Chromium has an atomic mass of 51.9961 u and consists of four isotopes, Cr50, Cr52, Cr53, and Cr54. The Cr52 isotope has a natural abundance of 83.79% and an atomic mass of 51.9405 u. The Cr54 isotope has a natural abundance of 2.37% and an atomic mass of 53.9389 u. The natural abundances of the Cr50 and Cr53 isotopes exist in a ratio of 1:0.4579, and the Cr50 isotope has an atomic mass of 49.9460 u. Determine the atomic mass of the Cr53 isotope.

Final answer:

To calculate the approximate enthalpy change for the given reaction, we need to consider the bond energies of the bonds in the reactants and products. The approximate enthalpy change is -3818 kJ.

Explanation:

To calculate the approximate enthalpy change (ΔH) for the given reaction, we need to consider the bond energies of the bonds in the reactants and products. The reaction is:

QX3 + 3H20 => Q(OH)3 + 3HX

We can calculate the energy absorbed or released by breaking and forming these bonds.

First, we calculate the energy required to break the bonds in the reactants:

Then, we calculate the energy released when the bonds in the products are formed:

Finally, we sum up the energy changes:

Therefore, the approximate enthalpy change for the reaction is -3818 kJ.

Answer:

P=atm

[tex]b=\frac{L}{mol}[/tex]

Explanation:

The problem give you the Van Der Waals equation:

[tex](P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT[/tex]

First we are going to solve for P:

[tex](P+\frac{n^{2}a}{V^{2}})=\frac{nRT}{(V-nb)}[/tex]

[tex]P=\frac{nRT}{(V-nb)}-\frac{n^{2}a}{v^{2}}[/tex]

Then you should know all the units of each term of the equation, that is:

[tex]P=atm[/tex]

[tex]n=mol[/tex]

[tex]R=\frac{L.atm}{mol.K}[/tex]

[tex]a=atm\frac{L^{2}}{mol^{2}}[/tex]

[tex]b=\frac{L}{mol}[/tex]

[tex]T=K[/tex]

[tex]V=L[/tex]

where atm=atmosphere, L=litters, K=kelvin

Now, you should replace the units in the equation for each value:

[tex]P=\frac{(mol)(\frac{L.atm}{mol.K})(K)}{L-(mol)(\frac{L}{mol})}-\frac{(mol^{2})(\frac{atm.L^{2}}{mol^{2}})}{L^{2}}[/tex]

Then you should multiply and eliminate the same units which they are dividing each other (Please see the photo below), so you have:

[tex]P=\frac{L.atm}{L-L}-atm[/tex]

Then operate the fraction subtraction:

P=[tex]P=\frac{L.atm-L.atm}{L}[/tex]

[tex]P=\frac{L.atm}{L}[/tex]

And finally you can find the answer:

P=atm

Now solving for b:

[tex](P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT[/tex]

[tex](V-nb)=\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}[/tex]

[tex]nb=V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}[/tex]

[tex]b=\frac{V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}}{n}[/tex]

Replacing units:

[tex]b=\frac{L-\frac{(mol).(\frac{L.atm}{mol.K}).K}{(atm+\frac{mol^{2}.\frac{atm.L^{2}}{mol^{2}}}{L^{2}})}}{mol}[/tex]

Multiplying and dividing units,(please see the second photo below), we have:

[tex]b=\frac{L-\frac{L.atm}{atm}}{mol}[/tex]

[tex]b=\frac{L-L}{mol}[/tex]

[tex]b=\frac{L}{mol}[/tex]

Answer:

0.8368 kiloJoules of an excess nutritional energy is consumed by an adult in a day.

It will take 17,447.41 days long to gain 1 lb of fat by this person.

Explanation:

Calorie intake of an adult in a day =[tex]2.2\times 10^3 calorie[/tex]

Calorie burnt by an adult in a day = [tex]2.0\times 10^3 calorie[/tex]

Excess  nutritional energy in day=

[tex](2.2\times 10^3 calorie)-(2.0\times 10^3 calorie)[/tex]

[tex]=2.0\times 10^2 calorie[/tex]

1 kcal = 4.184 kJ

So , 1000 cal = 4.184 kJ

[tex]1 cal = 4.184\times 10^{-3} kJ[/tex]

[tex]2.0\times 10^2 calorie=2.0\times 10^2\times 4.184\times 10^{-3} kJ[/tex]

[tex]=0.8368 kJ[/tex]

0.8368 kiloJoules of an excess nutritional energy is consumed by an adult in a day.

[tex]14.6\times 10^3[/tex] kilo joules of excess nutritional energy = 1 lb fat

The 1 kilo joule of excess nutritional energy  = [tex]\frac{1}{14.6\times 10^3}[/tex] lb fat

Excessive nutritional energy of an adult per day = 0.8368 kiloJoules

Amount of fat gained by an adult per day =

= [tex]0.8368 kiloJoules\times \frac{1}{14.6\times 10^3}=5.7315\times 10^{-5} lb[/tex] of fat

In 1 day an adult gains = [tex]5.7315\times 10^{-5} lb[/tex] of fat

Time taken to gain 1 lb fat:

[tex]\frac{1}{5.7315\times 10^{-5}} day=17,447.41 days[/tex]

It will take 17,447.41 days long to gain 1 lb of fat by this person.

Final Answer:

An adult who consumes an excess of 0.2 x 10³ Cal per day, equivalent to 0.8368 x 10³ kJ, will gain 1 lb of fat in approximately 17.5 days, as 1 lb of fat is stored for each 14.6 x 10³ kJ of excess nutritional energy consumed.

