Which of the following statistical metrics is an appropriate measure of the variability of data around the mean? standard deviation O mode width O median

Answer : All of the above are valid expressions of the reaction rate.

Explanation :

The given rate of reaction is,

[tex]4NH_3+7O_2\rightarrow 4NO_2+6H_2O[/tex]

The expression for rate of reaction for the reactant :

[tex]\text{Rate of disappearance of }NH_3=-\frac{1}{4}\times \frac{d[NH_3]}{dt}[/tex]

[tex]\text{Rate of disappearance of }O_2=-\frac{1}{7}\times \frac{d[O_2]}{dt}[/tex]

The expression for rate of reaction for the product :

[tex]\text{Rate of formation of }NO_2=+\frac{1}{4}\times \frac{d[NO_2]}{dt}[/tex]

[tex]\text{Rate of formation of }H_2O=+\frac{1}{6}\times \frac{d[H_2O]}{dt}[/tex]

From this we conclude that, all the options are correct.

Based on the equation of the reaction, all of the given expressions are valid expressions of the reaction rate.

The rate of a chemical reaction is the rate at which reactants are used up or the rate at which products are formed.

The rate of a reaction is obtained from the balanced equation of the reaction.

The equation of the reaction is given below:

Rate of disappearance of NH3 = - 1/4 × Δ[NH3]/Δt

Rate of disappearance of O2 = - 1/7 × Δ[O2]/Δt

Rate of formation of NO2 = 1/4 × Δ[NO2]/Δt

Rate of formation of H2O = 1/6 × Δ[H2O]/Δt

Therefore, all of the above are valid expressions of the reaction rate.

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Answer:

1-condensation of ethanol

2-evaporation of ethanol

Explanation:

According to the question ,

Answer:

The mean velocity is 13 ft/s.

The Reynolds number is 88,583 and it is dimensionless.

Explanation:

We have water flowing in a pipe of 1.05 in diameter.

The density is ρ=62.3 lb/ft and the viscosity is 1.2 cP.

The mean velocity can be calculated as

[tex]u=\frac{Q}{A}=\frac{Q}{\pi*D^2/4}=\frac{35gpm }{3.14*(1.05in)^2/4}\\\\  u=\frac{35}{0.865}*\frac{gal}{min}\frac{1}{in^2}*\frac{231in^3}{1gal}*\frac{1}{60s} \\\\    u=156\,in/s=13\,ft/s[/tex]

The Reynolds number now can be calculated for this flow as

[tex]Re=\frac{\rho*u*D}{\mu}[/tex]

being ρ: density, u: mean velocity of the fluid, D: internal diameter of the pipe and μ the dynamic viscosity.

To simplify the calculation, we can first make all the variables have coherent units.

Viscosity

[tex]\mu=1.2cP=\frac{1.2}{100}\frac{g}{cm*s}*\frac{1lb}{453.6g}*\frac{30.48cm}{1ft}= 0.0008\frac{lb}{ft*s}[/tex]

Diameter

[tex]D=1.05in*(\frac{1ft}{12in} )=0.0875ft[/tex]

Then the Reynolds number is

[tex]Re=\frac{\rho*u*D}{\mu}\\\\Re=62.3\frac{lb}{ft^3}*13\frac{ft}{s} *0.0875ft*\frac{1}{0.0008}*\frac{ft*s}{lb}\\\\Re=88,583[/tex]

The Reynolds number (Re) and mean velocity (u) for water flowing through a pipe can be calculated using the flow rate, pipe dimensions, fluid density, and viscosity. The Reynolds number is a dimensionless quantity that helps predict the flow pattern in the pipe.

The question relates to the calculation of the Reynolds number and the mean velocity (u) for a given flow rate of water through a pipe. First, to find the mean velocity u, the flow rate needs to be converted to cubic feet per second (ft³/s) and then divided by the cross-sectional area of the pipe. The Reynolds number Re is a dimensionless number used to predict flow patterns in different fluid flow situations. It is calculated using the formula Re = ρuD/μ, where ρ is the fluid density, u is the mean velocity, D is the pipe diameter, and μ is the dynamic viscosity of the fluid.

