Friction factor for fluid flow in pipe does not depend upon the A. pipe length. B. pipe roughness. C. fluid density & viscosity. D. mass flow rate of fluid.

Answer:

a) 7.1 x 10⁻⁵ : 2 significant figures

b) 0.00677 : 3 significant figures  

c) 750 : 2 significant figure  

Explanation:

The significant digits or figures refers to the digits of a given number that carry meaning and also contributes to precision of the given number.

a) 7.1 x 10⁻⁵ = 0.000071 : 2 significant figures, leading zeros are not significant.

b) 0.00677 : 3 significant figures, leading zeros are not significant.          

c) 750 : 2 significant figure, trailing zeros are not significant.        

Answer:

the equilibium changes since Q > Kc, by increasing the volume, therefore, the reaction will try to use some of the excess product and favor the reverse reaction to reach equilibrium.

Explanation:

CO(g) + 3H2(g) ↔ CH4(g) + H2O(g)

∴ Kc = [ H2O ] * [ CH4 ] / [ H2 ]³ * [ CO ] = 3.93.....equilibrium, V = 2.00L

PV = nRT; assuming P,T a standart conditions ( 1 atm, 298 K )

⇒ n / V = P / RT

∴ V = 8.00L

∴ R = 0.082 atm.L/K.mol

⇒ mol CO(g) = 0.327 mol = mol H2O = mol CH4

⇒ mol H2(g) = 0.327 mol CO * ( 3mol H2 / mol CO) = 0.981 mol

⇒ [ CO ] = 0.041 = [ CH4 ] = [ H2O]

⇒ [ H2 ] = 0.123 M

∴ Q =  [ H2O ] * [ CH4 ] / [ H2 ]³ * [ CO ]

⇒ Q = ( 0.041² ) / (( 0.123³ ) * ( 0.041 ))

⇒ Q = 22.03

⇒ we have more product present than we would have in the equilibrium.

Answer:

Q = 62.9

Q > Kc, so the reaction will proceed to the left so that Q takes the value of Kc.

Explanation:

Let's consider the following reaction.

CO(g) + 3 H₂(g) ⇄ CH₄(g) + H₂O(g)

The equilibrium constant (Kc) is:

[tex]Kc= 3.93 =\frac{[CH_{4}].[H_{2}O]}{[CO].[H_{2}]^{3} }[/tex]

If the volume is multiplied by 4 (2.00 L → 8.00 L), the concentrations will be divided by 4. The reaction quotient (Q) is:

[tex]Q=\frac{(0.25[CH_{4}]).(0.25[H_{2}O])}{(0.25[CO]).(0.25[H_{2}])^{3} }=16\frac{[CH_{4}].[H_{2}O]}{[CO].[H_{2}]^{3} }=16Kc=16 \times 3.93 = 62.9[/tex]

Q > Kc, so the reaction will proceed to the left so that Q takes the value of Kc.

Answer:

Monosaccharides are the simplest form of carbohydrates that cannot be hydrolyzed to smaller compounds. Monosaccharides are the basic units of carbohydrates and are also known as simple sugars.  

The monosaccharides are classified on the basis of number of carbon atoms present.

Triose is a type of monosaccharide molecule, which is composed of 3 carbon atoms.

Tetrose is a type of monosaccharide molecule, which is composed of 4 carbon atoms.

Pentose is a type of monosaccharide molecule, which is composed of 5 carbon atoms.

Hexose is a type of monosaccharide molecule, which is composed of 6 carbon atoms.

D-glucose is a hexose sugar and it is the most abundant monosaccharide in the nature.