Explanation:

An adult consumes approximately 2.2 x 10³ Calories (Cal) per day and burns 2.0 x 10³ Cal per day. To find the excess nutritional energy in kilojoules, we need to subtract the energy burned from the energy consumed and then convert the result from Calories to kilojoules:

Excess energy (Cal) = 2.2 x 10³ Cal - 2.0 x 10³ Cal = 0.2 x 10³ Cal

Excess energy (kJ) = 0.2 x 10³ Cal x 4.184 kJ/Cal = 0.8368 x 10³ kJ

To calculate how long it will take for the adult to gain 1 lb of fat, we divide the amount of kilojoules that corresponds to 1 lb of fat by the daily excess kilojoules:

Days to gain 1 lb = (14.6 x 10³ kJ per 1 lb) / (0.8368 x 10³ kJ/day)

Days to gain 1 lb = 17.4512 days

Therefore, it will take approximately 17.5 days for the adult to gain 1 lb of fat, given the daily excess of 0.2 x 10³ Cal.

Answer: The mass of air is 0.00260 lbs.

Explanation:

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

Or,

[tex]PV=\frac{m}{M}RT[/tex]

where,

P = pressure of the gas = 150 psia = 10.2 atm   (Conversion factor:  1 psia = 0.068 atm)

V = Volume of gas = [tex]10in^3=0.164L[/tex]    (Conversion factor:  [tex]1in^3=0.0164L[/tex] )

m = mass of air = ?

M = Average molar mass of air = 28.97 g/mol

R = Gas constant = [tex]0.0820\text{ L atm }mol^{-1}K^{-1}[/tex]

T = temperature of the gas = [tex]440^oF=499.817K[/tex]  (Conversion factor: [tex](T(K)-273.15)=(T(^oF)-32)\times \frac{5}{9}[/tex]  )

Putting values in above equation, we get:

[tex]10.2atm\times 0.164L=\frac{m}{28.97g/mol}\times 0.0820\text{ L atm }mol^{-1}K^{-1}\times 499.817K\\\\m=1.18g[/tex]

Converting this mass into lbs, we use the conversion factor:

1 lbs = 454 g

So, [tex]1.18g\times \frac{1lbs}{454g}=0.00260lbs[/tex]

Hence, the mass of air is 0.00260 lbs.

Answer:

Volume of HCl required = 28.4 mL

Explanation:

[tex]Zn(s) + 2HCl (aq) \rightarrow ZnCl_2(aq) + H_2(g)[/tex]

Mass of zinc ore = 4.65 g

% of zinc in zinc ore = 50 %

So, mass of zinc in zinc ore = 0.50 × 4.65 = 2.325 g

No. of moles of Zn = [tex]\frac{2.325}{65.40} = 0.0355\ mol[/tex]

So, as per the reaction coefficient,

1 mol of zinc reacts with 2 mol of HCl

0.0355 mol of zinc reacts with

                                    = 0.0355 × 2 = 0.071 mol of HCl

Molarity of HCl = 2.50 M

Volume of HCl = [tex]\frac{Moles}{Concentration}[/tex]

[tex]Volume\ of\ HCl = \frac{0.071}{2.50} = 0.0284\ L[/tex]

1 L = 1000 mL

0.0284 L = 1000 × 0.0284 = 28.4 mL

Answer:

[tex]93.43\times 10^{-9}[/tex]

Explanation:

Scientific notation is the way of writing numbers which are either large or small. The number is written in the scientific notation when the number is between 1 and 10 and then multiplied by the power of 10. Engineering notation is the same version of the scientific notation but the number can be between 1 and 1000 and in this exponent of the ten is divisible by three.

For example, [tex]1000^2[/tex] is to be written as [tex]10^6[/tex] in engineering notation.

The given number:

0.00000009345 can be written as [tex]93.425\times 10^{-9}[/tex]

Answer upto 4 significant digits = [tex]93.43\times 10^{-9}[/tex]

Answer:

0.7681

Explanation:

Given that the mass percentage of methanol = 85.5 %

which means that 85.5 g of methanol is present in 100 g of the solution.

Thus, mass of water = 100 g - 85.5 g = 14.5 g

The formula for the calculation of moles is shown below:

[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]

Mass of methanol = 85.5 g

Molar mass of methanol = 32.04 g/mol

Moles of methanol = 85.5 g / 32.04 g/mol = 2.6685 moles

Mass of water = 14.5 g

Molar mass of methanol = 18 g/mol

Moles of methanol = 14.5 g / 18 g/mol = 0.8056 moles

So, according to definition of mole fraction:

[tex]Mole\ fraction\ of\ methanol=\frac {n_{methanol}}{n_{methanol}+n_{water}}[/tex]

Applying values as:

[tex]Mole\ fraction\ of\ methanol=\frac {2.6685}{2.6685+0.8056}[/tex]

Mole fraction of methanol in the solution = 0.7681

Answer:

The total amount of heat required for the process is 76.86 KJ

Explanation:

We can divide the process in 5 parts, in which we can calcule each amount of heat required (see attached Heating curve):

(1) Ice is heated from -25ºC to 0ºC. We can calculate the heat of this part of the process as follows. Note that we must convert J in KJ (1 KJ= 1000 J).