To proceed with the calculation, the given flow rate of 35 gallons per minute (gpm) is converted to cubic feet per second, the pipe's internal diameter is converted to feet, the density of water (62.3 lb/ft³) is used, and the viscosity (1.2 cP) is converted to lb/(ft·s). The mean velocity u is then calculated, and subsequently, the Reynolds number Re is determined.

The units of the Reynolds number are indeed unitless, as demonstrated by the cancellation of units in its definition, ensuring it is a dimensionless quantity.

Answer:

0.1933 M

Explanation:

Data provided in the question:

Number of Moles of Na₂SO₄ initially taken = 0.145 mol

Volume of solution, V = 750 mL = 0.750 L

Now, the molarity is given as:

Molarity = [tex]\frac{\textup{moles of solute}}{\textup{volume of solution in liter}}[/tex]

or

Molarity = [tex]\frac{0.145}{0.750}[/tex]

or

Molarity = 0.1933 M

The molarity of the solution is 0.1933 M.

To calculate the molarity of a solution, you need to know the moles of solute and the volume of solution. In this case, we have 0.145 mol of Na2SO4 and a volume of 750 mL. First, convert the volume to liters by dividing by 1000: 750 mL = 0.75 L. Now, calculate the molarity using the formula: Molarity (M) = moles of solute / volume of solution in liters.

So, Molarity = 0.145 mol / 0.75 L = 0.1933 M

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Answer:

The answer is the option "A"

Explanation:

The reaction rate is the amount of substance formed or transformed per unit of time. The catalytic action allows to increase the reaction rate or the selectivity. The units are moles / time (mol / s)

mol = moles

s = seconds

So the correct answer is "A"

Light travels a distance of 3.69 meters within the given time frame of 12.3 nanoseconds.

Speed is the measure of how quickly an object changes its position concerning a specified frame of reference. It's a scalar quantity, indicating both magnitude and direction.

Common units for speed include meters per second (m/s) or kilometers per hour (km/h). This fundamental concept plays a pivotal role in various domains like travel, sports, and transportation.

If light travels at a constant speed of [tex]3.0 * 10^8 m/s[/tex] and continues for 12.3 nanoseconds, the distance covered can be calculated:

Distance = Speed × Time

Distance = [tex]3.0 * 10^8 m/s* 12.3 * 10^{-9} s[/tex]

Distance = 3.69 meters

In this instance, light travels a distance of 3.69 meters within the given time frame of 12.3 nanoseconds.

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Answer : The time taken will be 25 min.

Explanation :

First we have to determine the amount of fluid.

As, 1 mL contains 30 gtt

So, 500 mL contains [tex]\frac{500mL}{1mL}\times 30gtt=1500gtt[/tex]

Now we have to determine the time taken.

As, 60 gtt takes time = 1 min

So, 1500 gtt takes time = [tex]\frac{1500gtt}{60gtt}\times 1min=25min[/tex]

Therefore, the time taken will be 25 min.

Final answer:

To administer 500 mL of fluid with a drop factor of 30 gtt/mL at a drip rate of 60 gtt/min, the calculation reveals it will take 250 minutes.

Explanation:

To calculate the time required to administer exactly 500 mL of fluid through an IV with a drop factor of 30 gtt/mL at a drip rate of 60 gtt/min, we must first understand the given terms. The drop factor (gtt/mL) is a measure indicating the number of drops (gtt) that make up 1 mL of fluid. The drip rate (gtt/min) specifies how many drops of fluid are administered per minute.