Answer: 1. The molecular weight of the compound is 222.8 g/mol

2. The probable molecular formula of the solute is [tex]InCl_3[/tex]

Explanation:

Elevation in boiling point is given by:

[tex]\Delta T_b=i\times K_f\times m[/tex]

[tex]\Delta T_b=T_b-T_b^0=(116.3-114.1)^0C=2.2^0C[/tex] = elevation in boiling point

i= vant hoff factor = 1 (for non electrolyte)

[tex]K_b[/tex] = boiling point constant = [tex]9.43^0Ckg/mol[/tex]

m= molality

[tex]\Delta T_b=i\times K_b\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}[/tex]

Weight of solvent (tin chloride)= 50.0 g =0.05 kg

Molar mass of unknown solute = M g/mol

Mass of unknown solute = 2.6 g

[tex]2.2=1\times 9.43\times \frac{2.6g}{M g/mol\times 0.05kg}[/tex]

[tex]M=222.8g/mol[/tex]

The possible formula for the compound would be [tex]InCl_3[/tex] as indium has valency of 3 and chlorine has valency of 1 has molecular mass almost equal to 222.8.

Explanation:

It is given that diameter of the pipe is 10 cm which is also equal to [tex]10 \times 10^{-2}m[/tex].

Velocity of water = 5 m/s

(a)   Formula to calculate volumetric flow rate is as follows.

                       Q = Area of the pipe (A) × Velocity of water (V)

                           = [tex]\frac{\pi}{4} \times 10 \times 10^{-2} \times 5 m^{3}/sec[/tex]

                           = 0.039 [/tex]m^{3}/sec[/tex]

                           = [tex]\frac{0.039 m^{3}/sec \times 60 sec}{1 min}[/tex]

                           = 2.36 [tex]m^{3} min^{-1}[/tex]

Hence, the volumetric flow rate is 2.36 [tex]m^{3} min^{-1}[/tex].

(b)    Formula to calculate mass flow rate is as follows.

                      [tex]Q \times \rho[/tex]

                     = [tex]2.36 m^{3} min^{-1} \times 1000 kg m^{-3}[/tex]

                     = 2356.19 kg/min

Therefore, the mass flow rate is 2356.19 kg/min.

Answer:

It is written differently.

Explanation:

The answer as given is not written correctly.

First, we must consider significant figures. Any non-zero number is a significant figure, and zeroes between significant figures are also significant, so 0.3065 has four significant figures and 138.03 has five significant figures. (Leading zeroes before the decimal point are not significant)

When two numbers are multiplied or divided, the answer has the same number of significant figures as the number that had the least number of significant figures. So in this case, the result will have four significant figures.

The result is written in scientific notation:

(0.3065)/(138.03) = 2.221 x 10⁻³

Answer: The number of electrons for n = 0, 1 and 2 are 2, 6 and 10 respectively.

Explanation:

Huckel's rule is used to determine the aromaticity in a compound. The number of delocalized [tex]\pi-[/tex] electrons are calculated by using the equation:

[tex]\text{Number of delocalized }\pi-\text{ electrons}=4n+2[/tex]

where,

n = 0 or any whole number

Putting values in above equation, we get:

[tex]\text{Number of delocalized }\pi-\text{ electrons}=4(0)+2=2[/tex]

Putting values in above equation, we get:

[tex]\text{Number of delocalized }\pi-\text{ electrons}=4(1)+2=6[/tex]

Putting values in above equation, we get:

[tex]\text{Number of delocalized }\pi-\text{ electrons}=4(2)+2=10[/tex]

Hence, the number of electrons for n = 0, 1 and 2 are 2, 6 and 10 respectively.

Explanation:

The given data is as follows.

       n = 2 mol,         P = 1 atm,         T = 300 K

        Q = +34166 J,         W= -1216 J (work done against surrounding)

       [tex]C_{v}[/tex] = [tex]\frac{3R}{2}[/tex]

Relation between internal energy, work and heat is as follows.

      Change in internal energy ([tex]\Delta U[/tex]) = Q + W

                                   = [34166 + (-1216)] J

                                   = 32950 J

Also,  [tex]\Delta U = n \times C_{v} \times \Delta T[/tex]

                      = [tex]3R \times (T_{2} - T_{1})[/tex]

                 32950 J = [tex]3 \times 8.314 J/mol K \times (T_{2} - 300 K)[/tex]

                [tex]\frac{32950}{24.942} = T_{2} - 300 K[/tex]

                            1321.06 K + 300 K = [tex]T_{2}[/tex]    

                                       [tex]T_{2}[/tex] = 1621.06 K

Thus, we can conclude that the final temperature of the gas is 1621.06 K.