Heat (1) = mass ice x Specific heat ice x (Final temperature - Initial Temperature)

Heat (1) =[tex]25 g x 2.11 J/g.ºC x \frac{1 KJ}{1000 J} x (0ºC-(-25º)[/tex]

Heat (1) = 1.32 KJ

(2) Ice melts at ºC (it becomes liquid water). This is heating at constant temperature (ºC), so we use the melting enthalphy (ΔHmelt) and we must use the molecular weight of water (1 mol H₂O = 18 g):

Heat (2) = mass ice x ΔHmelt

Heat (2)= [tex]25 g  x  \frac{6.01KJ} {1 mol H2O} x \frac{1 mol H2O}{18 g}[/tex]

Heat (2)= 8.35 KJ

(3) Liquid water is heated from 0ºC to 100 ºC:

Heat (3)= mass liquid water x Specific heat water x (Final T - Initial T)

Heat (3)= 25 g x 4.18 J/gºC x 1 KJ/1000 J x (100ºC - 0ºC)

Heat (3)= 10.45 KJ

(4) Liquid water evaporates at 100ºC (it becomes water vapor). This is a process at constant temperature (100ºC), and we use boiling enthalpy:

Heat (4)= mass water x ΔH boiling

Heat (4)= 25 g x [tex]\frac{40.67 KJ}{mol H20}[/tex] x [tex]\frac{1 mol H20}{18 g}[/tex]

Heat (4)= 56.49 KJ

(5) Water vapor is heated from 100ºC to 105ºC. We use the specific capacity of water vapor:

Heat (5)= mass water vapor x Specific capacity vapor x (Final T - Initial T)

Heat (5)= 25 g x 2.00 J/g ºC x 1 KJ/1000 J x (105ºC - 100ºC)

Heat (5)= 0.25 KJ

Finally, we calculate the total heat involved in the overall process:

Total heat= Heat(1) + (Heat(2) + Heat(3) + Heat(4) + Heat(5)

Total heat= 1.32 KJ + 8.35 KJ + 10.45 KJ + 56.49 KJ + 0.25 KJ

Total heat= 76.86 KJ

Answer:

lets set the ratio -A/HA as R:

pH = pKa + log(R,10) => pKa + log10(R)

pH = 5.5

pKa = 4.76

R => 10^(pH - 4.76)

10^(pH - 4.76) => 5.4954

Given R (-A/HA) a number bigger than 1, then the concentration of  -A is bigger than HA

Explanation:

Answer:

There is more A⁻ than HA in the solution  

Explanation:

The equation for the ionization of a weak acid is

HA + H₂O ⇌H₃O⁺ + OH⁻

When HA and A⁻ are present in comparable amounts, we can use the Henderson-Hasselbalch equation:

[tex]\begin{array}{rcl}\text{pH} &=& \text{pK}_{\text{a}} + \log\dfrac{[\text{A}^{-}]}{\text{[HA]}}\\\\5.5 & = & 4.76 + \log\dfrac{[\text{A}^{-}]}{\text{[HA]}}\\\\0.74 & = & \log\dfrac{[\text{A}^{-}]}{\text{[HA]}}\\\end{array}[/tex]

0.74 is a positive number, and a number must be greater than one for its logarithm to be positive. That is,

[tex]\begin{array}{rcl}\dfrac{[\text{A}^{-}]}{\text{[HA]}} & > & 1\\\\\textbf{[A]}^{-} & > & \textbf{[HA]}\\\end{array}[/tex]

Explanation:

1.) 175 km to μm

[tex]1 km=10^9 \mu m[/tex]

[tex]175 km=175\times 10^9\mu m=1.75\times 10^{11} \mu m[/tex]

3.) 385 nm to dm

[tex]1 nm=10^{-8} dm[/tex]

[tex]385 nm=385\times 10^{-8} dm=3.85\times 10^{-6} dm[/tex]

5.) 492 μm  to m

1 μm =  [tex]10^{-6} m[/tex]

[tex]492 \μm=492\times 10^{-6} m=4.92\times 10^{-4} m[/tex]

7.) [tex]52\times 10^3[/tex] dm to mm

1 dm = 100 mm

[tex]52\times 10^3 dm=52\times 10^3\times 100 mm=5.2\times 10^{6}dm[/tex]

9.) [tex]321\times 10^{35}[/tex] mm to km

[tex]1 mm = 10^{-6} km[/tex]

[tex]321\times 10^{35} mm=321\times 10^{35}\times 10^{-6} km=3.21\times 10^{31} km[/tex]

11.) [tex]456\times 10^3[/tex] m to km

m = 0.001 km

[tex]456\times 10^3m =456\times 10^3 m\times 0.001 km=456 km[/tex]

13.) [tex]422\times 10^3[/tex] m to nm

[tex]1 m = 10^{9} nm[/tex]

[tex]422\times 10^3 m=422\times 10^3\times 10^{9} nm=4.22\times 10^{14} nm[/tex]

15.) [tex]4.87\times 10^{30}[/tex] m to pm

[tex]1 m = 10^{12} pm[/tex]

[tex]4.87\times 10^{30} m=4.87\times 10^{30}\times 10^{12} pm=4.82\times 10^{42} pm[/tex]

17.) [tex]5.26\times 10^3[/tex] m to um

1 m =  [tex]10^{6} \mu m[/tex]

[tex]5.26\times 10^3 m=5.26\times 10^3\times 10^6 \mu m=5.26\times 10^{9} \mu m[/tex]

19.) [tex]1.25\times 10^{35}[/tex]m to Mm

1 m =  [tex]10^{-6} Mm[/tex]

[tex]1.25\times 10^{35} m=1.25\times 10^{35}\times 10^{-6} Mm=1.25\times 10^{-29} Mm[/tex]

21.) [tex]4.22\times 10^3[/tex] Tm to nm

[tex]1 Tm = 10^{21} nm[/tex]

[tex]4.22\times 10^3 Tm=4.22\times 10^3\times 10^{21} nm=4.22\times 10^{24} nm[/tex]

Answer:

We need add acid to prepare 0.15 acetate buffer with pH 3.7.

