mL is a very small volume that is difficult to take, the best option in this case is to make serial dilutions.

Usually, we start from a concentrated solution and prepare a series of dilutions to the tenth (1:10) or half (1:2). In this way a series of solutions is obtained, related for example by a dilution factor of 10, i.e. 1/10; 1/100; 1/1000 and so on.

Here we can prepare a series of dilutions to the tenth, from 1M to 1.0x10^-6.

We start by doing a 1/10 dilution, using 100 µL of 1M solution into 900 µL of water. This is a 0.1M solution. Then we take 100 µL of it into 900 µL of water to get a 0.01M solution. We continue doing that until we get a 1.0x10^-6M (0.000001M) solution. This final solution is the desired concentration, however we need 10 mL of it, and actually we have 1 mL. So we can just take 1 mL of the 1.0x10^-5M solution into 9 mL of water:

1.0x10^-5M x 1 mL = 1.0x10^-6M x 10 mL.