Answer:

(CH₃)₃COCH3₃ and (CH₃)₂CHOCH₂CH₃

Explanation:

Isomers are compounds which have the same molecular formula. Constitutional isomers have different connectivity; the atoms are connected in different ways.

1. (CH₃)₃COCH₃

2. (CH₃)₂CHOCH3₃

3. (CH₃)₂CHOCH₂CH₃

Molecules 1 and 3 have the same formula (C₅H₁₂O) and are isomers. Molecule 2 is not an isomer. From the structural formula, it is clear that Molecules 1 and 3 have different connectivity.

Answer: Work done for the process is -390 J

Explanation:

The chemical equation for the reaction of zinc metal with sulfuric acid follows:

[tex]Zn+H_2SO_4\rightarrow ZnSO_4+H_2[/tex]

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

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

Given mass of zinc = 10.0 g

Molar mass of zinc = 65.38 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of zinc}=\frac{10.0g}{65.38g/mol}=0.153mol[/tex]

The equation given by ideal gas follows:

[tex]P\Delta V=nRT[/tex]

where, P = pressure of the gas

[tex]\Delta V[/tex] = Change in volume of the gas

T = Temperature of the gas = 298 K

R = Gas constant = 8.314 J/mol.K

n = number of moles of gas = 0.153 mol

Putting values in above equation, we get:

[tex]P\Delta V=0.153mol\times 8.314J/mol.K\times 298K\\\\P\Delta V=397J[/tex]

To calculate the work done, we use the equation:

[tex]\text{Work done}=-P\Delta V\\\\W=-390J[/tex]

Hence, work done for the process is -390 J

Answer:

Explanation:

From literature, we know that the typical pka of a lysine side is 10.3.

Then we use the Henderson-Hasselbalch equation:

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

In this case, [A⁻] is the concentration of deprotonated lysine, and [HA] is the concentration of protonated lysine.

We put the data from the problem in the equation and calculate the ratio of deprotonated to protonated lysine:

[tex]7.5=10.5+log\frac{[A^{-} ]}{[HA]}\\-3.0=log\frac{[A^{-} ]}{[HA]}\\10^{-3.0}=\frac{[A^{-} ]}{[HA]}\\0.001=\frac{[A^{-} ]}{[HA]}[/tex]

Answer:

50 mL

Explanation:

Explanation:

Aldonic Acid:

Aldonic acids are suger acids.

General formula of aldonic acid = [tex]HOOC-(CHOH)_n-CH_2OH[/tex]

Aldonic acids are obtained by the oxidation of aldehydic group of suger.

So, aldonic acids have hydroxyl group at one terminal and carboxylic group at another terminal.

Gluconic acid is an example of aldonic acid.

Uronic Acid:

It is also a type of suger acid having carbonyl functional group at one terminal and carboxylic group at other terminal.

It is obtained by oxidation of hydroxyl group of the sugar.

Aldaric Acid:

Aldaric acid is also a type of sugar acid having carboxylic acid functional group at both the ends.

Both the hydroxyl group and aldehydic group are oxidized to form class of compound, called aldaric acid.

Answer: This therapy is available for 15 days.

Explanation:

We are given:

Oral solution dosage = 75 mg/5 mL

To calculate the volume of oral situation for single dose per, we use unitary method:

The volume required for 75 mg of solution is 5 mL

So, the volume required for 300 mg of solution will be = [tex]\frac{5mL}{75mg}\times 300mg=20mL[/tex]

The total volume of the ranitidine bottle = 300 mL

To calculate the number of days, we divide the total volume of the bottle by the volume of dose taken per night, we get:

[tex]\text{Number of days}=\frac{\text{Total volume}}{\text{Volume of dose taken per night}}=\frac{300mL}{20mL}=15[/tex]

Hence, this therapy is available for 15 days.

