A 36.41-g sample of calcium carbonate (CaC O 3 )
contains 14.58 g of calcium and 4.36 g of carbon.

What is the mass of oxygen contained in the sam-
ple? What is the percent by mass of each element in

this compound?

Answers

Answer 1
m(O) = m(CaCO₃) - m(Ca) - m(C).
m(O) = 36.41 g - 14.58 g - 4.36 g.
m(O) = 17.47 g.
ω(Ca) = m(Ca) ÷ m(CaCO₃) · 100%.
ω(Ca) = 14.58 g ÷ 36.41 g · 100%.
ω(Ca) = 40 %; mass percent of calcium.
ω(C) = m(C) ÷ m(CaCO₃) · 100%.
ω(C) = 4.36 g ÷ 36.41 g · 100%
ω(C) = 12%; mass percent of carbon.
ω(O) = 100% - ω(Ca) - ω(C).
ω(O) = 100% - 40% - 12%.
ω(O) = 48%; mass percent of oxygen.

Related Questions

How much energy does an electric hair dryer use if it draws 8.3 amps of current when using 120 volts for 5.0 minutes?

Answers

0.083kwh is the answer your welcome

Answer: The energy drawn by electric hair dryer is 2,98,800 J

Explanation:

To calculate the power of the electric hair dryer, we use the equation:

[tex]P=I\times V[/tex]

where,

P = power of electric hair dryer

I = Current of electric hair dryer = 8.3 A

V = voltage of electric hair dryer = 120 V

Putting values in above equation, we get:

[tex]P=8.3\times 120=996W[/tex]

To calculate the energy of electric hair dryer, we use the equation:

[tex]E=P\times t[/tex]

where,

E = energy of electric hair dryer = ?

P = Power of electric hair dryer = 996 W

t = time taken = 5.0 min = 300 s     (Conversion factor:  1 min = 60 sec)

Putting values in above equation, we get:

[tex]E=996\times 300=2,98,800J[/tex]

Hence, the energy drawn by electric hair dryer is 2,98,800 J

Given that delta hvap is 58.2 kj/mol and the boiling point is 83.4 c 1atm if one mole of this substance is vaporized at 1atm calculate delta ssurr

Answers

Change in Gibb's free energy of system (ΔG) = ΔH - TΔS.........(Eq. 1)
Now, if magnitude of ΔG <0, then reaction is spontaneous.
         if magnitude of ΔG > 0, then reaction is non-spontaneous. 
         At equilibrium, ΔG = 0
When at boiling point, liquid state is in equilibrium with vapour state. Hence, it present case ΔG = 0

∴ Eq 1 becomes, ΔH = TΔS
here, ΔH = 58.2 kj/mol (Given),
∴ At T = 83.4 oC = 356.4 K, ΔS = 0.1633 kj/mol.K

Carl crumbled a piece of paper before throwing it in the trash can. Which is true about the paper after it has been crumbled

Answers

Answer is C: It's appearance changed.

Final answer:

Crumpling a piece of paper doesn't change its mass but alters its shape, leading to reduced air resistance.

Explanation:

When Carl crumbled a piece of paper before throwing it in the trash can, the physical shape of the paper changed, but not its mass. This action relates to a physics concept concerning gravity and air resistance.

An important observation is that when different objects are dropped from the same height, like a crumpled piece of paper and a textbook, they will hit the ground at approximately the same time if we ignore air resistance.

This is because all objects in a vacuum fall at the same rate regardless of mass, due to the uniform acceleration caused by gravity. In real-life conditions, air resistance affects how objects fall.

Crumpling the paper reduces its surface area, thereby reducing air resistance and allowing it to fall faster compared to an uncrumpled sheet of paper.

If you crumple one piece of paper into a small ball and then crumple two pieces of paper, making a bigger ball but with double the mass, they will still hit the ground at the same time if dropped from the same height because the acceleration due to gravity is constant.