First, we calculate the total number of drops in 500 mL of fluid, using the drop factor:

Next, to find out how long it will take to administer these 15000 drops at a rate of 60 drops per minute, we divide the total number of drops by the drip rate:

Therefore, it will take 250 minutes to administer exactly 500 mL of fluid through the IV at the specified conditions.

Answer: The cell potential of the cell is 0.31 V

Explanation:

We know that:

[tex]E^o_{(Br_2/2Br^-)}=1.07V\\E^o_{(MnO_4^-/Mn^{2+})}=1.51V[/tex]

The substance having highest positive [tex]E^o[/tex] potential will always get reduced and will undergo reduction reaction. Here, [tex]MnO_4^-[/tex] will undergo reduction reaction will get reduced. And, bromine will get oxidized.

Oxidation half reaction: [tex]2Br^-(aq.)+2e^-\rightarrow Br_2(l);E^o_{(Br_2^/2Br^-)}=1.07V[/tex]      ( ×  5)

Reduction half reaction: [tex]MnO_4^-(aq.)+8H^+(aq.)+5e^-\rightarrow Mn^{2+}(aq.)+4H_2O(l);E^o_{(MnO_4^-/Mn^{2+})}=1.51V[/tex]  ( ×  2)

The net reaction follows:

[tex]2MnO_4^-(aq.)+10Br^-(aq.)+16H^+(aq.)\rightarrow 2Mn^{2+}(aq.)+5Br_2(l)+8H_2O(l)[/tex]

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:

[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex]

Putting values in above equation, we get:

[tex]E^o_{cell}=1.51-(1.07)=0.44V[/tex]

To calculate the EMF of the cell, we use the Nernst equation, which is:

[tex]E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]^2}{[MnO_4^{-}]^2\times [Br^-]^{10}\times [H^+]^{16}}[/tex]

where,

[tex]E_{cell}[/tex] = electrode potential of the cell = ?V

[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = 0.44 V

n = number of electrons exchanged = 10

[tex][H^{+}]=1M[/tex]

[tex][Mn^{2+}]=0.15M[/tex]

[tex][MnO_4^{-}]=0.01M[/tex]

[tex][Br^{-}]=0.01M[/tex]

Putting values in above equation, we get:

[tex]E_{cell}=0.44-\frac{0.059}{10}\times \log(\frac{(0.15)^2}{(0.01)^2\times (0.01)^{10}\times (1)^{16}})\\\\E_{cell}=0.31V[/tex]

Hence, the cell potential of the cell is 0.31 V

Answer: Option (a) is the correct answer.

Explanation:

As the given situation shows that water is being added to water. This will liberate heat into the surround as the reaction will be exothermic in nature.

It is known that chemical reactions in which heat is released are known as exothermic reactions.

Hence, when more water is added to this tank the temperature of the system will increase due to release of heat.

Thus, we can conclude that if more water is added to this tank the temperature of the system will increase.

Answer:

(a) 17,178 mg/m3

(b) 11,625 mg/m3

Explanation:

The concentration of CO in mg/m3 can be calculated as

[tex]C (mg/m3) =(P/RT)*MW*C_{ppm}[/tex]

For standard conditions (1 atm and 25°C), P/RT is 0.0409.

Concentration of 1.5% percent by volume of CO is equivalent to 1.5*10,000 ppm= 15,000 ppm CO.

The molecular weigth of CO is 28 g/mol.

(1) For 25°C and 1 atm conditions

[tex]C=(P/RT)*MW*C_{ppm}\\\\C=0.0409*28*15,000=17,178[/tex]

(b) For 200°C and 1.1 atm,

[tex]P/RT=0.0409*(P/P_{std})*(T_{std}/T)\\P/RT=0.0409*(1.1atm/1atm)*(273+15K/273+200K)=0.0277[/tex]

Then the concentration in mg/m3 is

[tex]C=(P/RT)*MW*C_{ppm}\\\\C=0.0277*28*15,000=11,625[/tex]

Answer:

The final pH is 7

Explanation:

KOH + HCl ------> KCl + H2O

KOH ----> K+  + OH-

0,3 M        0,3 M  +  0,3 M

HCl -----> H+  +  Cl-

0,2 M         0,2M   +  0,2M

In both cases they are strong acid and base because they dissociate completely; The initial molarity is the same, at the end of the dissociation.