Answer:

a) pH = 2,23

b) pH = 3,26

c) pH = 3,74

d) pH = 7,98. Here we have the equivalence point of the titration

Explanation:

In a titration of a strong base (NaOH) with a weak acid (HCOOH) the reaction is:

HCOOH + NaOH → HCOONa + H₂O

a) Here you have just HCOOH, thus:

HCOOH ⇄ HCOO⁻ + H⁺ where ka =1,8x10⁻⁴ and pka = 3,74

When this reaction is in equilibrium:

[HCOOH] = 0,200 -x

[HCOO⁻] = x

[H⁺] = x

Thus, equilibrium equation is:

1,8x10⁻⁴ = [tex]\frac{[x][x] }{[0,200-x]}[/tex]

The equation you will obtain is:

x² + 1,8x10⁻⁴x - 3,6x10⁻⁵ = 0

Solving:

x = -0,006090675 ⇒ No physical sense. There are not negative concentrations

x = 0,005910674

As x = [H⁺] and pH = - log [H⁺]

pH = 2,23

b) Here, it is possible to use:

HCOOH + NaOH → HCOONa + H₂O

With adition of 5,00 mL of 1,00M NaOH solution the initial moles are:

HCOOH = [tex]0,100 L.\frac{0,200 mol}{L} =[/tex] = 2,0x10⁻² mol

NaOH = [tex]0,005 L.\frac{1,00 mol}{L} =[/tex] = 5,0x10⁻³ mol

HCOO⁻ = 0.

In equilibrium:

HCOOH = 2,0x10⁻² mol - 5,0x10⁻³ mol = 1,5x10⁻² mol

NaOH = 0 mol

HCOO⁻ = 5,0x10⁻³ mol

Now, you can use Henderson–Hasselbalch equation:

pH = 3,74 + log [tex]\frac{5,0x10^{-3} }{1,5x10^{-2} }[/tex]

pH = 3,26

c) With adition of 10 mL of 1,00M NaOH solution the initial moles are:

HCOOH = [tex]0,100 L.\frac{0,200 mol}{L} =[/tex] = 2,0x10⁻² mol

NaOH = [tex]0,010 L.\frac{1,00 mol}{L} =[/tex] = 1,0x10⁻² mol

HCOO⁻ = 0.

In equilibrium:

HCOOH = 2,0x10⁻² mol - 1,0x10⁻² mol = 1,0x10⁻² mol

NaOH = 0 mol

HCOO⁻ = 1,0x10⁻² mol

Now, you can use Henderson–Hasselbalch equation:

pH = 3,74 + log [tex]\frac{1,0x10^{-2} }{1,0x10^{-2} }[/tex]

pH = 3,74

d) With adition of 20 mL of 1,00M NaOH solution the initial moles are:

HCOOH = [tex]0,100 L.\frac{0,200 mol}{L} =[/tex] = 2,0x10⁻² mol

NaOH = [tex]0,020 L.\frac{1,00 mol}{L} =[/tex] = 2,0x10⁻² mol

HCOO⁻ = 0.

Here we have the equivalence point of the titration, thus, the equilibrium is:

HCOO⁻ + H₂O ⇄ HCOOH + OH⁻ kb = kw/ka where kw is equilibrium constant of water = 1,0x10⁻¹⁴; kb = 5,56x10⁻¹¹

Concentrations is equilibrium are:

[HCOOH] = x

[HCOO⁻] = 0,1667-x

[OH⁻] = x

Thus, equilibrium equation is:

5,56x10⁻¹¹ = [tex]\frac{[x][x] }{[0,01667-x]}[/tex]

The equation you will obtain is:

x² + 5,56x10⁻¹¹x - 9,27x10⁻¹³ = 0

Solving:

x = -9,628361x10⁻⁷⇒ No physical sense. There are not negative concentrations

x = 9,627806x10⁻⁷

As x = [OH⁻] and pOH = - log [OH⁻]; pH = 14 - pOH

pOH = 6,02

pH = 7,98

I hope it helps!