A 3.5 gram sample of a radioactive element was formed in a 1960 explosion of an atomic bomb at Johnson Island in the Pacific test site. The half-life of the radioactive element is 28 years. How much of the original sample will remain in the year 2030? Choose the closest.

DO NOT GUESS, ONLY COMMENT IF YOU KNOW

Answers

0.50 g

That should be the correct answer :)

Answer : The correct answer for amount of radioisotope remain in 2030 is 0.619 g .

Radioactive Decay is emission of radiations ( in form of alpha , beta particle etc ) by unstable atom .

Radioactive decay is FIRST ORDER reaction . So , the equation of first order can be used to find decay constant , amount of radioisotopes or half life .

The equation for radioactive decay is given as :

[tex] ln (\frac{N}{N_0}) = -k * t [/tex]

Where : N = amount of radioisotope at time t

N₀ = amount of radioisotope initially present

k = decay constant t = time

Half life :

It is time when amount of radioisotope decrease to 50 % of its original amount . Half life [tex] (T_\frac{1}{2} ) [/tex] and decay constant can be related :

[tex] T_\frac{1}{2} = \frac{ln 2 }{k} = \frac{0.693}{k} [/tex]

Following are the steps can be used to determine amount of radioisotope (N) :

1) To find decay constant :

Given : [tex] (T_\frac{1}{2} ) [/tex] = 28 yrs

Decay constant can be calculated using half life by plugging value in half life formula :

[tex] 28 yrs = \frac{0.693}{k} [/tex]

On multiplying both side by k

[tex] 28 yrs * k= \frac{0.693}{k} *k [/tex]

On dividing both side by 28 yrs

[tex] \frac{28 yrs * k}{28 yrs} = \frac{0.693}{28 yrs} [/tex]

k = 0.02475 yrs⁻¹

2) To find amount of radioisotope (N):

Given : Amount of radioisotope originally present = 3.5 g

Time = 2030 - 1960 = 70 yrs

decay constant = 0.02475 yrs⁻¹

Amount of radioisotope (N) = ?

Plugging these values in the formula as:

[tex] ln (\frac{N}{3.5 } ) = - 0.02475 yrs^-^1 * 70 yrs [/tex]

[tex] ln (\frac{N}{3.5 } ) = - 1.7325 [/tex]

[tex] ln\frac{N}{No} [/tex] can be converted using the formula ( [tex] ln (\frac{a}{b} ) = ln a - ln b [/tex] )

ln N - ln (3.5 ) = - 1.7325

(ln 3.5 = 1.253 )

ln N -1.253 = -1.7325

Adding both side 1.253

ln N -1.253 + 1.253 = -1.7325 + 1.253

ln N = -0.4795

Taking anti ln of -0.4795

N = 0.619 g

Hence amount of radioisotope remained in 2030 is 0.619 g



Calculate the freezing point of a 0.08500 m aqueous solution of nano3. the molal freezing-point-depression constant of water is 1.86°c/m. remember to include the value of i.

Answers

Depression in freezing point (Δ[tex] T_{f} [/tex]) = [tex] K_{f} [/tex]×m×i,
where, [tex] K_{f} [/tex] = cryoscopic constant = [tex] 1.86^{0} C/m[/tex],
m= molality of solution = 0.0085 m
i = van't Hoff factor = 2 (For [tex] NaNO_{3} [/tex])

Thus, (Δ[tex] T_{f} [/tex]) = 1.86 X 0.0085 X 2 = [tex] 0.03162^{0}C [/tex]

Now, (Δ[tex] T_{f} [/tex]) = [tex] T^{0} [/tex] - T
Here, T = freezing point of solution
[tex] T^{0} [/tex] = freezing point of solvent = [tex] 0^{0}C [/tex]
Thus, T = [tex] T^{0} [/tex] - (Δ[tex] T_{f} [/tex]) = -[tex] 0.03162^{0}C [/tex]

Answer : The freezing point of a solution is, [tex]0.32^oC[/tex]

Explanation :

First we have to calculate the Van't Hoff factor (i) for [tex]NaNO_3[/tex].