As we have an acid next to a strong base, the final solution will have a neutralization reaction where the OH- of the base react with the H + of the acid to give water. The excess of OH- or H + defines whether the solution will have an acid or alkaline pH.

Molarity = moles / volume (L)

Moles of OH- = 0,3 M x 0,020 L = 0,006 moles

(remember volume in L, so 20mL = 0,020 L)

Moles of H+ = 0,2 M x 0,030 L = 0,006 moles

As we have the same quantity for OH- and H+, the solution will have neutral pH.

Answer:

Explanation:

The equilibrium that takes place is:

H₂PO₄⁻ ↔ HPO₄⁻² + H⁺    pka= 7.21 (we know this from literature)

To solve this problem we use the Henderson–Hasselbalch (H-H) equation:

pH = pka + [tex]log\frac{[A^{-} ]}{[HA]}[/tex]

In this case [A⁻] is [HPO₄⁻²], [HA] is [H₂PO₄⁻], pH=7.0, and pka = 7.21

If we use put data in the H-H equation, and solve for [HPO₄⁻²], we're left with:

[tex]7.0=7.21+log\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]}\\ -0.21=log\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]}\\\\10^{-0.21} =\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]}\\0.616 * [H2PO4^{-}] = [HPO4^{-2}][/tex]

From the problem, we know that [HPO₄⁻²] + [H₂PO₄⁻] = 0.1 M

We replace the value of [HPO₄⁻²] in this equation:

0.616 * [H₂PO₄⁻] + [H₂PO₄⁻] = 0.1 M

1.616 * [H₂PO₄⁻] = 0.1 M

[H₂PO₄⁻] = 0.0619 M

With the value of [H₂PO₄⁻]  we can calculate [HPO₄⁻²]:

[HPO₄⁻²] + 0.0619 M = 0.1 M

[HPO₄⁻²] = 0.0381 M

With the concentrations, the volume and the molecular weights, we can calculate the masses:

Answer:

10,93% of a carrot is carbohydrate.

Explanation:

In order to solve this you just have to create a rule of three where 64g is the 100% and you want to calculate how much is 7g in percentage:

[tex]\frac{64}{100}= \frac{7}{x}\\ x=\frac{7*100}{64}\\ x=10,9375[/tex]

So there are 10.9375 % carbohydrates in a piece of carrot.

The percent, by mass, of carbohydrate in a medium-sized carrot is approximately 10.9%.

To find the percent, by mass, of carbohydrate in a medium-sized carrot, we can use the following formula:

Percent mass of carbohydrate = (mass of carbohydrate / total mass) × 100

In this case, the mass of carbohydrate is given as 7.0 g and the total mass of the carrot is 64 g. Plugging these values into the formula:

Percent mass of carbohydrate = (7.0 g / 64 g) × 100 = 10.9%

Therefore, approximately 10.9% of the carrot's mass is carbohydrate.

Answer:

No, it doesn't.

Explanation:

To convert ppb to mg/m³ we first need to convert to ppm, by just divide the amount by 1,000, so the concentration in the sample is 4.8 ppm.

mg/m³ = (ppm x molar mass)/molar volume

Using the molar mass in gram and the molar volume in liters, multiplying by the parts per million, we will get the concentration in mg/m³.