Answer:

salt mass=102.5gCrCl3.6H2O

Explanation:

Hello !

To know how much CrCl3.6H2O we need, we follow the steps below

First we have to know what the mass of the compound is

Cr = 52g / mol

Cl = 35.5g / mol

H = 1g / mol

O = 16g / mol

We calculate the mass

52+ (35.5 * 3) + 6 * (2 * 1 + 16) = 266.5g

We have 20g Cr, so we calculate the amount of salt:

20gCr * (266.5g salt / 52gCr) = 102.5gCrCl3.6H2O

salt mass=102.5gCrCl3.6H2O

Answer: The number of moles of nitrogen gas is 0.505 moles.

Explanation:

To calculate the number of moles of nitrogen gas, we use ideal gas equation, which is:

[tex]PV=nRT[/tex]

where,

P = pressure of the gas = 4.27 atm

V = Volume of the gas = 2.96 L

T = Temperature of the gas = [tex]32.0^oC=[32.0+273]K=305K[/tex]

R = Gas constant = [tex]0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]

n = number of moles of gas = ?

Putting values in above equation, we get:

[tex]4.27atm\times 2.96L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 305K\\n=0.505mol[/tex]

Hence, the number of moles of nitrogen gas is 0.505 moles.

Explanation:

The given data is as follows.

     Volume = 2.96 L,       Temperature = [tex]32.0^{o}C[/tex] = (32 + 273) K = 305 K,

        Pressure = 4.27 atm,       n = ?

And, according to ideal gas equation, PV = nRT

           [tex]4.27atm \times 2.96 L= n \times 0.0821 Latm/mol K \times 305K[/tex]    

                n = [tex]\frac{12.64 atm L}{25.04 Latm/mol}[/tex]

                         = 0.505 mol

Thus, we can conclude that the number of moles of nitrogen gas in the container is 0.505 mol.  

Answer:

The correct option is: E. No precipitate will form.

Explanation:

A solubility chart refers to the list of solubility of various ionic compounds. It shows the solubility of the various compounds in water at room temperature and 1 atm pressure.

Also, according to the solubility rules, the salts of chlorides, bromides and iodides are generally soluble and mostly all salts of sulfate are soluble.

Since, all the compounds formed in this double replacement reaction are soluble in water. Therefore, no precipitate will be formed.

ZnSO₄ (aq) + MgCl₂ (aq) → ZnCl₂ (aq) + MgSO₄ (aq)    

Answer:

0.971 grams

Explanation:

Given:

Temperature = 3.0° C = 3 + 273 = 276 K

Volume, V = 5.0 L

Pressure, P = 0.100 atm

Now, from the relation

PV = nRT

where,

n is the number of moles,

R is the ideal gas constant  = 0.082057 L atm/mol.K

thus,

0.1 × 5 = n ×  0.082057 × 276

or

n = 0.022 moles

Also,

Molar mass of the Dinitrogen monoxide gas (N₂O)

= 2 × Molar mass of nitrogen + 1 × Molar mass of oxygen

= 2 × 14 + 16 = 44 grams/mol

Therefore, Mass of 0.022 moles of N₂O = 0.022 × 44 = 0.971 grams

Answer:

Plausible structure has been given below

Explanation:

4.04 g of [tex]CO_{2}[/tex] = [tex]\frac{4.04}{44}moles[/tex] [tex]CO_{2}[/tex] = 0.0918 moles of [tex]CO_{2}[/tex]

1 mol of [tex]CO_{2}[/tex] contains 1 mol of C atom

So, 0.0918 moles of [tex]CO_{2}[/tex] contains 0.0918 moles of C atom

1.24 g of [tex]H_{2}O[/tex] = [tex]\frac{1.24}{18}moles[/tex] [tex]H_{2}O[/tex] = 0.0689 moles of [tex]H_{2}O[/tex]

1 mol of [tex]H_{2}O[/tex]  contain 2 moles of H atom

So, 0.0689 moles of [tex]H_{2}O[/tex] contain [tex](2\times 0.0689)moles[/tex] of [tex]H_{2}O[/tex] or 0.138 moles of [tex]H_{2}O[/tex]

Moles of C : moles of H = 0.0918 : 0.138 = 2 : 3

Empirical formula of hydrocarbon is [tex]C_{2}H_{3}[/tex]

So, molecular formula of one of it's analog is [tex]C_{4}H_{6}[/tex]

Plausible structure of [tex]C_{4}H_{6}[/tex] has been given below.