The dissociation of [tex]NaNO_3[/tex] will be,

[tex]NaNO_3\rightarrow Na^++NO_3^-[/tex]

So, Van't Hoff factor = Number of solute particles = [tex]Na^++NO_3^-[/tex] = 1 + 1 = 2

Now we have to calculate the freezing point of a solution.

Formula used for lowering in freezing point :

[tex]\Delta T_f=i\times k_f\times m[/tex]

or,

[tex]T_f^o-T_f=i\times k_f\times m[/tex]

where,

[tex]\Delta T_f[/tex] = change in freezing point

[tex]T_f[/tex] = temperature of solution = ?

[tex]T^o_f[/tex] = temperature of pure water = [tex]0^oC[/tex]

[tex]k_f[/tex] = freezing point constant  = [tex]1.86^oC/m[/tex]

m = molality  = 0.08500 m

i = Van't Hoff factor = 2

Now put all the given values in this formula, we get the freezing point of a solution.

[tex]0^oC-T_f=2\times (1.86^oC/m)\times 0.08500m[/tex]

[tex]T_f=0.32^oC[/tex]

Therefore, the freezing point of a solution is, [tex]0.32^oC[/tex]

what is the molarity of 10.0 g of kcl in 0.500 l of solution

Answers

Molarity is defined as number of moles of the solute in 1 liter of solution.

Molarity = Moles of solute (mol) / Volume of the solution (L)

moles of KCl = mass (g) / molar mass (g/mol)
                      = 10.0 g / 74.5 g/mol
                      = 0.134 mol

Hence molarity = 0.134 mol / 0.500 L
                         = 0.268 mol/L

In this lab, you will use the flame test to identify the metal ion in two boxes of unidentified fireworks. Write an investigative question that you can answer by doing this experiment.

Answers

The first one is chlorine ion, the second one is metal ion.

A sound is first produced by making something BLANK . The sound then travels through a BLANK to reach the ears, which are the parts of the body that allow for sounds to be heard.

Answers

A sound is first produced by making something VIBRATE. The sound then travels through a MEDIUM to reach the ears, which are the parts of the body that allow for sounds to be heard.

A sound is first produced by making something vibrates and then the sound travels through a medium and reach our ears.

Which are the parts of the body that allow for sounds to be heard?

The ear is responsible for hearing sounds as well as for balance in the human body. The ear has three parts i.e. the outer, middle and inner ears. We hear the sound when the body starts vibration.

So we can conclude that a sound is first produced by making something vibrates and then the sound travels through a medium and reach our ears.

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When 5-hydroxypentanal is treated with methanol in the presence of an acid catalyst, 2-methoxytetrahydropyran is formed. draw the structure of the second intermediate in this reaction?

Answers

The conversion of 5-Hydroxypentanal to 2-methoxytetrahydropyran takes place in two steps. In first step the carbonyl group of 5-hydroxypentanal gets protonated and is being attacked by the hydroxyl group of its own molecule and results in the formation of six membered hetrocyclic ring called tetrahydro-2H-pyran-2-ol
The tetrahydro-2H-pyran-2-ol on protonation looses the water molecule results in the formation of second intermediate called as 2,3,4,5-tetrahydropyrylium (Shown in red). This intermediate when attacked by Methanol molecule results in the formation of 2-methoxytetrahydropyran as shown below.

Among the alkali earth metals, the tendency to react with other substances

A) increases from bottom to top within the group.

B) is shown by the ways they react with water.

C) varies in an unpredictable way within the group.

D) does not vary among the members of the group.

