Which process is endothermic?a)b)c)d)The evaporation of water from the skin.The burning of candle wax.The oxidation of iron in a chemical hand warmer.The combustion of natural gas in a stove

Answer:

Explanation: M(PCL5)= 31 + 5(35.5)

=208.5g/mol

M(H20)= 18g/mol

n(PCL5) = 75.5÷208.5

= 0.362mol

n(HCl)/n(PCL5)= 5/1

n(HCl)= 5×0.362

=1.81mol of HCl

Answer:

i. n = 5

ii. ΔE = 7.61 × [tex]10^{-46}[/tex] KJ/mole

Explanation:

1. ΔE = (1/λ) = -2.178 × [tex]10^{-18}[/tex]([tex]\frac{1}{n^{2}_{final} }[/tex] - [tex]\frac{1}{n^{2}_{initial} }[/tex])

    (1/434 × [tex]10^{-9}[/tex]) = -2.178 × [tex]10^{-18}[/tex] ([tex]\frac{n^{2}_{initial} - n^{2}_{final} }{n^{2}_{final} n^{2}_{initial} }[/tex])

⇒ 434 × [tex]10^{-9}[/tex] = (1/-2.178 × [tex]10^{-18}[/tex])[tex]\frac{n^{2}_{final} *n^{2}_{initial} }{n^{2}_{initial} - n^{2}_{final} }[/tex]

But, [tex]n_{final}[/tex] = 2

434 × [tex]10^{-9}[/tex] = (1/2.178 × [tex]10^{-18}[/tex])[tex]\frac{2^{2} n^{2}_{initial} }{n^{2}_{initial} - 2^{2} }[/tex]

434 × [tex]10^{-9}[/tex]  × 2.178 × [tex]10^{-18}[/tex] = [tex](\frac{4n^{2}_{initial} }{n^{2}_{initial} - 4 })[/tex]

⇒ [tex]n_{initial}[/tex] = 5

Therefore, the initial energy level where transition occurred is from 5.

2. ΔE = hf

     = (hc) ÷ λ

    = (6.626 × 10−34 × 3.0 × [tex]10^{8}[/tex] ) ÷ (434 × [tex]10^{-9}[/tex])

    = (1.9878 × [tex]10^{-25}[/tex]) ÷ (434 × [tex]10^{-9}[/tex])

    = 4.58 × [tex]10^{-19}[/tex] J

    = 4.58 × [tex]10^{-22}[/tex] KJ

But 1 mole = 6.02×[tex]10^{23}[/tex], then;

energy in KJ/mole = (4.58 × [tex]10^{-22}[/tex] KJ) ÷ (6.02×[tex]10^{23}[/tex])

         = 7.61 × [tex]10^{-46}[/tex] KJ/mole

The initial energy level is 5  and the energy of this transition in units of kJ/mole is 7.57 * 10^-43 kJ/mole

We must first calculate ΔE as follows;

ΔE = hc/λ

h = Plank's constant = 6.6 * 10^-34 Js

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

λ = wavelength = 434 * 10^-9

ΔE =  6.6 * 10^-34 * 3 * 10^8/434 * 10^-9

ΔE = 0.0456 * 10^-17 J

ΔE = [tex]ΔE = -2.178 x10^-18 (\frac{1}{n^2final} - \frac{1}{n^2initial}) \\ΔE = -2.178 x10^-18 (\frac{1}{2^2} - \frac{1}{n^2initial} )\\\\4.56 * 10^-19/2.178 x10^-18 = (\frac{1}{2^2} - \frac{1}{n^2initial})\\0.210 = (\frac{1}{2^2} - \frac{1}{n^2initial})\\\frac{1}{n^2initial} = 0.25 - 0.210\\\frac{1}{n^2final} = 0.04\\n = (\sqrt{(0.04)^-1} \\n = 5[/tex]

Energy of this transition in units of kJ/mole = 4.56 * 10^-19/ 6.02 * 10^23

= 7.57 * 10^-43 kJ/mole

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Answer:

0.025 M

Explanation:

The following data were obtained from the question:

Initial volume (V1) = 5 mL

Initial concentration (C1) = 0.5 M

Final volume (V2) = 100 mL

Final concentration (C2) =..?