Molar mass of C = 12 g/mol; molar mass of Cl = 35.5 g/mol

Molar mass of CCl4 = 12 + 4x35.5 = 154 g/mol

Assuming, 25ºC and 1 atm, the molar volume of an ideal gas is 24.45 L, so:

mg/m³ = (4.8 x 154)/24.45

mg/m³ = 30.2

Which is higher than the limit of 12.6 mg/m³

Answer:

Sample is greater than the standard (32.96 mg/m³ > 12.6 mg/m³)

Explanation:

Please look at the solution in the attached Word file

Answer:

x = 0.01503 = 1.503 × 10⁻² , four significant figures

Explanation:

Significant figures refer to the digits of a number that have meaning and contribute to the precision of the given number.

Since the given equation is:

[tex]\frac{1}{x}=\:66.54[/tex]

Cross-multiplying to solve for the value of x:

[tex]x \times 66.54 = 1[/tex]

⇒ [tex]x = \frac{1}{66.54}[/tex]

⇒ x = 0.01503 = 1.503 × 10⁻² , has four significant figures.

Answer:

36.5 lbs weight of child able to take 413.91 mg/ day dose

Amoxicillin should be stored in temperature 0°C to 20°C, therefore it must be stores  in refrigerator as it provided temperature only between 0°C to 5°C.

Explanation:

Given data:

Dosage of Amoxicillin as prescribed is  25 mg/kg-day

Weight of the Child weight  = 36.5 lbs

As We know  1 lbs = 0.4536 kg

therefore, the weight of Child is  36.5\times 0.4536 kg = 16.5564 kg

From the information about  dosage,

1 kg of body takes = 25 mg/day

so, for 16.556 kg body [tex]16.5564\  kg \ body\  takes = 25\times 16.5564 = 413.91 mg/day[/tex]

Therefore 36.5 lbs weight of child able to take 413.91 mg/ day dose

Amoxicillin should be stored in temperature 0°C to 20°C, therefore it must be stored  in refrigerator as it provided temperature only between 0°C to 5°C.

Final answer:

To calculate the amoxicillin dosage for a child, convert the child's weight to kilograms, multiply by the prescribed mg/kg dosage, then divide by the tablet strength. Amoxicillin should be stored in a refrigerator, not a freezer.

Explanation:

The question relates to the prescription and administration of amoxicillin dosage based on a patient's weight. To determine the number of tablets to administer, you will need to convert the child's weight from pounds to kilograms (1 pound = 0.453592 kg), multiply the child's weight in kilograms by the prescribed dosage per kilogram and then divide the total dosage by the amount of medicine per tablet.

Using the information provided, if a doctor prescribes amoxicillin at 30mg/kg to a child weighing 73.5 lbs, first convert the weight: 73.5 lbs × 0.453592 = 33.3 kg approximately. Next, calculate the total dosage: 33.3 kg × 30 mg/kg = 999 mg. Since amoxicillin is available in 500 mg tablets, divide the total dosage by the tablet strength: 999 mg / 500 mg/tablet = about 2 tablets (always round to the nearest whole number when it involves whole tablets).

Amoxicillin should be stored between 0 °C and 20 °C, which is typically within the temperature range of a refrigerator, not a freezer. Therefore, amoxicillin should be stored in the refrigerator to maintain its efficacy.

Explanation:

Actual value = 102.0 mg/dL

Calculation of absolute error and relative error for the measured value, 98.4 mg/dL

Absolute error = |Actual value - measured value|

                         = | 102.0 - 98.4 | = 3.6

Relative error = [tex]\frac{Absolute\ error}{Actual\ value}[/tex]

                       = [tex]\frac{3.6}{102.0} = 0.03529[/tex]

Calculation of absolute error and relative error for the measured value, 104.3 mg/dL

Absolute error = |Actual value - measured value|

                         = | 102.0 - 104.3 | = 2.3

Relative error = [tex]\frac{Absolute\ error}{Actual\ value}[/tex]

                       = [tex]\frac{2.3}{102.0} = 0.02255[/tex]

Calculation of absolute error and relative error for the measured value, 97.4 mg/dL