Answers

The answer would be letter A.Among the alkali earth metals, the tendency to react with other substances increases from bottom to top within the group. We know that when there is an increase of metallic property, there will also be an increase of reactivity. 

if you have 100 grams of S8, how many moles of S8 is that

Answers

Given mass= m =100 g
Molar mass= M = (32.06 * 8)= 256.48 g/mol
No. of moles= n=?

n= Given mass/ Molar mass
n= 100/256.48
n=0.38989 mol

What is the ground-state electron configuration for the element cobalt (z = 27)?

Answers

Final answer:

The ground-state electron configuration for cobalt (Z = 27) is [Ar] 3d7 4s2.

Explanation:

The ground-state electron configuration for the element cobalt (Co), which has an atomic number (Z) of 27, is as follows:

Co: [Ar] 3d74s2

This configuration indicates that cobalt has 2 electrons in the outermost 4s subshell and 7 electrons in the 3d subshell after the [Ar] noble gas core.

The ground-state electron configuration for the element cobalt (Z = 27) is 1s²2s²2p⁶3s²3p⁶4s²3d⁷. Co has 27 protons, 27 electrons, and 33 neutrons. The electron configuration represents the distribution of electrons in the energy levels and orbitals of an atom.

what is the volume of 12.0 g of cl2 gas at stp

Answers

Hi I think I might know the answer to your question it is pretty tricky though!
Chlorine Gas = 70.8 grams/moles
Moles of Chlorine Gas = 57.8/70.8 = 0.816 mol
Volume Gas at STP = 22.4 L/mol
Volume of Chlorine Gas = .816*22.4=18.22 Litters
I hoped that helps if not so sorry!

Answer:

[tex]V=3.79L[/tex]

Explanation:

Hello,

In this case, we are talking about an ideal gas problem in which 12.0 g of chlorine gas is at standard both pressure (1atm) and temperature (0°C), therefore, to compute the corresponding volume, one applies the ideal gas law, then converts from grams to moles considering the diatomic chlorine and subsequently solves for volume as shown below:

[tex]PV=nRT\\V=\frac{nRT}{P}=\frac{12.0gCl_2*\frac{1molCl_2}{70.9gCl_2} *0.082 \frac{atm*L}{mol*K} * 273.15K}{1atm}  \\V=3.79L[/tex]

Best regards.

Define atom. list the subatomic particles, describe their relative masses,charges, and positions in the atom.

Answers

atom is a small indivisible particle that can only be seen under a microscope

Question 7 what is the volume of 28.0 g of nitrogen gas at stp? 44.8 l 33.6 l 11.2 l 22.4 l none of the above

Answers

At STP, 1 mol of gas corresponds to 22.4 L. The molar mass of nitrogen is 14.0g so the molar mass of nitrogen gas (N2) is 28.0g. The moles of nitrogen in this sample of nitrogen gas is
                           28.0 g / (1 mol / 28.0 g) = 1.00 mol
If we convert to liters, we get
                           (1.00 mol)(22.4 L / 1mol) = 22.4 L
Therefore, your answer is 22.4 L.

A mixture of 100 g of k2cr207 and 200 g of water is stirred at 60 °c until no more of the salt dissolves. the resulting solution is poured off, leaving the undissolved solid behind. the solution is now cooled to 20°c. what mass of k2cr207 crystallizes from the solution during the cooling?

Answers

One example of a solution is salt water which is a mixture of water and salt. You cannot see the salt and the salt and water will stay a solution if left alone. Parts of aSolution. Solute - The solute is the substance that is being dissolved by another substance.

No K₂Cr₂O₇ will crystallize out of the solution during cooling.

What is crystallization?

Crystallization is a separation technique used to purify a solid substance by selectively dissolving it in a suitable solvent at a high temperature and then cooling the solution to obtain pure crystals of the solute.

Given:

The solubility of K₂Cr₂O₇ in water at 60°C is 127 g/100 mL, and at 20°C is 13.9 g/100 mL.