Using the dilution formula, we can obtain the final concentration of the diluted solution as follow:

C1V1 = C2V2

0.5 x 5 = C2 x 100

Divide both side by 100

C2 = (0.5 x 5)/100

C2 = 0.025 M

Therefore, the final concentration of the diluted solution is 0.025 M

The concentration of the final diluted solution is 0.025M

The dilution formula is expressed according to the formula:

[tex]C_1V_1=C_2V_2[/tex]

Given the following parameters

[tex]C_1=0.5M\\V_1=5.00mL\\V_2=100.0mL\\C_2=?[/tex]

Substitute the given parameters into the formula:

[tex]C_1V_1=C_2V_2\\0.5(5)=100C_2\\2.5=100C_2\\C_2=\frac{2.5}{100}\\C_2= 0.025M[/tex]

Hence the concentration of the final diluted solution is 0.025M

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Answer:

here, as we have known the elements of group 3A(13) such as aluminium , boron has three valance electron and in perodic table the elements are kept with similar proterties in same place so, their valance electron is 3.

hope it helps...

The number of valence electrons are in the electron dot structures for the elements in group 3A(13) is three.

The periodic table is organized into groups (vertical columns), periods (horizontal rows), and families (groups of elements that are similar). Elements in the same group have the same number of valence electrons.

Groups are the columns of the periodic table, and periods are the rows. There are 18 groups, and there are 7 periods plus the lanthanides and actinides.

There are two different numbering systems that are commonly used to designate groups, and you should be familiar with both.

The traditional system used in the United States involves the use of the letters A and B. The first two groups are 1A and 2A, while the last six groups are 3A through 8A. The middle groups use B in their titles.

Therefore, The number of valence electrons are in the electron dot structures for the elements in group 3A(13) is three.

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Answer:

See explanation

Explanation:

1H NMR

In the 2-chloro-pentanal we have 4 different types of hydrogens. Therefore, we will have 4 different signals. (See figure 1)

Red hydrogen

For the red hydrogens we have only 1 neighbor. So, if we follow the n+1 rule we can calculate the multiplicity of this hydrogen. In this case a doublet.

Blue hydrogens

In this case, we have 3 neighbors (one in the right, two in the left). Therefore we will have a quartet.

Purple hydrogens

For these hydrogens, we have also will have a quartet, because we have 3 neighbors (one in the right, two in the left).

Green hydrogens

In the green hydrogen,s we have 5 neighbors (2 in the right 3 in the left). Therefore a sextet would be produced.

Orange hydrogens

Finally, in these hydrogens, we have 2 neighbors. Therefore a triplet is expected.

13C NMR

For the 13C NMR, we have again 4 different kinds of carbons. Therefore we will have 4 signals. The most deshielded carbon, in this case, is the red one (see figure 2), so this carbon would be on the left side (around 190). Then the next deshield carbon is the blue one, due to the "Cl" atom placed on this carbon.

I hope it helps!

Answer:

Edge length of the unit cell is 4.07x10⁻¹⁰m

Explanation:

In a face-centered cubic structure, the edge, a, could be obtained using pythagoras theorem knowing the hypotenuse of the unit cell, b, is equal to 4r:

a² + a² = b² = (4r)²

2a² = 16r²

a = √8 r

That means edge lenght is = √8 r

adius

As radius of Silver is 144pm = 144x10⁻¹²m:

a = √8 r

a = √8 ₓ 144x10⁻¹²m

a = 4.07x10⁻¹⁰m

Answer:

The answer is

Explanation:

Density = mass / volume

mass = density × volume

volume = 2cm³

density = 520g/cm³

mass = 2 × 520

= 1040g

Hope this helps you

Answer:

254.23 g/mol

Explanation:

Atomic mass for Osmium tetroxide would be 254.23 g/ml

Answer:254.2276

Explanation:

Answer:

Li

Explanation:

The phenomenon of wave particle duality was well established by Louis deBroglie. The wavelength associated with matter waves was related to its mass and velocity as shown below;

λ= h/mv

Where;

λ= wavelength of matter waves

m= mass of the particle

v= velocity of the particle

This implies that if the velocities of all particles are the same, the wavelength of matter waves will now depend on the mass of the particle. Hence; the wavelength of a matter wave associated with a particle is inversely proportional to the magnitude of the particle's linear momentum. The longest wavelength will then be obtained from the smallest mass of matter. Hence lithium which has the smallest mass will exhibit the longest DeBroglie wavelength

The atom that have the longest de Broglie wavelength is ; ( A ) Li

Wave particle duality is a phenomenon by de Broglie. that shows that The wavelength associated with matter waves is related to its mass and velocity .

