If a weak acid, HA, is 3% dissociated in a 0.25 M
solution,calculate the Ka and the pH of the solution.

Answers

Answer 1

Answer:

Ka = [tex]2.32 \times 10^{-4}[/tex]

pH = 2.12

Explanation:

Calculation of Ka:

% Dissociation = 3% = 0.03

Concentration of solution = 0.25 M

HA dissociated as:

       [tex]HA \rightarrow H^+ + A^{-}[/tex]

      C(1 - 0.03)    C×0.03   C×0.03

HA] after dissociation = 0.25×0.97 = 0.2425 M

[tex][H^+]= 0.25\times0.03 = 0.0075 M[/tex]

[tex][A^{-}]= 0.25 \times 0.03 = 0.0075 M[/tex]

[tex]Ka= \frac{[H^+][A^{-}]}{[HA]}[/tex]

[tex]Ka = \frac{(0.0075)^2}{0.2475} =2.32 \times 10^{-4}[/tex]

Calculation of pH of the solution

[tex]pH = -log [H^+][/tex]

[tex]H^+[\tex] = 0.0075 M[/tex]

[tex]pH = -log 0.0075 = 2.12[/tex]

Answer 2
Final answer:

The Ka of the weak acid is calculated using the concentrations of the dissociated and non-dissociated parts of the weak acid and the pH is derived from the concentration of H+ ions.

Explanation:

The subject in concern relates to the dissociation of a weak acid and the calculation of the Acid dissociation constant (Ka) and the pH of the solution.

We are given that 3% of the weak acid, HA, is dissociated. Meaning, 3% of 0.25 M HA dissociates into H+ and A-. This will be 0.03 * 0.25 M = 0.0075 M.

The equilibrium for the reaction is HA <--> H+ + A-. Due to water autoionization, the concentration of water is approximately taken as constant in the denominator of Ka calculation. Hence, Ka for HA can be approximated as [H+][A-] / [HA]. We know that [H+] = [A-] = x, and [HA] = 0.25 - x. We approximate x as 0.0075 since only 3% of HA dissociates. Hence Ka=~(0.0075)^2/(0.25-0.0075).

The pH of the solution is -log[H+] = -log(0.0075).

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Related Questions

What is the Ka of a weak acid (HA) if the initial concentration of weak acid is 4.5 x 10-4 M and the pH is 6.87? (pick one)

5.5 x 10-5

4.0 x 10-6

6.9 x 10-4

3.5 x 10-10

4.0 x 10-11

Answers

Answer:

Ka = [tex]4.04 \times 10^{-11}[/tex]

Explanation:

Initial concentration of weak acid = [tex]4.5 \times 10^{-4}\ M[/tex]

pH = 6.87

[tex]pH = -log[H^+][/tex]

[tex][H^+]=10^{-pH}[/tex]

[tex][H^+]=10^{-6.87}=1.35 \times 10^{-7}\ M[/tex]

HA dissociated as:

[tex]HA \leftrightharpoons H^+ + A^{-}[/tex]

(0.00045 - x)    x     x

[HA] at equilibrium = (0.00045 - x) M

x = [tex]1.35 \times 10^{-7}\ M[/tex]

[tex]Ka = \frac{[H^+][A^{-}]}{[HA]}[/tex]

[tex]Ka = \frac{(1.35 \times 10^{-7})^2}{0.00045 - 0.000000135}[/tex]

0.000000135 <<< 0.00045

[tex]Therefore, Ka = \frac{(1.35 \times 10^{-7})^2}{0.00045 } = 4.04 \times 10^{-11}[/tex]

Draw the Lewis Structure for NI3.

Answers

Explanation:

Nitrogen triiodide (NI₃)

Valence electrons of nitrogen = 5

Valence electrons of iodine = 7

The total number of the valence electrons  = 5 + 3(7) = 26

The Lewis structure is drawn in such a way that the octet of each atom and duet for the hydrogen in the molecule is complete. So,  

The Lewis structure is shown in image below.

Nitrogen posses one lone pair and thus the geometry is pyramidal.

Final answer:

The Lewis structure of NI3 is drawn by counting the valence electrons, creating single bonds between nitrogen and iodine, completing octets for iodine atoms, and placing remaining electrons on the nitrogen to complete its octet. This results in a structure where nitrogen is single-bonded to three iodine atoms, each surrounded by three lone pairs, and nitrogen has one lone pair.

Explanation:

To draw the Lewis structure for NI3 (nitrogen triiodide), follow these steps:

Count the total number of valence electrons. Nitrogen has 5 valence electrons, and each iodine has 7 valence electrons, totaling (5 + 3*7) = 26 valence electrons.

Draw a single bond between the nitrogen atom and each iodine atom. This will use up 6 of the valence electrons (2 for each bond).

Complete the octets for the iodine atoms by adding six more electrons to each iodine in the form of electron pairs, using up 18 of the remaining valence electrons.

Place any remaining electrons (2) on the nitrogen atom to complete its octet.

Examine the structure. Each iodine has 8 electrons, and nitrogen has 8 electrons, making the structure complete.