Dissolved 100 g of  K₂Cr₂O₇ in 200 g of water, which is 200 mL of water.

At 60°C, the solution can dissolve 127 g/100 mL × 2 L

= 254 g of  K₂Cr₂O₇.

100 g of  K₂Cr₂O₇, the solution is saturated and no more of the salt can dissolve.

When the solution is cooled to 20°C, the solubility of K₂Cr₂O₇ is only 13.9 g/100 mL.

The amount of water in the solution at 20°C is 200 mL. The maximum amount of K₂Cr₂O₇ that can remain in solution at this temperature is:

13.9 g/100 mL × 2 L = 278 g

Dissolved 100 g of K₂Cr₂O₇ in the solution, the amount that will crystallize out is:

100 g - 278 g = -178 g

This result is negative, indicating that all the K₂Cr₂O₇ will remain in solution at 20°C.

Therefore, no K₂Cr₂O₇ will crystallize out of the solution during cooling.

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which of the following is a balanced chemical equation for the reaction of magnesium with nitrogen gas to form magnesium nitride

Answers

So the unbalanced equation would be Mg + N^2 --> Mg^3N^2

Which means the balanced equation would be 3Mg + N^2 --> Mg^3N^2

This is balance the equation out since you now has 3 magnesium and 2 nitrogen on the left side, and 3 magnesium on 2 nitrogen on the right. Double check my work though, it's been awhile.

Find the new concentration of a solution if 25.0 mL of water is added to 125.0 mL of 0.150 M NaCl solution. What is the final volume? mL

Answers

Answer:

150.0 is the final volume

next question is a

and the next is 0.125 M

Explanation:

got it right in edge in 2020

Final answer:

To calculate the new concentration of NaCl solution after dilution, the final volume of the solution is determined to be 150.0 mL by adding the initial volume and the volume of water added. Using the molarity equation, the new concentration is found to be 0.125 M.

Explanation:

To find the new concentration of the NaCl solution after dilution, we use the concept of molarity and the principle of conservation of mass. The amount of solute (NaCl in this case) stays the same, but the volume increases when more solvent (water) is added.

The initial molarity (Mi) of the NaCl solution is 0.150 M and the initial volume (Vi) is 125.0 mL. The volume of water added is 25.0 mL.

The final volume (Vf) is the sum of the initial volume and the volume of water added:

Vf = Vi + volume of water added

Vf = 125.0 mL + 25.0 mL

Vf = 150.0 mL

The new concentration Cf can be calculated using the formula:

Ci × Vi = Cf × Vf

(0.150 M) × (125.0 mL) = Cf × (150.0 mL)

Cf = (0.150 M × 125.0 mL) / 150.0 mL

Cf = 0.125 M

The final volume of the solution is 150.0 mL.

What crossed-aldol product results when butanal is heated in the presence of excess benzaldehyde and sodium hydroxide? draw the molecule on the canvas by choosing buttons from the tools (for bonds), atoms, and advanced template toolbars. the single bond is active by defa?

Answers

Final answer:

When butanal is heated with excess benzaldehyde and sodium hydroxide, the crossed-aldol product formed is 4-phenyl-2-butanol, which is obtained through the aldol condensation reaction.

Explanation:

Crossed-aldol product formation occurs when two different carbonyl compounds, such as aldehydes or ketones, react in an aldol condensation. The products are a mixture of different aldol adducts, resulting from the nucleophilic addition of one carbonyl compound to the other. This reaction allows for the creation of complex molecules with varied functionalities.

When butanal is heated in the presence of excess benzaldehyde and sodium hydroxide, the crossed-aldol product formed is 4-phenyl-2-butanol. The product is obtained by the aldol condensation reaction between butanal and benzaldehyde. In this reaction, a molecule of water is eliminated, and the resulting product contains both the aldehyde and the alcohol functional groups.