Wave particle duality is represented as ;  λ = h / mv

λ= wavelength of matter waves

m= mass of the particle

v= velocity of the particle

Given that the elements have the same velocity the atom that would have the longest de Broglie wavelength is Li

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Answer:

a. boiling

Explanation:

Answer:

The element potassium forms a cation with the charge +1 . The symbol for this ion is K⁺, and the name is potassium ion. The number of electrons in this ion is 18.

Explanation:

Potassium is a metal. It belongs to the group 1 elements. Metals form cations by losing electrons. Since potassium is a group element, it forms a cation by losing one electron. The charge it has is +1 due to the excess of the protons compared t the electrons by 1.

Potassium has  19 electrons. Potassium io on the other hand has 19-1 = 18 electrons.

Answer:

427°C .

Explanation:

Step 1:

Data obtained from the question. This include the following:

Initial temperature (T1) = 77°C

Initial pressure (P1) = P

Final pressure (P2) = 2P

Final temperature (T2) =?

Step 2:

Conversion of celsius temperature to Kelvin temperature.

This is illustrated below:

T(K) = T (°C) + 273

Initial temperature (T1) = 77°C

Initial temperature (T1) = 77°C+ 273 = 350K

Step 3:

Determination of the new temperature. The new temperature can be obtained as follow:

P1/T1 = P2/T2

P/350 = 2P/T2

Cross multiply

P x T2 = 350 x 2P

Divide both side by P

T2 = (350 x 2P ) / P

T2 = 700K

Step 4:

Conversion of Kelvin temperature to celsius temperature.

This can be obtained as follow:

T(°C) = T(K) – 273

T(K) = 700K

T(°C) = 700 – 273

T(°C) = 427°C

Therefore, the new temperature of the gas is 427°C

Answer:

the mass of HBr that would react is 25.41 g of HBr

Explanation:

attached is the calculations.

Answer:

pH is a measure of the concentration of H+ ions in a solution of an acid or base. The pH plots the concentration of solutions in a range from 0–14.

Explanation:

The pH is a measure of the hydrogen ion(H^+) concentration in an acid or base. It can be obtained mathematically by the formula:

pH = —Log [H^+]

The pH scale ranges from 0 to 14

Answer:

it really is A

Explanation:

just got wrong answer because i put 16 and clearly b and c makes no sence : )

Answer:

See explanation

Explanation:

In this question, we have to follow the IUPAC rules. Lets analyze each compound:

a. 1-methylbutane

In this compound we have a chain of 5 carbons, so the correct name is Pentane.

b. 1,1,3-trimethylhexane

In this compound, we longest chain is made of 7 carbons, so, we have to use the name "heptane". Carbon one would be the closet one to the methyl group, so the correct name is  2,4-dimethylheptane.

c. 5-octyne

In this case, carbon 1 would be the closet one to the triplet bond. With this in mind, the correct name is oct-3-yne.

d. 2-ethyl-1-propanol

In this compound, we longest chain is made of 4 carbons, so, we have to use the name "butane". Carbon one would be the carbon with the "OH" group, so the correct name is  2-methylbutan-1-ol.

e. 2.2-dimethyl-3-butanol

In this case, carbon 1 would be the closet one to the "OH". With this in mind, the correct name is 3,3-dimethylbutan-2-ol.

See figure 1

I hope it helps!

The approximate age of the cloth is 17190 years.

We'll begin by calculating the number of half-lives that has elapsed. This can be obtained as follow:

2ⁿ = 100 / 13

2ⁿ = 8

2ⁿ = 2³

n = 3

Finally, we shall determine the age of the cloth.

t = n × t½

t = 3  × 5730

t = 17190 years

Thus, the approximate age of the cloth is 17190 years

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Answer:

H2O

Electron geometry-tetrahedral

Molecular geometry bent

CH2Cl2

Electron geometry- tetrahedral

Molecular geometry-tetrahedral

OPCL3

Electron geometry- tetrahedral

Molecular geometry- tetrahedral

CO3^2-

Electron geometry- trigonal planar

Molecular geometry- trigonal planar

ALCL6^3-

Electron geometry-octahedral

Molecular geometry- octahedral

SO2

Electron geometry-tetrahedral

Molecular geometry-bent

PCL5

Electron geometry-trigonal bipyramidal

Molecular geometry- trigonal bipyramidal

Explanation:

Water contains four electron domains this corresponds to a tetrahedral electron geometry. How ever, there are two lone pairs in the molecule hence it is bent.