Explanation:

As we know that buffer solution is the combination of weak acid with strong base or strong acid with weak base.

We know that CH3COOH is weak acid .Acetate buffer is the combination of weak acid CH3COOH and conjucate base CH3COO- (from salt).

So we have to add acid to achieve the proper final pH value of mixture.We need add acid to prepare 0.15 acetate buffer with pH 3.7.

Answer:

The vapor pressure at 298 K for this solution is 50,49 Torr.

Explanation:

We have to apply Raoult's law to solve this which its formula is:

P pure sv° - P sl = P pure sv° . X (molar fraction)

ΔP = P pure sv° . X (molar fraction)

58,9 Torr - P sl = 58,9 Torr . X

Molar mass urea CO(NH2)2: 60,06 g/m

Molar mass ethanol: 46,07 g/m

Moles for urea: 42,4 g /  60,06 g/m = 0,705 moles

Moles for ethanol: 195 g / 46,06 g/m = 4,233 moles

X (molar fraction): moles from solute / moles from solute + moles from solvent

X (molar fraction): 0,705 / 0,705 + 4,233 = 0,142

58,9 Torr - P sl = 58.9 Torr . 0,142

58,9 Torr - 8,40 Torr = P sl

50,49 Torr = P sl

Answer:

1) 0.022 lbm

2) 276.253 RPM

3) 0.287 lbf

Explanation:

Given data:

mass = 10 kg

acceleration - [tex]13 g = 13\times 9.81 m/s^2 = 12[/tex]

7.53 m/s^2

r =6 inches = 0.1524 m

1) mass in lbm  [tex]= 0.01\times 2.2 = 0.022 lbm[/tex]

as 1 kg = 2.2 lbm

2) acceleration [tex] =  r \omega ^2[/tex]

[tex]127.53 = 0.1524 \times \omega^2[/tex]

[tex]\omega^2 = 836.811[/tex]

[tex] \omega = 28.927 rad/s[/tex]

[tex]1 rad/s  = 9.55 RPM[/tex]    

[tex][ 1 revolution = 2\pi,    1 rad/s = 1/2\pi RPS = \frac{60}{2\pi} RPM][/tex]

SO IN [tex]28.927 rad/s = \frac{60}{2\pi} \times 28.297 = 276.253 RPM[/tex]

3) Force in [tex]N = mass \times a = 0.01\times 127.53 = 1.2753 N[/tex]

                                                 [tex]=  1.2753\times 0.225 lbf = 0.287 lbf[/tex]

By calculating the moles of Na2CO3, we can infer the moles of HNO3 due to their 1:1 ratio in the reaction. Dividing the moles of HNO3 with the volume of the solution in liters gives the molarity, which is 0.39 M. The absolute uncertainty is calculated using a specific formula and found to be 0.001 M.

The molarity, M, of a solution is the amount of solute divided by the volume of solution (in Liters). We'll start by calculating the moles of Na2CO3, which is obtained by dividing the given mass of the Na2CO3 by its molar mass.

Calculate moles of Na2CO3 = mass (g) / molar mass (g/mol) = 0.9616 g / 105.988 g/mol = 0.00907 mol.

Since the reaction between HNO3 and Na2CO3 is a 1:1 ratio, the moles of HNO3 is equal to the moles of Na2CO3. Now we know amount of solute (HNO3) is 0.00907 mol. To express the volume of the HNO3 solution in liters, we divide by 23.45 mL by 1000 mL/L, giving 0.02345 L.

Molarity of HNO3 = moles of solute / volume of solution in L = 0.00907 mol / 0.02345 L = 0.387 mol/L or M.

The absolute uncertainty can be found using the following formula: (Delta M/M) = sqrt((Delta m / m)^2 + (Delta V / V) ^2) = sqrt((0.0009 / 0.9616)^2 + (0.05 / 23.45)^2) = 0.002. Therefore, absolute uncertainty = M * (Delta M/M) = 0.387 M * 0.002 = 0.000774 M.

The final answer with significant figures: The molarity of HNO3 = 0.39 ± 0.001 M.

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Answer : The rate law is 4.232 mol/L.s

Explanation :

First we have to calculate the value of rate constant.

According to the question, the expression for rate law will be:

[tex]Rate=k[A]^2[B][/tex]

where,

k = rate constant

As we are given :

Rate law = 0.23 mol/L.s

Initial concentration of A = 1.0 mol/L

Initial concentration of B = 1.0 mol/L

Now put all the given values in the above rate law expression, we get:

[tex]0.23mol/L.s=k\times (1.0mol/L)^2\times (1.0mol/L)[/tex]

[tex]k=0.23M^{-2}s^{-1}[/tex]

Now we have to calculate the rate law when initial concentration of A and B is 2.0 mol/L and 4.6 mol//L respectively.

The expression for rate law will be:

[tex]Rate=k[A]^2[B][/tex]

where,

k = rate constant = [tex]0.23M^{-2}s^{-1}[/tex]

Rate law = ?

Initial concentration of A = 2.0 mol/L

Initial concentration of B = 4.6 mol/L

Now put all the given values in the above rate law expression, we get:

[tex]Rate=(0.23M^{-2}s^{-1})\times (2.0mol/L)^2\times (4.6mol/L)[/tex]

[tex]Rate=4.232mol/L.s[/tex]

Therefore, the rate law is 4.232 mol/L.s

The study of chemicals and bonds is called chemistry. When the amount of reactant and product are equal then is said to be an equilibrium state.