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Why are Noble gases inert?
A. They cannot form bonds because they have 0 electrons in their valence shell.
B. They are actually so reactive that they bond instantly with any substance they touch. This means they cannot be found in their pure form in nature.
C. They have a full shell of valence electrons.
D. They bond only with atoms of their own element.

Answers

Correct answer is C. They have a full shell of valence electrons. 

Sucrose is very soluble in water. at 25◦c, 211.4 grams of sucrose will dissolve in 100 g of water. given that the density of the saturated sucrose solution is 1.34 g/ml, what is the molarity of the solution

Answers

Final answer:

The molarity of the sucrose solution is calculated by finding the number of moles of sucrose and the volume of the solution in liters. By using the provided data, we came up with a molarity of 2.67 M.

Explanation:

To calculate the molarity of the solution, we need to know the number of moles of sucrose and the volume of the solution in liters. Given that the molecular weight of sucrose (C12H22O11) is approximately 342 g/mol, the number of moles of sucrose in 211.4 g can be calculated as (211.4 g) / (342 g/mol) = 0.618 moles.

To get the volume of the solution, first calculate the total mass of the solution which is mass of water + mass of sucrose = 100 g + 211.4 g = 311.4 g. Since the density of the solution is provided as 1.34 g/ml, convert this to kg/L to get the volume. So, 311.4 g / 1.34 g/ml = 232.39 ml or 0.232 L.

Finally, we calculate the molarity which is moles/volume in liters. So, molarity = 0.618 moles / 0.232 L = 2.67 M.

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Answer:

2.66 M

from ut quest

n comparison to molecules that interact by London dispersion forces only, the melting point of similar-sized molecules forming hydrogen bonds would most likely be about the same. unpredictable. lower. higher.

Answers

Hydrogen bonds are the strongest intermolecular bonds, so the melting points would be much higher. 

Answer:

d

Explanation:

Consider 2NH3(g)+3CuO(s)→N2(g)+3Cu(s)+3H2O(g) 2NH 3 ​ (g)+3CuO(s)→N 2 ​ (g)+3Cu(s)+3H 2 ​ O(g). What volume (in mL) of NH3 NH 3 ​ is required to completely react 45.2 g of CuO at STP? Enter your solution as a numerical value with no units.

Answers

the volume of  in Ml  of NH3  required  to completely  react  with  45.2 g  of CUO   at  STP  is 8489.6 Ml

 calculation
calculate the moles  of CUO  used
moles= mass/molar mass
45.2 g/79.5 g/mol =0.569 moles

Reacting  equation 
2NH3  + 3CUO = N2  +3CU  +3H2O
by  use of reacting  ratio  between  NH3 to CUO  which jis  2:3 the moles of  NH3  = 0.569 x2/3 = 0.379 moles

At STP  1  mole  = 22.4 L
what about  0.379 moles

=0.379  x22.4  = 8.4896  L

in  Ml  = 8.4896  x1000 =8489.6 ml

help!!
The electron sea is responsible for which of the following properties?
A. All of these
B. Malleable
C. Ductile
D. Conducting electricity and heat

Answers

A. All of these, because these are properties of metals, and the electron sea presets in metals.

Answer:

D.Conducting electricity and heat

Explanation:

Metal : It is defined as that substance which is good conductor of electricity and heat. Metal have  ductile and malleable property.In metal, atoms are held together by metallic bonds .The valence electrons from s and p orbital are delocalised and they form sea of electrons that surround the positively charged nuclei of the interacting metal ion.Then , the electrons are freely move throughout the space between atomic nuclei.

Due to availability of free  electrons, metal conduct electricity and heat.

Therefore, the electron sea is responsible for conducting electricity and heat .

Select all that apply.

Which statements concerning this diagram are correct?



X = -ΔH
X = ΔH
Z = A.E.