CH2Cl2 is shows a tetrahedral molecular geometry and a tetrahedral electron geometry. This can only be observed from the structure of the compound.

OPCL3 is bonded to four groups making it a tetrahedral molecule. There are non lone pairs on phosphorus so the molecule is not bent.

CO3^2- is bonded to three groups which leads to a trigonal planar geometry.

ALCL6^3- contains six bonding groups which arrange themselves at the corners of a regular octahedron at a bond angle of 90°.

SO2 has four electron domains leading to a tetrahedral electron domain geometry according to valence shell electron pair repulsion theory. However, the lone pairs on the central atom in the molecule leads to a bent molecular geometry.

PCL5 has five electron domains without lone pairs of electrons on its central atom. Hence the molecule possess a trigonal bipyramidal geometry.

The electron geometry and molecular geometry of the molecule are as follows:

A. H₂O: The electron geometry is tetrahedral because it has four electron domains (two bonding pairs and two lone pairs). However, due to the presence of two lone pairs, the molecular geometry is bent or V-shaped.

B. CH2Cl₂: The electron geometry is tetrahedral. However, the molecular geometry is trigonal planar because two of the electron domains are occupied by chlorine atoms, resulting in a bent shape.

C. OPCl₃: The electron geometry is tetrahedral. However, the molecular geometry is trigonal pyramidal because one of the electron domains is occupied by a lone pair on phosphorus.

D. CO3⁻²: The electron geometry is trigonal planar because it has three electron domains (three single bonds). The molecular geometry is also trigonal planar.

E. AlCl6⁻³: The electron geometry is octahedral because it has six electron domains. The molecular geometry is also octahedral.

F. SO₂: The electron geometry is trigonal planar because it has three electron domains (two single bonds and one lone pair). The molecular geometry is bent or V-shaped due to the presence of a lone pair on sulfur.

G. PCl₅: The electron geometry is trigonal bipyramidal because it has five electron domains. The molecular geometry is also trigonal bipyramidal.

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Answer:

0.109 mol/tablespoon

Explanation:

6.37 g/ 58.5 mol = 0.10888888 mol (0.109 significantly)

Answer:

A: 0.109

Explanation:

Edge 2020

Answer:

[tex]\Delta _cH=-1328.3kJ/mol[/tex]

Explanation:

Helllo,

In this case, for the given chemical reaction in gaseous state:

[tex]CH_3OCH_3+3O_2\rightarrow 2CO_2+3H_2O[/tex]

We comoute the combustion enthalpy as the reaction enthalpy for one mole of fuel (dimethyl ether) considering the formation enthalpy of each given substance and whether they are reactants (subtracting) or products (adding), therefore we write:

[tex]\Delta _cH=2*\Delta _fH_{CO_2}+3*\Delta _fH_{H_2O}-\Delta _fH_{CH_3OCH_3}-3*\Delta _fH_{O_2}[/tex]

Whereas the formation enthalpies for carbon dioxide, water, dimethyl ether and oxygen are -393.5, -241.8, -184.1 and 0 kJ/mol respectively, thereby, the combustion enthalpy turns out:

[tex]\Delta _cH=2(-393.5)+3*(-241.8)-(-184.1)-3(0)\\\\\Delta _cH=-1328.3kJ/mol[/tex]

Notice that enthalpy of formation of oxygen is zero since forming an element has no chemical sense, it just exists as it has been early demonstrated.

Regards.

Answer:

Following are the answer to this question:

Explanation:

The answer are:

1) ketopentose

2) Triose

3) Aldose

4) Ketose

5) Glucose

6) Aldohexose

Answer:

The number of moles of Cl₂ present at equilibrium is 3.94x10⁻⁴ moles.