The correct answer is 4.232.

According to the question, the expression for rate law will be:

[tex]Rate =k[A]^2[B][/tex]

where,

The data is given as follows:-

Place all the values in the equation and solve it

[tex]0.23= k*1^2*1\\\\k= 0.23[/tex]

Now we have to calculate the rate law when the initial concentration of A and B is 2.0 mol/L and 4.6 mol//L respectively.

The expression for rate law will be:

[tex]Rate =k[A]^2[B][/tex]

where,

[tex]Rate = 0.23*2^2*4.6[/tex]

Hence, the rate will be 4.232.

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Answer:

1,2 dichlorobenzene was used as the solvent for the diels alder reaction: because the co elimination part of the reaction needs high temp and a high boiling point solvent such as 1,2 dichlorobenzene

Explanation:

Diels-Alder Reaction is a useful synthetic tool to prepare cyclohexane rings. It is a process, which occurs in a single step that consists of a cyclic redistribution of its electrons. The two reagents are bond together through a cyclic transition state in which the two new C-C bonds are formed at the same time. For this to occur, most of the time, it is necessary a high temperature and high-pressure conditions. Since 1,2 dichlorobenzene has a boiling point of 180ºC is a good solvent for this type of reactions.

Answer:

Sodium laurate, also known as sodium dodecanoate, is a soap. It is the salt of lauric acid. It is an amphiphilic organic molecule which is composed of a hydrophilic head (polar ) and a hydrophobic tail (non-polar fatty acid).

In a aqueous solution, it leads to the formation of a micelle. The hydrophilic head of the molecule interacts with the surrounding polar solvent molecules. Thereby, making the micelle soluble in the solution. Whereas, the hydrophobic tails present in the core of micelle, interacts with the non-polar oil particles.

Answer:

The least amount of energy emitted in this case is 0.6 eV.

The corresponding quantum number n would be n=4.

The wavelenght asociated to the emitted photon would be 2.06 [tex]\mu[/tex]m, corresponding to the Infrared spectrum.

Explanation:

For calculating the energy of an electron emitted/absorbed in an electronic transition of the hydrogen atom, the next equation from the Bohr model can be used:

[tex]E=E_{0} Z^2 [\frac{1}{n_1^2}-\frac{1}{n_2^2}][/tex]

, where E is the photon energy, [tex]E_0[/tex] is the energy of the first energy level (-13.6 eV), Z is the atomic number, [tex]n_1[/tex] is the quantum number n of the starting level and [tex]n_2[/tex] the quantum number n of the finishing level. In this case, [tex]n_2=3[/tex], and [tex]n_1=4[/tex], because this excited level is the next in energy to n=3.

Considering that [tex]1 eV= 1.60217662x10^{-19} J[/tex], and using the Planck equation [tex]E=h\nu=\frac{hc}{\lambda}[/tex], you can calculate the wavelenght or the frequency associated to that photon. Values in the order of [tex]\mu[/tex]m in wavelenght belong to the Infrared spectrum, wich can not being seen by humans.

Answer:

1. [tex]E=1.059x10^{-19}J[/tex]

2. [tex]n=4[/tex]

3. The associated wavelength belongs to the infrared spectrum which is invisible for humans.

Explanation:

Hello,

1. At first, according to the equation:

[tex]E=E_oZ^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})[/tex]

Whereas [tex]E[/tex] is the emitted energy, [tex]E_o[/tex] the first level energy, [tex]Z[/tex] the atomic number, [tex]n_1[/tex] the first level and [tex]n_2[/tex] the second level. In such a way, the closest the [tex]n[/tex]'s are, the least the amount of emitted energy, therefore, [tex]n_1=3[/tex] (based on the statement) and [tex]n_2=4[/tex], thus, the least amount of energy turns out being:

[tex]E=(2.179x10^{-18}J)(1^2)(\frac{1}{3^2}-\frac{1}{4^2})\\E=1.059x10^{-19}J[/tex]

2. Secondly, and based on the first question, the quantum number for the excited state is mandatorily [tex]n=4[/tex]

3. Finally, to substantiate why we cannot observe the emitted photons we apply the following equation:

[tex]E=\frac{hc}{\lambda} \\\lambda=\frac{hc}{E}=\frac{(6.62607004x10^{-34} m^2 kg / s)(299 792 458m/s)}{1.059x10^{-19}m^2 kg / s^2}  \\\lambda=1.88x10^{-6}m[/tex]

Whereas the obtained wavelength corresponds to the infrared spectrum which is not observable by humans.

Best regards.

Answer:

16 minutes

Explanation:

First, we need to calculate the amount of heat needed to cool the beef stew:

Q = mcΔT

Where m is the mass, c is the heat capacity and ΔT is the variation of the temperature.

Q = 10x4x(40 - 90)

Q = -2000 kJ

So, the beef stew needs to lost 2000 kJ to cool.

With the initial temperature at 90ºC, the rate of cooling(r) will be:

r = 1.955x(90 - 25)

r = 127.075 kJ/min

So, to lose 2000 kJ, will be necessary:

t = Q/r

t = 2000/127.075

t = 16 minutes

Answer : The reaction must shift to the product or right to be in equilibrium.  The equilibrium concentration of [tex]H_2[/tex] is 7.0 M

Explanation :

Reaction quotient (Qc) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

First we have to determine the concentration of [tex]CH_4,H_2S,CS_2\text{ and }H_2[/tex].