Answers

Correct answer is X = ΔH

Reason:
1) The graph of enthalpy Vs reaction coordinate  suggest the reaction is endothermic in nature. For endothermic reaction, energy if product is more than that of reactant. Hence,  option 1 i.e. X = -ΔH cannot be correct.
2) Since the reaction is endothermic in nature, energy if product is more than that of reactant. Hence,  option 2 i.e. X = ΔH is correct.
3) Activation energy is energy difference between Reactant (A) and transition state (B). However, as per option C, activation energy (A.E.) is energy difference between product (C) and transition state (B), which is incorrect. 

Answer:

Correct answer is X = ΔH

Explanation:

If 0.158 g of a white, unknown solid carbonate of a group 2A metal (M) is heated and the resulting CO2 is transferred to a 285 ml sealed flask and allowed to cool to 25 degrees Celsius, the pressure in the flask is 69.8 mmHg. What is the identity of the carbonate?

Answers

using the ideal gas law equation we can find the number of moles of CO₂ formed 
PV = nRT
where 
P - pressure - 69.8 mmHg x 133 Pa/mmHg = 9 283 Pa
V - volume - 285 x 10⁻⁶ m³
n - number of moles 
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in Kelvin - 25 °C + 273 = 298 K
substituting these values in the equation 
9283 Pa x 285 x 10⁻⁶ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 298 K
n = 1.067 x 10⁻³ mol

decomposition of metal carbonate is as follows 
MCO₃ ---> MO + CO₂
stoichiometry of MCO₃ to CO₂ is 1:1
therefore number of moles of MCO₃ heated = number of CO₂ moles formed 
number of MCO₃ moles = 1.067 x 10⁻³ mol
molar mass = mass / number of moles 
molar mass = 0.158 g / 1.067 x 10⁻³ mol = 148 g/mol 
since carbonate molar mass is known -
 (molar mass of C x 1 C atom) + (molar mass of O x 3  O atoms)  
12 + 16x 3 = 12 + 48 = 60 
then mass of metal M - 148 - 60 = 88
group II metal with molar mass of 88 is Ra - Radium 

How many carbon-carbon sigma bonds are present in each of the following molecules: benzene, cyclobutane?

Answers

The first bond which forms between two atoms is known as Sigma Bond. It is considered as the strongest covalent bond which forms by the head-on overlapping between two orbitals.

In Benzene there are six sp² hybridized carbon atoms. Each carbon atom is bonded to two carbons atoms and one hydrogen atom. Each carbon atom is bonded to one carbon atom via a double bond and through single bond to another carbon atom. Hence, there are Six sigma bonds between carbon atoms in Benzene.

In Cyclobutane there are four sp³ hybridized carbon atoms. Each carbon atom is singly bonded to two other carbon atoms and two hydrogen atoms. Therefore, there are FOUR sigma bonds between the carbon atoms of cyclobutane.

The sigma bonds are highlighted BLUE in given structures,

Final answer:

Benzene has six carbon-carbon sigma bonds because each carbon atom forms two sigma bonds with its adjacent carbon atoms. Cyclobutane has four carbon-carbon sigma bonds since it is a four-membered ring with each carbon bonded to two other carbons.

Explanation:

The question asks about the number of carbon-carbon sigma bonds in benzene and cyclobutane. Let's examine each molecule separately.

Benzene (C₆H₆)

In benzene, the carbons form a hexagonal ring and each carbon atom is bonded to two other carbon atoms. The structure of benzene is often depicted with alternating single and double bonds. However, due to resonance, the electrons are delocalized and each carbon-carbon bond is equivalent, resembling a structure with one-and-a-half bonds. Despite the presence of alternating double bonds, each carbon is involved in two sigma bonds with adjacent carbons and one sigma bond with a hydrogen atom, making a total of six carbon-carbon sigma bonds in benzene.

Cyclobutane (C₄H₈)

Cyclobutane is a four-membered ring where each carbon atom is bonded to two other carbons forming single bonds. Thus, for cyclobutane, there are four carbon-carbon sigma bonds, one between each pair of adjacent carbon atoms.