Explanation:

The reaction is:

CO(g) + Cl₂(g) ⇄ COCl₂(g)  

The equilibrium constant of the above reaction is:

K = 1.2x10³

To find the moles of Cl₂ present at equilibrium, let's evaluate the reverse reaction:

COCl₂(g) ⇄ CO(g) + Cl₂(g)  

The equilibrium constant for the reverse reaction is:

[tex] K_{r} = \frac{1}{1.2 \cdot 10^{3}} = 8.3 \cdot 10^{-4} [/tex]

Now, we need to calculate the concentration of CO and COCl₂:

[tex] C_{CO} = \frac{\eta_{CO}}{V} = \frac{0.3500 moles}{3.050 L} = 0.115 M [/tex]

[tex] C_{COCl_{2}} = \frac{\eta_{COCl_{2}}}{V} = \frac{0.05500 moles}{3.050 L} = 0.018 M [/tex]

Now, from the reaction we have:

COCl₂(g) ⇄ CO(g) + Cl₂(g)  

0.018 - x       0.115+x   x    

The concentration of Cl₂ is:

[tex] K_{r} = \frac{[CO][Cl_{2}]}{[COCl_{2}]} [/tex]

[tex] 8.3 \cdot 10^{-4} = \frac{(0.115 + x)(x)}{0.018 - x} [/tex]  

[tex] 8.3 \cdot 10^{-4}*(0.018 - x) - (0.115 + x)(x) = 0 [/tex]  

By solving the above equation for x we have:

x = 1.29x10⁻⁴ M = [Cl₂]

Finally, the number of moles of Cl₂ present at equilibrium is:

[tex] \eta_{Cl_{2}} = C_{Cl_{2}}*V = 1.29 \cdot 10^{-4} mol/L*3.050 L = 3.94 \cdot 10^{-4} moles [/tex]

Therefore, the number of moles of Cl₂ present at equilibrium is 3.94x10⁻⁴ moles.

I hope it helps you!

Answer:

Product: propan-1-ol

Explanation:

IIn this case, we have to remember that [tex]LiAlH_4[/tex]  is a reduction agent.  So, this is a reduction reaction. The [tex]LiAlH_4[/tex] has the ability to produce hydride ions [tex]H^-[/tex]. This ion can attack the carbonyl group generating a negative charge in the oxygen. In the next step, the negative charge in the oxygen can attack a water molecule to protonate the molecule and produce propan-1-ol.

See figure 1

I hope it helps!

Answer:

The reaction is not in equilibrium

Explanation:

For the reaction:

PCl₅ ⇄ PCl₃ + Cl₂

Equilibrium constant, Kp, is defined as:

[tex]Kp = \frac{P_{PCl_3}P_{Cl_2}}{P_{PCl_5}} = 0.497[/tex]

When this ratio is = 0.497, the reaction is in equilibrium. Replacing the pressures of the problem, reaction quotient, Q, is:

[tex]Q =\frac{1.322atm*0.911atm}{0.820atm} = 1.469[/tex]

As Q ≠ Kp, the reaction is not in equilibrium

To reach the equilibrium, the reaction will shift to the left producing more reactant and decreasing amount of products.

Answer:

The balanced equation is given below: C2H6O + 3O2 —> 2CO2 + 3H2O

The coefficients are: 1, 3, 2, 3

Explanation:

C2H6O + O2 —> CO2 + H2O

The above equation can be balance as follow:

There are 2 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 2 in front of CO2 as shown below:

C2H6O + O2 —> 2CO2 + H2O

There are 6 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 3 in front of H2O as shown below:

C2H6O + O2 —> 2CO2 + 3H2O

There are a total of 3 atoms of O on the left side and a total of 7 atoms on the right side. It can be balance by putting 3 in front of O2 as shown below:

C2H6O + 3O2 —> 2CO2 + 3H2O

Now the equation is balanced.

The coefficients are: 1, 3, 2, 3.

Answer:

The balanced equation is given below: C2H6O + 3O2 —> 2CO2 + 3H2O

The coefficients are: 1, 3, 2, 3

Explanation:

C2H6O + O2 —> CO2 + H2O

The above equation can be balance as follow:

There are 2 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 2 in front of CO2 as shown below:

C2H6O + O2 —> 2CO2 + H2O

There are 6 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 3 in front of H2O as shown below:

C2H6O + O2 —> 2CO2 + 3H2O

There are a total of 3 atoms of O on the left side and a total of 7 atoms on the right side. It can be balance by putting 3 in front of O2 as shown below:

C2H6O + 3O2 —> 2CO2 + 3H2O

Now the equation is balanced.

The coefficients are: 1, 3, 2, 3.

Explanation:

Answer:

A) One that occurs on its own

Answer:

See explanation

Explanation:

Hydration of alkenes is a common reaction in organic chemistry. Hydration is simply the addition of water to an alkene. This is an acid catalysed reaction as we can see from the mechanism attached.