[tex]\text{Concentration of }CH_4=\frac{\text{Moles of }CH_4}{\text{Volume of solution}}=\frac{1mol}{400mL}\times 1000=5M[/tex]

[tex]\text{Concentration of }H_2S=\frac{\text{Moles of }H_2S}{\text{Volume of solution}}=\frac{2mol}{400mL}\times 1000=2.5M[/tex]

[tex]\text{Concentration of }CS_2=\frac{\text{Moles of }CS_2}{\text{Volume of solution}}=\frac{1mol}{400mL}\times 1000=2.5M[/tex]

[tex]\text{Concentration of }H_2=\frac{\text{Moles of }H_2}{\text{Volume of solution}}=\frac{2mol}{400mL}\times 1000=5M[/tex]

Now we have to determine the value of reaction quotient (Qc).

The given balanced chemical reaction is,

[tex]CH_4(g)+2H_2S(g)\rightarrow CS_2(g)+4H_2(g)[/tex]

The expression for reaction quotient will be :

[tex]Q_c=\frac{[CS_2][H_2]^4}{[CH_4][H_2S]^2}[/tex]

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

[tex]Q_c=\frac{(2.5)\times (5)^4}{(2.5)times (5)^2}=25[/tex]

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

There are 3 conditions:

When [tex]Q>K[/tex] that means product > reactant. So, the reaction is reactant favored.

When [tex]Q<K[/tex] that means reactant > product. So, the reaction is product favored.

When [tex]Q=K[/tex] that means product = reactant. So, the reaction is in equilibrium.

The given equilibrium constant value is, [tex]K_c=225[/tex]

From the above we conclude that, the [tex]Q<K[/tex] that means reactant > product. So, the reaction is product favored that means reaction must shift to the product or right to be in equilibrium.

Now we have to calculate the concentration of [tex]H_2[/tex] at equilibrium.

The given balanced chemical reaction is,

                     [tex]CH_4(g)+2H_2S(g)\rightarrow CS_2(g)+4H_2(g)[/tex]

Initial conc.   2.5            5               2.5          5

At eqm.       (2.5-x)       (5-2x)       (2.5+x)    (5+4x)

The concentration of [tex]CH_4[/tex] at equilibrium = 2.0 M

As we know that, at equilibrium

(2.5-x) = 2.0 M

x = 0.5 M

The concentration of [tex]H_2[/tex] at equilibrium = (5+4x) = 5 + 4(0.5) = 7.0 M

Therefore, the equilibrium concentration of [tex]H_2[/tex] is 7.0 M

Final answer:

The reaction CH4 (g) + 2 H2S (g) ⇋ CS2 (g) + 4 H2 (g) will proceed to the left to reach equilibrium since the reaction quotient Qc is greater than the given Kc. The equilibrium concentration of H2 when the concentration of methane, CH4, at equilibrium is 2.0 M is 7.0 M.

Explanation:

To determine in which direction the reaction CH4 (g) + 2 H2S (g) ⇄ CS2 (g) + 4 H2 (g) will proceed to reach equilibrium, we need to calculate the reaction quotient Qc and compare it with the equilibrium constant Kc. The provided Kc is 225. The initial concentrations are the moles of each substance divided by the volume of the container. Given initial amounts: CH4 = 1.0 mol, H2S = 2.0 mol, CS2 = 1.0 mol, and H2 = 2.0 mol in a 0.4 L container, at 960°C.

To find the concentrations, we divide the moles by the volume (in liters): [CH4] = 1.0 mol/0.4 L = 2.5 M, [H2S] = 2.0 mol/0.4 L = 5.0 M, [CS2] = 1.0 mol/0.4 L = 2.5 M, and [H2] = 2.0 mol/0.4 L = 5.0 M. The reaction quotient Qc is calculated using the expression Qc = [CS2][H2]⁴ / ([CH4][H2S]²). Plugging in the initial concentrations, we get Qc = (2.5 × 5.0⁴) / (2.5 × 5.0²).
As calculated, Qc turns out to be higher than the given Kc, indicating the reaction will shift to the left to reach equilibrium.

When it comes to the equilibrium concentration of H2, the student provided that the concentration of methane (CH4) at equilibrium is 2.0 M. The ratio of H2 to CH4 in the balanced chemical equation is 4:1, so for every 1 M of CH4 that reacts, 4 M of H2 is produced. Thus, if CH4 decreases to 2.0 M from 2.5 M (a decrease of 0.5 M), then H2 would have to increase by 4 times that amount (2.0 M). The equilibrium concentration of H2 would be the initial concentration plus this change: H2(eq) = 5.0 M + 2.0 M = 7.0 M.

Answer:

1384 kJ/mol

Explanation:

The heat absorbed by the calorimeter is equal to the heat released due to the combustion of the organic compound. C is the total heat capacity of the calorimeter and Δt is the change in temperature from intial to final:

Q = CΔt = (3576 J°C⁻¹)(30.589°C - 25.000°C) = 19986.264 J

Extra significant figures are kept to avoid round-off errors.

We then calculate the moles of the organic compound:

(0.6654 g)(mol/46.07) = 0.0144432 mol

We then calculate the heat released per mole and convert to the proper units. (The conversion between kJ and J is infinitely precise and is not involved in the consideration of significant figures)

(19986.264 J)(1kJ/1000J) / (0.0144432 mol) = 1384 kJ/mol

Answer:

[tex]n=2.20x10^{3} mmol[/tex]

Explanation:

Hello,

To compute such moles, one must identify that the concentration unit in this problem is molarity, which is defined by:

[tex]M=\frac{n}{V}[/tex]

Since we are inquired to compute the moles, we solve for them as follows:

[tex]n=M*V=11.0mol/L*200.0*\frac{1L}{1000mL}\\n=2.2mol[/tex]

Finally, this value in millimoles turns out into:

[tex]n=2.2mol*\frac{1000mmol}{1mol}\\n=2.20x10^{3} mmol[/tex]

Best regards.