If you follow these steps, the resulting Lewis structure for NI3 will show a central nitrogen atom single-bonded to three iodine atoms, with each iodine atom surrounded by three lone pairs of electrons, and one lone pair of electrons on the nitrogen atom.

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the standard change in Gibbs free energy is Δ????°′=7.53 kJ/molΔG°′=7.53 kJ/mol . Calculate Δ????ΔG for this reaction at 298 K298 K when [dihydroxyacetone phosphate]=0.100 M[dihydroxyacetone phosphate]=0.100 M and [glyceraldehyde-3-phosphate]=0.00300 M[glyceraldehyde-3-phosphate]=0.00300 M .

Answers

Answer:

ΔG = 16.218 KJ/mol

Explanation:

dihydroxyacetone phosphate ↔ glyceraldehyde-3-phosphateΔG = ΔG° - RT Ln Q

∴ ΔG° = 7.53 KJ/mol * ( 1000 J / KJ ) = 7530 J/mol

∴  R = 8.314 J/K.mol

∴ T = 298 K

∴ Q = [glyceraldehyde-3-phosphate] / [dihydroxyacetone phosphate]

⇒ Q = 0.00300 / 0.100 = 0.03

⇒ ΔG = 7530J/mol - (( 8.314 J/K.mol) * ( 298 K ) * Ln ( 0.03 ))

⇒ ΔG = 16217.7496 J/mol ( 16.218 KJ/mol )

how is the use of magnetic fields to plasma related to trying to generate energy using nuclear fusion?

Answers

Answer:

See explanation

Explanation:

Fusion is the process of fusing two isotopes of Hydrogen namely Tritium  and Deuterium to produce Helium. To achieve this, tremendous heat required (about a million degrees Celsius). The same for the pressure. To achieve this, the hydrogen is made into a plasma through rarefaction so it becomes susceptible to magnetic fields. Magnetic confinement fusion is an approach to generating thermonuclear fusion power that uses magnetic fields to confine the hot fusion fuel in the form of a plasma. Electro-Magnets surround the chamber/reactor and are pulsed adiabatically (as in a bicycle pump) and the gas becomes extremely hot that may melt the surroundings.

As the ions in the plasma are charged (the plasma is so hot all the negatively-charged electrons are stripped off the atoms, leaving them with a positive charge) they respond to magnetic fields. Extra fields help shape the plasma and hold it stable.

Calculate the volume of a 0.200 M KCl solution containing 5.00 10-2 mol of solute. Enter your answer in the provided box. IL

Answers

Answer:

The volume of a 0.200 M KCl solution containing 5.00 10-2 mol of solute is 0,25 L

Explanation:

Molarity (M) means: moles of solute which are contained in 1 L of solution.  

In this case we have 0,2 moles which are in 1 L, so, as we have 5x10*-2 moles we have to apply a rule of three to find out the volume.

0,2 moles ........... 1 L

5x10*-2 moles ........... x

x= (5x10*-2 moles . 1 L) / 0,2 moles = 0.25L

(we can also say 250 mL)

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.

Answers

Answer:

C. fluid density & viscosity

Explanation:

In 1850, Darcy-Weisbach experimentally deduced an equation to calculate shear losses ("friction"), in a tube with permanent flow and constant diameter:

hf = (f x L x V^2) / (D x 2g)

where:

hf: shear losses

f:  shear loss factor (pipe roughness)

g: gravity acceleration

D: tube diameter

L: tube length  

V: fluid average speed in the tube

To calculate the loss factor “f” in the Poiseuille laminar region, he proposed in 1846 the following equation:

f = 64 / Re

Where:

Re: Reynolds number

The influence of the parameters on f is quantitatively different according to the characteristics of the current.

In any straight pipeline that transports a liquid at a certain temperature, there is a critical speed below which the regimen is laminar. This critical value that marks the transition between the two regimes, laminar and turbulent, corresponds to a Re = 2300, although in practice, between 2000 and 4000 the situation is quite inaccurate. Thus:

Re <2000: laminar regimen

2000 <Re <4000: critical or transition zone

Re> 4000: turbulent regime

Final answer:

The friction factor for fluid flow in a pipe does not depend upon the pipe length, pipe roughness, fluid density & viscosity, or mass flow rate of fluid.

Explanation:

The friction factor for fluid flow in a pipe does not depend upon the pipe length (A), pipe roughness (B), fluid density & viscosity (C), or mass flow rate of fluid (D).

This is because the friction factor, also known as the Darcy-Weisbach factor, is determined by the characteristics of the flow itself, such as the Reynolds number, which is a dimensionless quantity that relates the inertia of the fluid to the viscous forces acting on it.

The friction factor can be calculated using the Colebrook-White equation or obtained from Moody's diagram based on the relative roughness of the pipe and the Reynolds number.

A sample of oxygen gas has a volume of 3.24 L at 29°C. What volume will it occupy at 104°C if the pressure and number of mol are constant? Enter your answer in the provided box.

Answers

Answer: The final volume of the oxygen gas is 4.04 L

Explanation:

To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure and number of moles.