Answer:

1000m2 / s2

Explanation:

Hello! In order to verify this, we have to do unit conversion. We also have to know that J (Joule) = kg * m2 / s2

Then we can start with the test.

1kJ / kg * (1000J / 1kJ) = 1000J / kg

1000J / kg = 1000kg * m2 / kg * s2

In this step we can simplify "kg".

So the result is

1000m2 / s2

Final answer:

To demonstrate that 1 kJ/kg equals 1000 m²/s², we recognize that the joule (J) is defined as kg-m²/s², and by converting kJ to J and canceling out the kg units, we affirm the equality.

Explanation:

To show that 1 kJ/kg is equal to 1000 m²/s², we start by recognizing that the unit of energy, the joule (J), is defined as 1 kilogram-meter²/second² (kg-m²/s²). Therefore, when we talk about energy per unit mass, we are effectively dividing energy by mass, leaving us with units of m²/s².

Given that 1 joule is 1 kg·m²/s², 1 kJ is 1000 joules (since the prefix 'kilo' means 1000). So when we have 1 kJ/kg, it's the same as saying 1000 J/kg. When we divide each term (kg·m²/s²) by kg, the kilograms cancel out, leaving us with m²/s². Thus, 1 kJ/kg is indeed equivalent to 1000 m²/s².

Answer:

At the equivalence point of the titration of a monoprotic weak acid with a strong base: B. the moles of strong base added must equal the moles of weak acid.

Explanation:

In every titration, the equivalence point is defined as the point where the moles of the titrant and analyte are equal. For every acid-base titration, the equivalence point is defined as the point where the moles of the base is equal to the moles of the acid.

If the solutions of the acid and base are at a different concentration the volume added from the buret will not be the same as the volume of the analyte.

Answer:

8 drops of 1.00 M NaOH will be needed.

Explanation:

Concentration of [tex][OH^-][/tex] in bleach solution = 0.02 M

[tex]NaOH\rightarrow OH^-+Na^+[/tex]

[tex]NaOH=[OH^-]=0.02M[/tex]

Concentration of bleach solution we want ,[tex]M_1[/tex] = 0.02 M

Volume of the bleach solution,[tex]V_1[/tex] = 20 ml

Concentration of NaOH solution,[tex]M_2[/tex] = 1.00 M

Volume of the NaOH solution required ,[tex]V_2[/tex] = ?

[tex]M_1V_1=M_2V_2[/tex]

[tex]0.02 M\times 20 mL=1.00 M\times V_2[/tex]

[tex]V_2=\frac{0.02 M\times 20 mL}{1.00 M}=0.4 mL[/tex]

1 mL = 20 drops

0.4 mL = 0.4 × 20 drops = 8 drops

8 drops of 1.00 M NaOH will be needed.

Answer:

-0.93 °C

Explanation:

Hello,

The freezing-point depression is given by:

[tex]T_f-T_f^*=-iK_{solvent}m_{solute}[/tex]

Whereas [tex]T_f[/tex] is the freezing temperature of the solution, [tex]T_f^*[/tex] is the freezing temperature of the pure solvent (0 °C since it is water), [tex]i[/tex] the Van't Hoff factor (1 since the solute is covalent), [tex]K_{f,solvent}[/tex] the solvent's freezing point depression point constant (in this case [tex]1.86 C\frac{kg}{mol}[/tex]) and [tex]m_{solute}[/tex] the molality of the glucose.

As long as the unknown is [tex]T_f[/tex], solving for it:

[tex]T_f=T_f^*-iK_fm\\T_f=0C-1*1.86C\frac{kg}{mol}*0.5\frac{mol}{kg}  \\T_f=-0.93C[/tex]

Best regards.