Answer : The unit of (D) in metric system is [tex]m^2/s[/tex]

Explanation :

The given expression is:

[tex](dm/dt)=(-2\pi)\times (d)\times (D)\times (\frac{M}{RT})\times P[/tex]

where,

m = mass

t = time

d = diameter

D = diffusion coefficient

M = molar mass

R = universal gas constant

T = temperature

P = partial pressure

In metric system,

The unit of mass is, kg

The unit of time is, s

The unit of diameter is, m

The unit of molar mass is, kg/mol

The unit of universal gas constant is, [tex]Nm/^oC.mol[/tex]

The unit of temperature is, [tex]^oC[/tex]

The unit of partial pressure is, [tex]N/m^2[/tex]

The unit of diffusion coefficient will be:

[tex]D=\frac{(dm/dt)}{(-2\pi)\times (d)\times (\frac{M}{RT})\times P}[/tex]

or,

[tex]D=\frac{(dm)\times (R)\times (T)}{2\pi \times (d)\times (dt)\times (M)\times (P)}[/tex]

[tex]D=\frac{(dm)\times (R)\times (T)}{(d)\times (dt)\times (M)\times (P)}[/tex]

Now put all the unit in this expression, we get:

[tex]D=\frac{(kg)\times (Nm/^oC.mol)\times (^oC)}{(m)\times (s)\times (kg/mol)\times (N/m^2)}[/tex]

[tex]D=m^2/s[/tex]

Therefore, the unit of (D) in metric system is [tex]m^2/s[/tex]

Answer: The volume of 0.10 M NaOH required to neutralize 30 ml of 0.10 M HCl is, 30 ml.

Explanation:

According to the dilution law,

[tex]M_1V_1=M_2V_2[/tex]

where,

[tex]M_1[/tex] = molarity of stock [tex]HCl[/tex] solution = 12.0 M

[tex]V_1[/tex] = volume of stock [tex]HCl[/tex]solution = 5.0 ml

[tex]M_2[/tex] = molarity of dilute [tex]HCl[/tex] solution = 0.2 M

[tex]V_2[/tex] = volume of  dilute [tex]HCl[/tex]  solution = ?

Putting in the values we get:

[tex](12.0M)\times 5.0=0.2\times V_2[/tex]

[tex]V_2=300ml[/tex]

Therefore, 300 ml of 0.2M HCl can be made from 5.0mL of a 12.0M [tex]Hcl[/tex] solution.

Answer: The mass of air present in the room is 37.068 kg

Explanation :  Given,

Length of the room = 10.0 ft

Breadth of the room = 11.0 ft

Height of the room = 10.0 ft

To calculate the volume of the room by using the formula of volume of cuboid, we use the equation:

[tex]V=lbh[/tex]

where,

V = volume of the room

l = length of the room

b = breadth of the room

h = height of of the room

Putting values in above equation, we get:

[tex]V=10.0ft\times 11.0ft\times 10.0ft=1100ft^3=31148.53L[/tex]

Conversion used : [tex]1ft^3=28.3168L[/tex]

Now we have to calculate the mass of air in the room.

[tex]Density=\frac{Mass}{Volume}[/tex]

[tex]1.19g/L=\frac{Mass}{31148.53L}[/tex]

[tex]Mass=37066.7507g=37.068kg[/tex]

Conversion used : (1 kg = 1000 g)

Therefore, the mass of air present in the room is 37.068 kg

Answer: There are 37 kg of air in the room.

Explanation:

To calculate the volume of cuboid (room), we use the equation:

[tex]V=lbh[/tex]

where,

V = volume of cuboid

l = length of room = 11 ft

b = breadth of room =  10 ft

h = height of room= 10 ft

Putting values in above equation, we get:

[tex]V=10\times 11\times 10=1100ft^3=1100\times 28.3L=31130L[/tex]  (Conversion factor: [tex]1ft^3=28.3L[/tex]

To calculate mass of a substance, we use the equation:

[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]

We are given:

Density of air = 1.19 g/L

Volume of air = volume of room =  31130 L

Putting values in above equation, we get:

[tex]1.19g/L=\frac{\text{Mass of air}}{31130L}\\\\\text{Mass of air}=37000g=37.0kg[/tex]    (1kg=1000g)

Hence, the mass of air is 37 kg.

Answer: The mass of hydrogen gas and nitrogen gas produced is 0.45 g and 3.15 g respectively.

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]      .....(1)

Given mass of water = 4.05 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of water}=\frac{4.05g}{18g/mol}=0.225mol[/tex]

For the given chemical reaction:

[tex]2H_2O\rightarrow 2H_2+O_2[/tex]

By Stoichiometry of the reaction:

2 moles of water is producing 2 moles of hydrogen gas

So, 0.225 moles of water will produce = [tex]\frac{2}{2}\times 0.225=0.225mol[/tex] of hydrogen gas.

Now, calculating the mass of hydrogen gas by using equation 1, we get:

Moles of hydrogen gas = 0.225 mol

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:

[tex]0.225mol=\frac{\text{Mass of hydrogen gas}}{2g/mol}\\\\\text{Mass of hydrogen gas}=0.45g[/tex]

By Stoichiometry of the reaction:

2 moles of water is producing 1 mole of nitrogen gas

So, 0.225 moles of water will produce = [tex]\frac{1}{2}\times 0.225=0.1125mol[/tex] of nitrogen gas.