Mathematically,

[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]

where,

[tex]V_1\text{ and }T_1[/tex] are the initial volume and temperature of the gas.

[tex]V_2\text{ and }T_2[/tex] are the final volume and temperature of the gas.

We are given:

[tex]V_1=3.24L\\T_1=29^oC=(29+273)K=302K\\V_2=?\\T_2=104^oC=(104+273)K=377K[/tex]

Putting values in above equation, we get:

[tex]\frac{3.24L}{302K}=\frac{V_2}{377K}\\\\V_2=4.04L[/tex]

Hence, the final volume of the oxygen gas is 4.04 L

All final to the right of the decimal place are significant

Answers

The answer is: Zeros
The Answer is: zero

Tracy is studying for her chemistry exam and she reads that a nitric acid molecule is made of one hydrogen atom, one nitrogen atom and 3 oxygen atoms. The chemical formula for nitric acid is a. HNO3 Oьно OCHN O d.HO3 Oe. HN30

Answers

Answer:

a. HNO₃

Explanation:

Nitric acid -

The chemical formula for nitric acid is - HNO₃ , with molecular mass of 63 g /mol , having one hydrogen atom, one nitrogen atom and 3 oxygen atoms .

Nirtic acid is a colorless solution , but the left over solution turns yellow as it get decomposed to nitrogen oxide and water .

The commercially used nitric acid is 68 % in water .

and more than 86% nitric acid is fuming nitric acid .

The reagent nitric acid is used a source of nitration .

Water (10 kg/s) at 1 bar pressure and 50 C is pumped isothermally to 10 bar. What is the pump work? (Use the steam tables.) O -7.3J/s O 7.3 kJ/s O -210 kJ/s O 3451 kJ/s

Answers

Explanation:

For an isothermal process equation will be as follows.

                W = nRT ln[tex]\frac{P_{1}}{P_{2}}[/tex]

It is given that mass is 10 kg/s or 10,000 g/s (as 1 kg = 1000 g). So, calculate number of moles of water as follows.

                    No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]

                                           = [tex]\frac{10000 g/s}{18 g/mol}[/tex]

                                           = 555.55 mol/s

                                           = 556 mol/s (approx)

As T = [tex]50^{o}C[/tex] or (50 + 273.15) K = 323.15 K. Hence, putting the given values into the above formula as follows.

                  W = nRT ln[/tex]\frac{P_{1}}{P_{2}}[/tex]

                      = [tex]556 mol/s \times 8.314 J/ K mol K \times 323.15 K \times ln\frac{1}{10}[/tex]    

                     = [tex]556 mol/s \times 8.314 J/ K mol K \times 323.15 K \times -2.303[/tex]    

                     = -3440193.809 J/s

Negative sign shows work is done by the pump. Since, 1 J = 0.001 kJ. Therefore, converting the calculated value into kJ as follows.

                     [tex]3440193.809 J/s \times \frac{0.001 kJ}{1 J}[/tex]

                          = 3440.193 kJ/s

                          = 3451 kJ/s (approx)

Thus, we can conclude that the pump work is 3451 kJ/s.

How many grams of precipitate can be formed in the reaction between 6.00 mL of 0.10 M silver nitrate and 5.00 mL of 0.15 M potassium chloride? Hint: Start with a balanced chemical equation. Do not enter units with your answer.

Answers

Answer:

0.09 g of precipitate can be formed.

Explanation:

The chemical equation can be written as follows:

AgNO₃ + KCl → AgCl(↓) + NO₃⁻ + K⁺

From the equation, we know that 1 mol AgNO₃ reacts with 1 mol KCl to produce 1 mol AgCl.

The problem gives us data to calculate the initial number of moles of silver nitrate and potassium chloride:

n° of moles of silver nitrate = concentration * volume

n° of moles of silver nitrate = 0.10 mol/l * 0.006 l = 6 x 10⁻⁴ mol AgNO₃

n° of moles of KCl = 0.15 mol/l * 0.005 l = 7.5 x 10⁻⁴ mol KCl

Since 1 mol AgNO₃ reacts with 1 mol KCl, 6 x 10⁻⁴ mol AgNO₃ will react with 6 x 10⁻⁴ mol KCl and produce 6 x 10⁻⁴ mol AgCl.

1.5 x 10⁻⁴ mol KCl is in excess.

The molar mass of AgCl is 143.32 g/mol, then, 6 x 10⁻⁴ mol AgCl will have a mass of (6 x 10⁻⁴ mol AgCl * 143.32 g / 1 mol) 0.09 g.

Water (25◦C) flows through 1-inch Schedule 40 steel pipe at 2.0 gpm. What is the Reynolds number of the flow? What is the friction factor? Is the flow laminar or turbulent?

Answers

Answer:

Re=8561.79

Friction factor is 0.014

Flow is turbulent flow.