Absolute error = |Actual value - measured value|

                         = | 102.0 - 97.4 | = 4.6

Relative error = [tex]\frac{Absolute\ error}{Actual\ value}[/tex]

                       = [tex]\frac{4.6}{102.0} = 0.04509[/tex]

Calculation of absolute error and relative error for the measured value, 106.7 mg/dL

Absolute error = |Actual value - measured value|

                         = | 102.0 - 106.7 | = 4.7

Relative error = [tex]\frac{Absolute\ error}{Actual\ value}[/tex]

                       = [tex]\frac{4.7}{102.0} = 0.04607[/tex]

Calculation of absolute error and relative error for the measured value, 93.0 mg/dL

Absolute error = |Actual value - measured value|

                         = | 102.0 - 93 | = 9

Relative error = [tex]\frac{Absolute\ error}{Actual\ value}[/tex]

                       = [tex]\frac{9}{102.0} = 0.08823[/tex]

Answer:

a) k = [tex]\frac{[C][D]}{[A][B]}[/tex]

b) A value of 45000 means that tendency of your reaction is to have more products in a ratio of 45000:1

c) No, the products are more stable than reactants.

Explanation:

a) For a reaction:

A + B → C + D

The equlibrium constant (k) is:

k = [tex]\frac{[C][D]}{[A][B]}[/tex]

Where [x] is the molar concentration of x. Always the expression of equilibrium constant is molar concentration of products over molar concentration of reagents.

b) Having in mind the expression of equilibrium constant when k>1 the concentration of products is higher than concentration of reagents. Thus, when k<1 concentration of reagents is higher than concentration of products.

A value of 45000 means that tendency of your reaction is to have more products in a ratio of 45000:1

c) Again, a value of 45000 means that tendency of your reactants is react to produce products. Thus, the products are more stable than reactants.

I hope it helps!

Answer: An aqueous solution will contains both hydrogen ion and hydroxide ion.An here we are talking about free hydrogen ions, which means the description will be related to the pH. If the ratio of ion are equal then the pH will be 7 i.e. neutral. But if there are free hydrogen atoms in the solution we have, the pH of solution will be less than 7 and the solution will be described ad acidic. So the inaccurate description for a solution containing free hydrogen ion is that it is basic in nature.

Answer:

Yes, there is a loss of 8350 J of energy in the form of heat

Explanation:

The principle of energy conservation is described mathematically

as the energy conservation equation as follows:

                            ΔK+ΔU=Q+W        

where:

Analyzing each term of the equation:

   [tex]ΔK=\frac{1}{2}mv_{f} ^{2}  - \x]frac{1}{2}mv_{i} ^{2}=\frac{1}{2}*100*15^{2} - \frac{1}{2}*100*0^{2}=11250 J[/tex]

ΔU=[tex]mgh_{f}-mgh_{i}=100*9,8*0-100*9,8*20=-19600[J][/tex]

Q=?

W=0 [J]

Replacing in the main equation:

11250-19600=Q+0

Q= -8350 [J]

So, the answer is YES, there is a loss of 8350 J of  energy in the form of heat.

Answer:

a) HCN  - hybridization sp

b) C(CH₃)₄ - hybridization sp³

c) H₃O⁺  - hybridization sp³

d) - CH₃ - hybridization sp³

Explanation:

Hybridization occurs to allow an atom to make more covalent bonds than the original electronic distribution would allow or to allocate ligands in an energetically stable geometry.

Carbon can have thre hybridization states: sp³ , sp² and sp.

Oxygen usualluy has an sp³ hybridization.

In order to determine the hybridization, we need to consider the number of atoms attached to the central atom and the number of lone pairs.

The figure attached shows the species and the hybridization of their central atoms.

Answer:

True

Explanation:

Carlos Santana

Carlos Santana is a famous american - mexico musician and even a guitarist  with some unique music categories like , jazz , Latin , salsa . He started off in late late 1960's with the brand name Santana , which consists of a fusion of rock and roll and even Latin American jazz . And then after his career grows and he became famous .