Answer:

691.84g

Explanation:

I'm assuming that by CzHg, you mean C3H2

First, use the mole ratio in the equation to find the moles of CO2

n (CO2)= n ( C3H2) × 3

= 5.24 × 3

= 15.72

To find the mass of CO2 produced in grams, complete the following calculation

m= n × MM

where

m = mass

n= moles

MM= molecular mass

m= 15.72 × (12.01 +( 16×2))

m =691.8372

m= 691.84g

Answer:

The ratio [G1P]/[G6P] = 5.7 . 10⁻².

Explanation:

Let us consider the reaction G1P ⇄ G6P, with ΔG° = -7.1 kJ/mol. According to Hess's Law, we can write the inverse reaction, and Gibbs free energy would have an opposite sign.

G6P ⇄ G1P        ΔG° = 7.1 kJ/mol

This is the reaction for which we want to find the equilibrium constant (the equilibrium ratio of [G1P] to [G6P]):

[tex]Kc=\frac{[G1P]}{[G6P]}[/tex]

The equilibrium constant and Gibbs free energy are related by the following expression:

[tex]Kc=e^{-\Delta G\si{\textdegree}/R.T } } =e^{-7.1kJ/mol/8.314.10^{-3}kJ/mol.K.298K} } }=5.7.10^{-2}[/tex]

where,

R is the ideal gas constant (8.314 . 10⁻3 kJ/mol.K)

T is the absolute temperature (in kelvins)

Final answer:

To calculate the equilibrium ratio of [G1P] to [G6P], the Gibbs free energy equation is used with ΔG', universal gas constant R, and the temperature T substituted. The result is that the concentration of G6P is approximately 1.331 times that of G1P at equilibrium and at 25°C.

Explanation:

The student is asking about the equilibrium ratio of concentrations of glucose-1-phosphate ([G1P]) to glucose-6-phosphate ([G6P]) at 25°C when the standard free energy change (ΔG') for the isomerization reaction is given as -7.1 kJ/mol. To calculate this ratio, we can use the Gibbs free energy equation for the equilibrium constant (Keq):

ΔG' = -RT ln(Keq)

Where ΔG' is the standard free energy change, R is the universal gas constant (8.314 J/mol K), T is the temperature in Kelvin (25°C + 273.15 = 298.15 K), and Keq is the equilibrium constant which for this reaction is [G6P]/[G1P].

Substituting the values into the equation we get:

-7100 J/mol = -(8.314 J/mol K)(298.15 K) ln([G6P]/[G1P])

Now, we solve for ln([G6P]/[G1P]):

ln([G6P]/[G1P]) = ΔG' / (-R * T)

ln([G6P]/[G1P]) = -7100 J/mol / (-(8.314 J/mol K)(298.15 K))

ln([G6P]/[G1P]) = 0.286

Exponentiating both sides to remove the natural logarithm, we get:

[G6P]/[G1P] = e0.286 = 1.331

Therefore, at equilibrium and at 25°C, the concentration of G6P is approximately 1.331 times that of G1P.

Answer:

397 g

Explanation:

From the information given in the question ,

the atomic mass of X = 12 ,

and , the atomic mass of Y = 1 .

For the molecular formula , X₇Y₁₁ , the molecular mass is given as -

X₇Y₁₁ = 7 * atomic mass of X  +  11 *  atomic mass of Y

         = 7 * 12 + 11 * 1

         = 84 + 11  = 95 g / mol

Hence , molecular mass of X₇Y₁₁ = 95 g / mol .

Moles is denoted by given mass divided by the molecular mass ,

Hence ,

n = w / m

n = moles ,

w = given mass ,

m = molecular mass .

From question ,

moles n = 4.18 mol

and ,

m = molecular mass of X₇Y₁₁ = 95 g / mol .