Recall that our task is to carry out the synthesis of 2-butanol using an alkene starting material in which there will be no rearrangement of the intermediate carbocation. If we start with the compound shown in the image (but-2-ene), the first step is the formation of the secondary carbocation. This is followed by the addition of water. Subsequently, the added water is deprotonated by another water molecule to yield 2-butanol and the acid catalyst. All these steps have been clearly outlined in the image attached.

Answer:

Ba(NO₃)₂(s) → Ba²⁺ + 2NO₃⁻

Explanation:

A strong electrolyte is a salt (A compound that has an anion and a cation and are neutral) that, in water, dissociates completely in its ions.

In Barium nitrate, Ba(NO₃)₂, the cation is Ba²⁺ (Alkaline earth metal), and the anion is the nitrate ion, NO₃⁻.

Thus, when Ba(NO₃)₂ (s) is dissolved in water, its transformation is:

When solid barium nitrate (Ba(NO₃)₂) dissolves in water, it undergoes a dissociation process where the compound breaks apart into its constituent ions.

Dissociation refers to the process in which a compound breaks apart into its constituent ions when dissolved in a solvent, typically water. In this process, the chemical bonds within the compound are disrupted, resulting in the separation of positive and negative ions.

The dissociation occurs due to the interaction between the solute particles and the solvent molecules, leading to the formation of hydrated ions.

The transformation can be represented as follows:

Ba(NO₃)₂(s) → Ba²⁺(aq) + 2NO₃⁻(aq)

In this process, the barium nitrate compound dissociates into barium ions (Ba²⁺) and nitrate ions (NO₃⁻) in the aqueous solution. The resulting ions are free to move and conduct electricity, indicating that barium nitrate is a strong electrolyte when dissolved in water.

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Answer:

5.42

Explanation:

Step 1: Consider the dissociation of NH₄Br

NH₄Br(aq) ⇒ NH₄⁺(aq) + Br⁻(aq)

Br⁻ is the conjugate base of HBr, a strong acid, so it doesn´t react with water. NH₄⁺ is the conjugate acid of NH₃, so it does react with water.

Step 2: Consider the acid reaction of NH₄⁺

NH₄⁺(aq) + H₂O(l) ⇄ NH₃(aq) + H₃O⁺(aq)

Step 3: calculate the acid dissociation constant for NH₄⁺

We will use the following expression.

[tex]K_a \times K_b = K_w\\K_a = \frac{K_w}{K_b} = \frac{1.00 \times 10^{-14} }{1.76 \times 10^{-5}} = 5.68 \times 10^{-10}[/tex]

Step 4: Calculate the concentration of H₃O⁺

We will use the following expression.

[tex][H_3O^{+} ]= \sqrt{K_a \times C_a } = \sqrt{5.68 \times 10^{-10} \times 0.0255 } = 3.81 \times 10^{-6}M[/tex]

Step 5: Calculate the pH

We will use the following expression.

[tex]pH = -log [H_3O^{+} ] = -log (3.81 \times 10^{-6}) = 5.42[/tex]

The pH of 0.0255 M solution should be 5.42.

Since we know that

ka * kb = kw

So,

ka = kw/kb

= 1.00*10^-14 / 1.76*10^-5

= 5.68*10^-10

Now the concentration of H3O should be

= √ka * Ca

= √5.68*10^-10 * 0.0255

= 3.81*10^-6M

Now the pH value should be

= -log(H3O+)

= -log(3.81*10^-6)

= 5.42

hence, The pH of 0.0255 M solution should be 5.42.

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Answer:

Q = 1.5 kJ

Explanation:

It is given that,

The specific heat for aluminum is 0.900 J/g°C

Mass of sample, m = 3.8 g

Initial temperature, [tex]T_i=450^{\circ} C[/tex]

Final temperature, [tex]T_f=25^{\circ} C[/tex]

We need to find the heat released. The amount of heat released is given by the formula:

[tex]Q=mc\Delta T\\\\Q=mc(T_f-T_i)\\\\Q=3.8\times 0.9\times (25-450)\\\\Q=1453.5\ J\\\\Q=1.45\ kJ[/tex]

or

[tex]Q=1.5\ kJ[/tex]

So, the correct option is (A) i.e. 1.5 kJ.

3.97×1023 molecules C2H6          1 mol  C2H6  

------------------------------------------ x ------------------------------------   = 0.66 mol C2H6

                                                    6.022 x 1023 molec. C2H6