Hence , the statement is correct .

Answer:

Binding affinity measures the strength of the interaction between a molecule to its ligand; it is expressed in terms of the equilibrium dissociation constant; and the higher value of this constant, the more weaker the binding between the molecule and the ligand is. On the other hand, small constans means that the interaction is tight. So "C" binds most tightly to the enzyme and "D" binds least tightly.

Answer: The concentration of water at equilibrium is 0.3677 mol/L

Explanation:

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric coefficients. It is represented by [tex]K_{c}[/tex]

For a general chemical reaction:

[tex]aA+bB\rightarrow cC+dD[/tex]

The [tex]K_{c}[/tex] is written as:

[tex]K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}[/tex]

The chemical equation for the conversion of methane to carbon monoxide and hydrogen gas follows:

[tex]CH_4+H_2O\rightleftharpoons 3H_2+CO[/tex]

The [tex]K_{c}[/tex] is represented as:

[tex]K_{c}=\frac{[H_2]^3[CO]}{[CH_4][H_2O]}[/tex]      ....(1)

To calculate the concentration, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]

We are given:

[tex][CO]=\frac{0.145mol}{0.379L}=0.383mol/L[/tex]

[tex][H_2]=\frac{0.218mol}{0.379L}=0.575mol/L[/tex]

[tex][CH_4]=\frac{0.25mol}{0.379L}=0.660mol/L[/tex]

[tex]K_c=0.30[/tex]

Putting values in equation 1, we get:

[tex]0.30=\frac{(0.575)^3\times 0.383}{0.660\times [H_2O]}[/tex]

[tex][H_2O]=\frac{(0.575)^3\times 0.383}{0.660\times 0.30}=0.3677[/tex]

Hence, the concentration of water at equilibrium is 0.3677 mol/L

Answer: The wavelength of radiation is [tex]4.34\times 10^{21}m[/tex]

Explanation:

The relationship between wavelength of light and frequency of light is given by the equation:

[tex]\nu=\frac{c}{\lambda}[/tex]

where,

[tex]\nu[/tex] = frequency of radiation = [tex]6.912\times 10^{-14}s^{-1}[/tex]

c = speed of light = [tex]3\times 10^8m/s[/tex]

[tex]\lambda[/tex] = wavelength of radiation = ?

Putting values in above equation, we get:

[tex]6.912\times 10^{-14}s^{-1}=\frac{3\times 10^8m/s}{\lambda}\\\\\lambda=4.34\times 10^{21}m[/tex]

Hence, the wavelength of radiation is [tex]4.34\times 10^{21}m[/tex]

Answer:  Final temperature of the gas will be 330 K.

Explanation:

Gay-Lussac's Law: This law states that pressure is directly proportional to the temperature of the gas at constant volume and number of moles.

[tex]P\propto T[/tex]     (At constant volume and number of moles)

[tex]{P_1\times T_1}={P_2\times T_2}[/tex]

where,

[tex]P_1[/tex] = initial pressure of gas   = 1.00 atm

[tex]P_2[/tex] = final pressure of gas  = 1.13 atm

[tex]T_1[/tex] = initial temperature of gas  = [tex]100^0C=(100+273)K=373K[/tex] K

[tex]T_2[/tex] = final temperature of gas  = ?

[tex]{1.00\times 373}={1.13\times T_2}[/tex]

[tex]T_2=330K[/tex]

Therefore, the final temperature of the gas will be 330 K.

Answer:

HCI (aq) + NaOH (aq) --> H2O (l) + NaCl (aq)

Explanation:

A double displacement reaction is a type of reaction where two reactants exchange ions to form two new compounds. Here we can see that both the acid and the base exchanged an ion each to for salt and water, 2 new products.