The equilibrium constant for the reaction sr(s) + mg2+(aq) ⇌ sr2+(aq) + mg(s) is 3.69 × 1017 at 25°c. calculate e o for a cell made up of sr / sr2+ and mg / mg2+ half-cells.

Answers

we are going to use this formula:

E° = 0.0592/ n * ㏒K

when E° is the standard state cell potential

and n is number of moles of electrons transferred in the balanced equation

for the reaction of the cell.

Sr(s) + Mg2+ (aq) ↔ Sr2+(aq) + Mg(s)

we can get it by using the given balanced equation, we here have 2 electrons

transferred so, n = 2 

and K the equilibrium constant = 3.69 x 10^17

so, by substitution:

∴ E° = 0.0592 / 2 * ㏒ (3.69 x 10^17)

        = 0.52 V

To calculate the standard electrode potential (E°) for a cell composed of Sr/Sr²⁺ and Mg/Mg²⁺ half-cells, use the Nernst equation with the given equilibrium constant value. The resulting E° is approximately 1.038 V.

To find the standard electrode potential (E°) for the cell, we can use the Nernst equation and the relationship between the equilibrium constant (Keq) and the standard electrode potential:

Nernst equation:

E° = (RT / nF) * ln(Keq)

Where:

R = 8.314 J/(mol·K)T = 298 K (since the temperature is 25°C)n = 2 (number of electrons transferred, as each Sr replaces an Mg)F = 96485 C/mol

Plugging in the values, we get:

E° = (8.314 * 298 / (2 * 96485)) * ln(3.69 × 1017)

Simplify the constants:

E° = (0.0257 V) * ln(3.69 × 1017)

Evaluating the natural logarithm:

E° = 0.0257 * 40.38

E° ≈ 1.038 V

Therefore, the standard electrode potential (E°) for the cell is approximately 1.038 V.

In the lewis structure of ch3oh, how many bonding pairs of electrons are there?

Answers

Answer:
            There are five bonding pairs of electrons in Methanol.

Explanation:
                   Those electron pairs which are being shared between two atoms in molecule are called as bonding pair electrons. While, those electron pairs which are not involved in bond formation and are not shared between two atoms are called as Non-Bonding electron pairs.
                   In molecule of Methanol as shown below, it can be seen that carbon atom is forming four bonds with three hydrogen atoms and one oxygen atom by sharing four electron pairs and oxygen is forming two bonds, one with carbon atom and one with hydrogen atom. There are two lone pair of electrons present on oxygen atom which are not taking part in and formation.

the number of atoms bonded to the central atom is 4

Explanation:

The solubility product for chromium(iii) fluoride is ksp = 6.6 × 10–11. what is the molar solubility of chromium(iii) fluoride?

Answers

Final answer:

To determine the molar solubility of chromium(III) fluoride, we need to use the solubility product constant (Ksp) of 6.6 × 10⁻¹¹. The molar solubility of chromium(III) fluoride is 8.184 × 10⁻⁴M.

Explanation:

To determine the molar solubility of chromium(III) fluoride, we need to use the solubility product constant (Ksp) of 6.6 × 10⁻¹¹ for chromium(III) fluoride. The Ksp expression is Ksp = [Cr³⁺][F⁻]³.

Let's assume the molar solubility of chromium(III) fluoride is x M. Since the stoichiometry of the balanced equation for its dissolution is 1:3 (one chromium ion per three fluoride ions), the equilibrium expression becomes Ksp = x(3x)³.

Now, we can set up the expression and calculate the molar solubility by solving the equation: Ksp = 108x⁴ = 6.6 × 10⁻¹¹. Solving for x gives x = 8.184 × 10⁻⁴ M. Therefore, the molar solubility of chromium(III) fluoride is 8.184 × 10⁻⁴ M.

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