Now, calculating the mass of nitrogen gas by using equation 1, we get:

Moles of nitrogen gas = 0.1125 mol

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

[tex]0.1125mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=3.15g[/tex]

Hence, the mass of hydrogen gas and nitrogen gas produced is 0.45 g and 3.15 g respectively.

Answer:

See explanation

Explanation:

Sea water = a mixture of different substances in water. If the water evaporates, some of those substances remain as salts. It's not a compound because it's a mixture of compounds and substances. An example of a compound is water ( because it has hydrogen atoms and an oxygen atom)

Calcium = Calcium is found on the periodic table so it cannot be a compound or mixture. It's a metallic element because we can find it on the left side of the periodic table, in group 2.  It's an alkaline earth metal, what makes that calcium is a reactive metal.

Argon = Argon is found on the periodic table, so it can't be a mixture or compound. It is a non-metallic element. We can find it in group 18 on the periodic table. Argon is a noble gas, so non-metallic.

Water = a compound because its only made of 2 atoms : oxygen and hydrogen.  Reasons why water is a compound and not a mixture are:

 - The ormation of a compound is a chemical change which is followed by absorption of energy or evolution of energy, in case of water, electricity is required.

- Mixtures can be separated by physical separation techniques ,Water can not be separated into it its elements by physical separation techniques. But by the absorption of chemical energy.

Air = Mixture because it can be separated into different atoms, molecules,..  like oxygen, nitrogen etc. by the physical separation techniques (fractional distillation).

Carbon Monoxide = Carbon Monoxide is not found on the periodic table so it cannot be an element. It's made of 2 elements, this means caron monoxide is a compound. It's not a mixture since the elements cannot be separated by physical separating techniques.

Iron = Iron is found on the periodic table so it cannot be a compound or mixture. It's a metallic element because:

-High melting point

-  Some metals form a dull oxide layer, this explains the shiny luster surface

- Electrical conductivity and thermal conductivity

Sodium chloride = NaCl cannot be found on the periodic table, so it isn't an element. It's a compound because it only has 2 atoms (elements). Those elements cannot be seperated by physical separating techniques, but would require electricity. So it's a compound.

Diamond = is a solid form of the element Carbon.  It's an allotrope of carbon. They have the same physical state but in distinct form. Technically diamond is a non-metallic element. Since it's seen as carbon.

Brass = Brass is a mixture of the elements of copper and zinc. Those elements can be separated by physical techniques.

Copper = Copper is found on the periodic table so it cannot be a compound or mixture. It's a metallic element because:

-High melting point

-  Some metals form a dull oxide layer, this explains the shiny luster surface

- Electrical conductivity and thermal conductivity

Dilute sulphuric acid = This is a mixture. Sulphuric acid is a compound but to dilute it's added in water, what is another compound. So it's a mixture of different compounds.

Sulphur = Can be found in the periodic table so it cannot be a mixture or compound. It's part of the metalloids, therefore, it can be concluded that sulfur is a non-metal. It belongs to the oxygen family.

Oil = Oil is a mixture of hydrocarbon compounds which varies in lengths.

Nitrogen = Nitrogen is found on the periodic table, so it can't be a mixture or compound. It can be found as a gas so it is a non-metallic element.

Ammonia = a compound of nitrogen and hydrogen with the formula NH3. Those elements cannot be separated with physical separating techniques.

Answer:

NCL3

Explanation:

NCL3

- nitrogen trichloride.

The chemical formula for a compound containing one nitrogen atom for every three chlorine atoms is NCl₃.

The chemical formula NCl₃ represents a compound consisting of one nitrogen atom and three chlorine atoms. In this formula, "N" represents nitrogen, and "Cl" represents chlorine. The subscript "3" indicates that there are three chlorine atoms bonded to each nitrogen atom in the compound.

This arrangement reflects the stoichiometry of the compound, specifying the relative ratios of the elements. Chemical formulas provide a concise way to describe the composition of substances, allowing scientists to understand the types and quantities of atoms present in a compound.

NCl₃, nitrogen trichloride, is a covalent compound with notable chemical properties, and its formula encapsulates the balanced proportion of nitrogen and chlorine within its molecular structure.

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Answer:

a) n-butane has a higher boiling point

b) 1-butanol has a higher boiling

Explanation:

Given the molecules, propane (C3H8) and n-butane (C4H10), n-butane has a higher boiling point mainly due to greater molar mass and longer chain (more interactions between each molecule).

Given the molecules, diethyl ether (CH3CH2OCH2CH3) and 1-butanol (CH3CH2CH2CH2OH), 1-butanol has a higher boiling point mainly due to hydrogen bonding forces. The OH group is more electronegative than ether group.

The larger and more complex n-butane has a higher boiling point due to London dispersion forces, while 1-butanol has a higher boiling point due to hydrogen bonding.

Given the molecules, propane (C3H8) and n-butane (C4H10), n-butane has a higher boiling point mainly due to its larger size and increased London dispersion forces, a type of intermolecular force.

London dispersion forces increase with increased molecular size and shape because larger and more complex molecules tend to have more electrons, leading to larger temporary fluctuations in charge distribution.

Given the molecules diethyl ether (CH3CH2OCH2CH3) and 1-butanol (CH3CH2CH2CH2OH), 1-butanol has the higher boiling point mainly due to the presence of hydrogen bonding, another type of intermolecular force. Hydrogen bonding is generally stronger than other types of intermolecular forces, leading to higher boiling points.

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