Answer : The value of the rate constant 'k' for this reaction is [tex]1.9375\times 10^{-3}M^{-2}s^{-1}[/tex]

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

[tex]A+B+C\rightarrow D+E[/tex]

Rate law expression for the reaction:

[tex]\text{Rate}=k[A]^a[B]^b[C]^c[/tex]

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

[tex]1.24\times 10^{-4}=k(0.40)^a(0.40)^b(0.40)^c[/tex] ....(1)

Expression for rate law for second observation:

[tex]3.6\times 10^{-4}=k(0.40)^a(0.40)^b(1.20)^c[/tex] ....(2)

Expression for rate law for third observation:

[tex]4.8\times 10^{-4}=k(0.80)^a(0.40)^b(0.40)^c[/tex] ....(3)

Expression for rate law for fourth observation:

[tex]4.8\times 10^{-4}=k(0.80)^a(0.80)^b(0.40)^c[/tex] ....(4)

Dividing 1 from 2, we get:

[tex]\frac{3.6\times 10^{-4}}{1.24\times 10^{-4}}=\frac{k(0.40)^a(0.40)^b(1.20)^c}{k(0.40)^a(0.40)^b(0.40)^c}\\\\3=3^c\\c=1[/tex]

Dividing 1 from 3, we get:

[tex]\frac{4.8\times 10^{-4}}{1.24\times 10^{-4}}=\frac{k(0.80)^a(0.40)^b(0.40)^c}{k(0.40)^a(0.40)^b(0.40)^c}\\\\4=2^a\\a=2[/tex]

Dividing 3 from 4, we get:

[tex]\frac{4.8\times 10^{-4}}{4.8\times 10^{-4}}=\frac{k(0.80)^a(0.40)^b(0.40)^c}{k(0.80)^a(0.80)^b(0.40)^c}\\\\1=2^b\\b=0[/tex]

Thus, the rate law becomes:

[tex]\text{Rate}=k[A]^2[B]^0[C]^1[/tex]

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

[tex]1.24\times 10^{-4}=k(0.40)^2(0.40)^0(0.40)^1[/tex]

[tex]k=1.9375\times 10^{-3}M^{-2}s^{-1}[/tex]

Hence, the value of the rate constant 'k' for this reaction is [tex]1.9375\times 10^{-3}M^{-2}s^{-1}[/tex]

Answer:

[tex]Mg(s) + HCl(aq) \rightarrow MgCl_2(aq) + H_2(g)[/tex]

[tex]MgO(s) + HCl(aq) \rightarrow MgCl_2(aq) + H_2O(l)[/tex]

Explanation:

When Mg reacts with HCl, magnesium chloride and hydrogen is formed. Mg is an active element and displaces hydrogen from HCl. So, this is a type of single displacement reaction.

[tex]Mg(s) + HCl(aq) \rightarrow MgCl_2(aq) + H_2(g)[/tex]

When magnesium oxide (MgO) reacts with HCl, magnesium chloride and water is formed. This reaction is a type of neutralization reaction. MgO is a water insoluble base and HCl is acid. So. in this reaction, acid reacts with base to form salt [tex]MgCl_2[/tex]

[tex]MgO(s) + HCl(aq) \rightarrow MgCl_2(aq) + H_2O(l)[/tex]

The equation for the reaction between HCl and MgO is 2HCl(aq) + MgO(s) → MgCl2(aq) + H2O(l). The equation for the reaction between HCl and Mg is 2HCl(aq) + Mg(s) → MgCl2(aq) + H2(g). The reactions differ in terms of the reactants involved.

The equation for the reaction between HCl and MgO is:

2HCl(aq) + MgO(s) → MgCl2(aq) + H2O(l)

The equation for the reaction between HCl and Mg is:

2HCl(aq) + Mg(s) → MgCl2(aq) + H2(g)

The reactions differ in terms of the reactants involved. In the reaction with MgO, the reactant is an oxide compound (MgO) while in the reaction with Mg, the reactant is a pure metal (Mg).