To find the mass of the compound X₇Y₁₁   , the above formula can be used , and putting the respective values ,

n = w / m  

4.18 = w / 95

w = 4.18 * 95 = 397 g

Answer:

The air density in at points 1 is [tex]60.7 kg/m^3[/tex] and 2 is [tex]2060 kg/m^3[/tex].

Explanation:

Average molecular weight of an air ,M= 28.97 g/mol

[tex]PV=nRT[/tex]

or [tex] PM=dRT[/tex]

P = Pressure of the gas

n = moles of gas

T = Temperature of the gas

d = Density of the gas

M = molar mass of the gas

R = universal gas constant

Density at point-1 = [tex]d_1[/tex]

[tex]P_1 = 10,000 Pa=0.0986 atm[/tex]

[tex]1 Pa=9.86923\times 10^{-6} atm[/tex]

[tex]T_1 = 300^oC = 573.15 K[/tex]

M = 28.97 g/mol

[tex]d_1=\frac{PM}{RT}=\frac{0.0986 atm\times 28.97 g/mol}{0.0821 atm L/ mol K\times 573.15 K}[/tex]

[tex]d_1 =0.0607 g/ml[/tex]

1 g = 0.001 kg

[tex]1 mL = 10^{-6} m^3[/tex]

[tex]d_1=\frac{0.0607\times 0.001 kg}{10^{-6} m^3}=60.7 kg/m^3[/tex]

Density at point-2 = [tex]d_2[/tex]

[tex]P_2 = 575,000 Pa=5.67 atm[/tex]

[tex]T_2 = 700^oC = 973.15 K[/tex]

M = 28.97 g/mol

[tex]d_2=\frac{PM}{RT}=\frac{5.67 atm\times 28.97 g/mol}{0.0821 atm L/ mol K\times 973.15 K}[/tex]

[tex]d_2 =2.06 g/ml=2060 kg/m^3[/tex]

Explanation:

It is known that for [tex]NO_{2}[/tex], ppm present in 1 [tex]mg/m^{3}[/tex] are as follows.

                      1 [tex]\frac{mg}{m^{3}}[/tex] = 0.494 ppm

So, 150 [tex]pg/m^{3}[/tex] = [tex]\frac{150}{1000} mg/m^{3}[/tex]

                       = 0.15 [tex]mg/m^{3}[/tex]

Therefore, calculate the equivalent concentration in ppm as follows.

             [tex]0.15 \times 0.494 ppm[/tex]

              = 0.074 ppm

Thus, we can conclude that the equivalent concentration in ppm at STP is 0.074 ppm.  

Answer:

Ea = 177x10³ J/mol

ko = [tex]1.52x10^{19}[/tex] J/mol

Explanation:

The specific reaction rate can be calculated by Arrhenius equation:

[tex]k = koxe^{-Ea/RT}[/tex]

Where k0 is a constant, Ea is the activation energy, R is the gas constant, and T the temperature in Kelvin.

k depends on the temperature, so, we can divide the k of two different temperatures:

[tex]\frac{k1}{k2} = \frac{koxe^{-Ea/RT1}}{koxe^{-Ea/RT2}}[/tex]

[tex]\frac{k1}{k2} = e^{-Ea/RT1 + Ea/RT2}[/tex]

Applying natural logathim in both sides of the equations:

ln(k1/k2) = Ea/RT2 - Ea/RT1

ln(k1/k2) = (Ea/R)x(1/T2 - 1/T1)

R = 8.314 J/mol.K

ln(2.46/47.5) = (Ea/8.314)x(1/528 - 1/492)

ln(0.052) = (Ea/8.314)x(-1.38x[tex]10^{-4}[/tex]

-1.67x[tex]10^{-5}[/tex]xEa = -2.95

Ea = 177x10³ J/mol

To find ko, we just need to substitute Ea in one of the specific reaction rate equation:

[tex]k1 = koxe^{-Ea/RT1}[/tex]

[tex]2.46 = koxe^{-177x10^3/8.314x492}[/tex]

[tex]1.61x10^{-19}ko = 2.46[/tex]

ko = [tex]1.52x10^{19}[/tex] J/mol