Explanation:

Given that

Diameter ,d=1 in

d=0.0254 m     (1 in =0.0254 m )

Volume flow rate,Q = 2 gpm

We know that

[tex]1\ gpm=7.5\times 10^{-5}\ m^3/s[/tex]

[tex]2\ gpm=2\times 7.5\times 10^{-5}\ m^3/s[/tex]

[tex]2\ gpm=15\times 10^{-5}\ m^3/s[/tex]

We know that

Q= A x V

[tex]A=\dfrac{\pi}{4}\times 0.0254^2\ m^2[/tex]

[tex]A=0.00050\ m^2[/tex]

So

[tex]V=\dfrac{15\times 10^{-5}}{0.00050}[/tex] m/s

V=0.3 m/s

So Reynolds number(Re)

[tex]Re=\dfrac{\rho VD}{\mu }[/tex]

Properties of water at 25 C

[tex]\mu=8.9\times 10^{-4}\ Pa.s[/tex]

[tex]Re=\dfrac{1000\times 0.0254\times 0.3}{8.9\times 10^{-4}}[/tex]

Re=8561.79

Re>4000 ,It means that flow is turbulent flow.

Friction factor

If we assume that pipe is smooth

[tex]f=\dfrac{0.136}{Re^{0.25}}[/tex]

[tex]f=\dfrac{0.136}{8561.79^{0.25}}[/tex]

f=0.014.

Friction factor is 0.014

Final answer:

To identify whether the flow is laminar or turbulent and to determine the friction factor, you should first find the Reynolds number using the given parameters, such as the flow rate of 2.0 gpm and then refer to a Moody chart or similar source.

Explanation:

The Reynolds number (NR) and friction factor of the flow can be determined using the given parameters (Flow rate, pipe diameter, and water temperature). The Reynolds number is an indicator that can reveal whether flow is laminar or turbulent. For flow in a tube of uniform diameter, the Reynolds number is defined as an equation related to the properties of the fluid and the characteristics of the flow.

In this case, if the Reynolds number (NR) is below 2000, the flow is considered laminar. If NR is above 3000, the flow becomes turbulent. For values of NR between 2000 and 3000, it may be either or both, depending on factors such as the roughness of the pipe's internal surface and the flow velocity.

The friction factor is a measure of the total resistance created by the force on a fluid as it moves through a pipe. The calculation of this factor also considers the Reynolds number and pipe roughness.

To determine whether your given conditions (2.0 gpm water flow through a 1-inch Schedule 40 steel pipe at 25◦C) will result in laminar or turbulent flow and the corresponding friction factor, please consider calculating the Reynolds number first using the suitable formula and the given conditions and then consulting a Moody chart or a similar source to find the corresponding friction factor.

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Calculate the volume of 38.0 g of carbon dioxide at STP. Enter your answer in the box provided. L

Answers

Answer:

19.3 L

Explanation:

V= n × 22.4

where V is volume and n is moles

First, to find the moles of CO2, divide 38.0 by the molecular weight of CO2 which is 44.01

n= m/ MM

n= 38/ 44.01

n= 0.86344012724

V= 0.86344012724 × 22.4

V= 19.3410588502 L

V= 19.3 L

Magnesium reacts with hydrochloric acid to produce magnesium chloride and hydrogen gas. Write a balanced chemical equation for the reaction. If 0.510 g of magnesium reacts with 0.500 L of hydrochloric acid, determine the molarity and pH of the hydrochloric acid solution. Determine the volume of hydrogen gas that will be produced if the reaction takes place at 25 °C and 1.00 atm.

Answers

Answer:

The balanced chemical equation is Mg + 2HCl ⇒ MgCl2 + H2

The molarity of the hydrochloric acid solution is HCl 0.04 M and the pH = 1.4.

The volume of hydrogen gas produced by the reaction of 0.510 g of Mg will be 0.482 L.

Explanation:

First, for the balanced equation you have to consider the oxidation state of the elements to find subscripts. Then you can find the correct coeficients. Mg= +2, Cl = -1.

Mg + HCL ⇒ MgCl2 + H2

For the molarity of the solution you have to notice tha if 0.510 grams of Mg reacts with 0,5 L of hydroclhoric acid, and from the previous equation 1 mol of Mg reacts with 2 mol HCl.

The atomic mass of Mg = 24.31 grs/mol

24.31 grs------------ 1 mol Mg

0.510 grs------------ x=0.02 mol Mg.

If 1 mol of Mg reacts with 2 mol HCl, then 0.02 mol of Mg will react with

0.04 mol HCl. So, the molarity of the solution is 0.04 M HCl.

Then to calculate the pH we use the formula pH = - log [H+]

⇒ pH = -log [0.04]⇒ pH=1.4.

Finally, from the balanced equation and the findings described, and considering that at 25°C and 1.00 atm 1 mol of gas has volume of 24.1 L.

1 mol H2----------- 24.1 L

0.02 mol H2----- x= 0.482L.

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.

Answers

Final answer:

The atomic mass of Cr53 can be calculated by establishing an equation that takes into account the atomic masses and abundances of all isotopes of Chromium. Solve the equation for 'x' (representing the atomic mass of Cr53) to find your answer.

Explanation:

The atomic mass of a given isotope of an element is determined by the weighted average of the masses of its isotopes, each multiplied by the abundance of that isotope. Since the atomic masses and abundances of Cr50, Cr52, and Cr54 are already provided, we only need to account for Cr53's mass. Using Chromium's atomic mass (51.9961 u), we can establish an equation to solve for Cr53's atomic mass.

Here's our equation:

51.9961 u = (49.9460 u * Cr50's abundance) + (51.9405 u * 0.8379) + (x * Cr53's abundance) + (53.9389 u * 0.0237)

First calculate the contribution of Cr50, Cr52, and Cr54 to the atomic weight, then subtract this from the total atomic weight (51.9961 u). Divide this value by the Cr53's abundance to get the atomic weight of Cr53.

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The atomic mass of the Cr53 isotope is approximately 23.83 u, calculated based on its natural abundance and atomic mass in the given chromium isotope mixture.

To determine the atomic mass of the Cr53 isotope, we can use the information provided about the natural abundances and atomic masses of the chromium isotopes.

Let's denote the natural abundance of Cr50 as x. Since the ratio of Cr50 to Cr53 is 1:0.4579, the natural abundance of Cr53 would be 0.4579x. The total natural abundance of Cr50 and Cr53 is 1, so we have the equation:

x + 0.4579x = 1

Solving for x, we find that x is approximately 0.6852.

Now, we can calculate the total contribution of Cr53 to the atomic mass:

Atomic mass of Cr53 = Natural abundance of Cr53 * Atomic mass of Cr53

Atomic mass of Cr53 = 0.4579 * 51.9961

Calculating this gives us the atomic mass of the Cr53 isotope.

Atomic mass of Cr53 = 23.8256 u

In summary, the atomic mass of the Cr53 isotope is approximately 23.83 u.

Calculate the amount of CO2 (in kg) released when 1 kg coal is burnt. Assume that carbon content of the coal is 50% by mass. 4. Calculate the CO2 production in kg/MJ if a coal fired power plant (efficiency - 30%) is used to produce electricity. Assume energy density of coal - 24 MJ/kg and assume the coal is

Answers

Explanation:

(a)   Here, the assumption is that complete combustion of carbon is taking place. This means that there is no CO formation

Reaction equation when carbon is burning is as follows.

              [tex]C + O_{2} \rightarrow CO_{2}[/tex]

Molar mass of C = 12 kg/kmol

Molar mass of [tex]CO_{2}[/tex] = 44 kg/kmol

Hence, according to the stoichiometry,

1 Kmol of C reacted = 1 kmol of [tex]CO_{2}[/tex] produced

Therefore, 12 kg of C reacted = 44 kg [tex]CO_{2}[/tex] produced

It is given that coal contains 50 mass % Carbon

So, 1 kg of coal contains (0.50 × 1)kg Carbon

           Carbon in 1 kg Coal = 1 × 0.5 = 0.5 kg

As per the stoichiometry,

12 kg of C reacted = 44 kg [tex]CO_{2}[/tex] produced

0.5 kg of C reacted = x kg [tex]CO_{2}[/tex] produced

Therefore, value of x can be calculated as follows.

                    x = [tex]\frac{\text{44 kg of CO_{2} produced} \times \text{0.5 kg of C reacted}}{\text{12 kg of C reacted}}[/tex]

                       = 1.83 kg

This means that amount of [tex]CO_{2}[/tex] released is 1.83 kg.

(b)   It is assumed that coal contains 50 mass % carbon and 1 kg of coal burnt.

Since, it is given that energy density of coal is 24 Mj and efficiency of the power plant is 30%.

After burning 1 kg of coal amount of energy released = 24 Mj

Amount of energy converted to electricity = [tex]24 Mj \times 0.3[/tex] = 7.2 Mj

It is calculated that amount of [tex]CO_{2}[/tex] released per 1 kg of coal = 1.83 kg

Therefore, calculate the amount of [tex]CO_{2}[/tex] released in kg/Mj as follows.

        amount of [tex]CO_{2}[/tex] released in kg/Mj = [tex]\frac{\text{Amount of CO_{2} released in kg}}{\text{amount of energy converted to electricity}}[/tex]        

                        = [tex]\frac{1.83 kg}{7.2 mJ}[/tex]

                        = 0.2541 kg/Mj

Hence, the production of [tex]CO_{2}[/tex] in kg/MJ is 0.2541 kg/Mj.

Final answer:

To calculate the amount of CO2 produced from 1 kg of coal, first determine the mass of carbon, apply the conversion for the mass of CO2 produced per mass of carbon, and use the plant's efficiency and the coal's energy density to determine CO2 production per MJ.

Explanation:

The question involves consulting basic chemical stoichiometry and energy conversion to solve problems related to carbon dioxide emissions and energy efficiency in the context of coal combustion. Knowing the percentage of carbon in coal, we can determine the mass of carbon dioxide produced from burning coal. Additionally, we'll use the energy density of coal to calculate CO2 production per unit energy output of a coal-fired power plant.

To find the mass of CO2 produced from burning 1 kg of coal with 50% carbon content, we first calculate the mass of carbon that would burn:

Mass of carbon in 1 kg of coal = 1 kg * 50% = 0.5 kg.

The molecular weight of carbon is approximately 12 g/mol, and for CO2 it is approximately 44 g/mol. This implies that every 12 kg of carbon produces 44 kg of CO2. For 0.5 kg (or 500 g) of carbon, the mass of CO2 produced would be:

(44 kg CO2 / 12 kg C) * 0.5 kg C = 1.833 kg CO2.

For the second part, to calculate the CO2 production in kg/MJ for a coal-fired power plant with 30% efficiency:

Energy content per kg coal = 24 MJ/kg.Effective energy per kg coal (due to 30% efficiency) = 24 MJ/kg * 30% = 7.2 MJ/kg.Thus, for 1 kg of coal combusted, 1.833 kg of CO2 is released. This amount of CO2 is produced for every 7.2 MJ of energy output.CO2 production per MJ = 1.833 kg CO2 / 7.2 MJ = 0.2546 kg CO2/MJ.

By increasing the length of heat exchanger, the effectiveness and cold stream output temperature of the heat exchanger

a- Decreases

b- Increases

c- Does not change

Answers

Answer:

Effectiveness and cold stream output temperature of the heat exchange Increases. So, Answer is b) Increases.

Explanation:

We have a heat exchanger, and it is required to compare the effectiveness and cold stream output if the length is increased.

Heat exchangers are engineering devices used to transfer energy. Thermal energy is transferred from Fluid 1 - Hot fluid (HF) to a Fluid 2 - Cold Fluid (CF). Both fluids 1 and 2 can flow with different values of mass flow rate and different specific heat. When the streams go inside the heat exchanger Temperature of Fluid 1 (HF) will decrease, at the same time Temperature of the Fluid 2 (CF) will increase.

In this case, we need to analyze the behavior taking into account different lengths of heat exchangers. If the length of the heat exchanger increases, it means the transfer area will increases. Heat transfer will increase if the transfer area increases. In this sense, the increasing length is the same than increase heat transfer.

If the heat transfer increases, it means Fluid 1 (HF) will reduce its temperature, and at the same time Fluid 2 (CF) will increase its temperature.

Finally, Answer is b) Effectiveness and cold stream output temperature increases when the length of the heat exchanger is increased.

1826.5g of methanol (CH3OH), molar mass = 32.0 g/mol is added to 735 g of water, what is the molality of the methane 0.0348 m 1.13m 2.03 m 3.61 m 36.1 m Navigator F10 Delete Backspace

Answers

Answer:

Molality = 1.13 m

Explanation:

Molality is defined as the moles of the solute present in 1 kilogram of the solvent.

Given that:

Mass of [tex]CH_3OH[/tex] = 26.5 g

Molar mass of [tex]CH_3OH[/tex] = 32.04 g/mol

The formula for the calculation of moles is shown below:

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

Thus,

[tex]Moles= \frac{26.5\ g}{32.04\ g/mol}[/tex]

[tex]Moles\ of\ CH_3OH= 0.8271\ moles[/tex]

Mass of water = 735 g = 0.735 kg ( 1 g = 0.001 kg )

So, molality is:

[tex]m=\frac {0.8271\ moles}{0.735\ kg}[/tex]

Molality = 1.13 m

This patient has a bone density of 820mg/cm3. What is the volume of a 25g sample?

Answers

Answer: The volume of bone for given sample is [tex]30.49cm^3[/tex]

Explanation:

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

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

Density of bone = [tex]820mg/cm^3=0.82g/cm^3[/tex]      (Conversion factor:  1 g = 1000 mg)

Mass of bone = 25 g

Putting values in above equation, we get:

[tex]0.82g/cm^3=\frac{25g}{\text{Volume of bone}}\\\\\text{Volume of bone}=30.49cm^3[/tex]

Hence, the volume of bone for given sample is [tex]30.49cm^3[/tex]

How many significant figures does the rounded ans 101 cm - 99.1 cm = 1.9 cm (unrounded) After rounding, the answer should be reported to the and thus has 0.5 atm + 9.8 atm = 10.3 atm (unrounded) After rounding, the answer should be reported to the and thus has

Answers

Explanation:

Rules for counting significant figures:

Digits from 1 to 9 are always significant and have infinite number of significant figures.All zero’s preceding the first integers are never significant. For example: 0.0078 has two significant figures.All zero’s after the decimal point are always significant. For example: 4.500, 45.00 and 450.0 all have four significant figures.All zero’s between integers are always significant. For example: 5005, 5.005 and 50.05 all have four significant figures.

1) 101 cm - 99.1 cm = 1.9 cm ≈ 2.0 cm

1.9 cm ≈ 2.0 cm

There are two significant figures that is 2 and 0.

2) 0.5 atm + 9.8 atm = 10.3 atm ≈ 10.0 atm

There are three significant figures that is 3 , 0 and 0.

How do you calculate the pH of a 1.6 M KOH solution?

Answers

Answer:

14.20

Explanation:

The pH can be calculated as:

KOH is a strong base and it will dissociate completely in the solution as:

[tex]KOH\rightarrow K^++OH^-[/tex]

According to the equation and also mentioned that KOH will dissociate completely. So,

Molarity of the hydroxide ions furnished = Molarity of the KOH = 1.6 M

So,

pOH = -log[OH⁻] = -log(1.6) = -0.20

Also,  

pH + pOH = 14  

So, pH = 14 - ( - 0.20 ) = 14.20

Attem A liquid solvent is added to a flask containing an insoluble solid. The total volume of the solid and liquid together is 91.0 mL. The liquid solvent has a mass of 26.6 g and a density of 0.865 g/mL. Determine the mass of the solid given its density is 1.75 g/mL. mass: 38.852

Answers

Final answer:

The mass of the insoluble solid is determined by first calculating the volume of the liquid solvent, then using that information to find the volume of the solid, and finally multiplying the volume of the solid by its density.

Explanation:

To determine the mass of the solid, we first calculate the volume of the liquid solvent. The volume of a substance can be calculated using the formula volume = mass/density. Thus, the volume of the liquid solvent is 26.6 g / 0.865 g/mL = approximately 30.8 mL.

Now, knowing that the total volume of the solid and liquid together is 91.0 mL, we can find out the volume of the solid as follows: 91.0 mL (total volume) - 30.8 mL (volume of the liquid) = 60.2 mL.

We can then use the given density of the solid (1.75 g/mL) to calculate its mass. According to the formula mass = volume x density, the mass of the solid is therefore 60.2 mL x 1.75 g/mL = 105.35g.

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Assume that the complete combustion of one mole of glucose to carbon dioxide and water liberates 2870 kJ/mol2870 kJ/mol ( Δ????°′=−2870 kJ/molΔG°′=−2870 kJ/mol ). If one contraction cycle in muscle requires 67 kJ67 kJ , and the energy from the combustion of glucose is converted with an efficiency of 39%39% to contraction, how many contraction cycles could theoretically be fueled by the complete combustion of one mole of glucose? Round your answer to the nearest whole number.

Answers

Answer:

Number of contraction cycles that could theoretically be fueled by the complete combustion of one mole of glucose is around 17

Explanation:

Energy released during the complete combustion of 1 mole glucose = 2870 kJ

Energy required/muscle contraction cycle = 67 kJ/contraction

Energy conversion efficiency = 39%

Actual amount of energy converted to contraction per mole of glucose is:

[tex]=\frac{39}{100} *2870 kJ=1119.3 kJ[/tex]

Total contraction cycles fueled by the above energy is:

[tex]=\frac{1119.3\ kJ}{67\ kJ/contraction} =16.7\ i.e.\ around\ 17\ contractions[/tex]

Suppose you dissolve 0.1 moles of 1-aminobutane (NH2-CH2-CH2-CH2-CH3) in 1.0 liter of water. a. What are all of the molecules and ions you would expect to find in the solution? b. Which two of the above will be found in the greatest total amount? Hint: what is the approximate pKa of an amino group?) c. Which one of the above will be found in the least total amount?

Answers

Answer:

a) The two compounds you will expect in these solution are 1-aminobutane and its conjugate acid.

b) The greatest total amount is of 1-aminobutane.

c) The least total amount is of the conjugate acid.

Explanation:

The equilibrium in water of 1-aminobutane is:

CH₃(CH₂)₃NH₃⁺ ⇄ CH₃(CH₂)₃NH₂ + H⁺

a) The two compounds you will expect in these solution are 1-aminobutane and its conjugate acid.

b) The equlibrium constant is: K = 1,66x10⁻¹¹.

That means you will have 1-aminobutane:Conjugate acid in a ratio of 6x10¹⁰ : 1 .

The greatest total amount is of 1-aminobutane

c) Thus, The least total amount is of the conjugate acid.

I hope it helps!

When a_ ( sublimes, a converts to a_ (ii) This phase change is_ because energy is_(iv) when the intermolecular forces_ [M]__ (i) (ii) (iii) (iv) (v) a. solid Liquid endothermic absorbed form b. solid Gas endothermic absorbed break c. solid Liquid exothermic released break solid Gas exothermic released form e gas Solid exothermic released form d

Answers

Answer: Option (b) is the correct answer.

Explanation:

Sublimation is defined as the process in which a solid substance changes directly into gaseous phase without undergoing into liquid phase.

For example, when naphthalene balls are kept in a trunk or cupboard then after some days or months it changes into gas. Hence, they become consumed.

Also during this process energy is being absorbed by the solid substance because for breaking bonds energy is always absorbed by a substance. Only then it can change into liquid or vapor state.

So, a chemical process in which energy is being absorbed by a reaction is known as endothermic reaction. And, if heat energy is being released by the reaction then it is known as an exothermic reaction.

Therefore, we can conclude that when a solid sublimes, a solid converts to a gas. This phase change is endothermic because energy is absorbed when the intermolecular forces break.      

A decomposition reaction has a rate constant of 0.0012 yr^-1

(a) What is the half-life of the reaction?

(b) How long does it take for [reactant] to reach 15% of its original value?

Answers

Answer:

A) 577.6 years

B) 1580.9 years

Explanation:

You can know the order of a reaction given the units of the rate constant:

For a 0 order reaction are M/s=(mol/L*s)For a 1 order reaction are 1/sFor a 2 order reaction are 1/M*s=(L/mol*s)

Then we know that this is a first order reaction because years is a unit of time as well as seconds. The half life of a first order reaction is given by:

[tex]t_{1/2}=\frac{ln(2)}{k}[/tex]

Here you can solve for the half life:

[tex]t_{1/2}=\frac{ln(2)}{0.0012 years^-1}=577.6 years[/tex]

Now for the rate law for a first order reaction is:

[tex][A]=[A_{o} ]*e^{-kt}[/tex]

Then, if you want to know how long does it take to reach a certain value you solve for time:

[tex]t=-ln([A]/[A]_{o}) *1/k[/tex]

The 15% of its original value is [tex]0.15*[A_{o}][/tex]

You solve for time:

[tex]t=-ln(0.15) *1/0.0012 years^-1=1580.9 years[/tex]

Hope it helps!

In the Decomposition reaction, a reactant gets split into two or more products. The half-life of the reaction is 577.6 years and the time taken to reach 15% of the original is 1580.9 years.

What is half-life?

Half-life is the time taken by the radioactive isotope to decay one-half of its total amount.

From the units of the rate of the reaction, the order can be estimated as, for a first-order reaction unit is 1/s.

The half-life of the reaction can be given as,

[tex]\rm t\dfrac{1}{2} = \dfrac{\rm ln (2)}{k}[/tex]

The rate constant (k) for the decomposition reaction is 0.0012 per year.

Substituting values in the above equation:

[tex]\begin{aligned} \rm t\dfrac{1}{2} &= \dfrac{\rm ln (2)}{0.0012}\\\\&= 577.6\;\rm years\end{aligned}[/tex]

The rate law of the first-order reaction is given as,

[tex]\begin{aligned}\rm [A] = [A_{o}] \times e^{-kt}\\\\\rm t = -ln ([A][A_{o}]) \times \dfrac{1}{k}\end{aligned}[/tex]

Solving for time (t):

[tex]\begin{aligned}\rm t &= \rm -ln (0.15) \times \dfrac{1}{0.0012}\\\\& = 1580.9\;\rm years\end{aligned}[/tex]

Therefore, the half-life is 577.6 years and the time taken is 1580.9 years.

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Consider water at 1400 kPa and 200 C. What is the specific volume (in m3 /kg)?

Answers

Answer:

Vw = 2.80907 E-3 m³/Kg

Explanation:

specific volume is the inverse of density:

⇒ Vw = 1 / ρ

∴ water at:

⇒ P = 1400 KPa * ( 1000 Pa / KPa ) = 1400000 Pa

⇒ T = 200°C = 473 K

Water at these conditions is found as saturated steam, specific volume would be:

⇒ Vw = R*T / P

∴ R = 8.3144 Pa.m³/ Kg.K

⇒ Vw = (( 8.3144 ) * ( 473)) / 1400000

⇒ Vw = 2.80907 E-3 m³/Kg

Maressa has 150 g of a NaCl solution that has concentration of 15.0 % NaCl (mm). How much NaCl is dissolved in that solution? O a. 0.05 O b. 1.08 Oc2.25 O d. 225 Oe. None of the above.

Answers

Answer:

I don't know if option e. is 22,5 g. because that is the correct answer for this. 15% NaCl (mm) gives you the information that in 100 g of solution, you have 15 g of solute. So, if in 100 g of solution you have 15 g of NaCl, in 150 g of solution, how much. Try the rule of three.

Explanation:

If a microwave oven emits 1,100 J of energy every second, how long will it take for the microwave to emit 4650 J of energy? Express your answer in seconds using the correct number of significant figures. Do not enter your answer using scientific notation.

Answers

It will take the microwave approximately 4.2 seconds to emit a total of 4650 J of energy, calculated by dividing the total energy by the power it emits per second, using two significant figures.

To calculate how long it will take for a microwave emitting 1,100 J of energy every second to emit a total of 4650 J, we can use the formula:

Time (s) = Total Energy (J) / Power (J/s).

Substituting the given values:

Time (s) = 4650 J / 1,100 J/s = 4.22727272727... seconds

Since the microwave emits energy at a rate of 1,100 J every second, we want to express the answer with the same number of significant figures as the least precise measurement, which is two significant figures. Therefore, the time should be rounded to 4.2 seconds.

Components A and B form ideal solution. At 350 K, a liquid mixture containing 40% (mole) A is in equilibrium with a vapour containing 70% (mole) A. If the vapour pressure of A at 350 K is 70 kPa, what is the vapour pressure of B? (b) 20 kPa (d) 12 kPa (а) 25 kPa (c) 40 kPa

Answers

Answer : The correct option is, (b) 20 kPa

Explanation :

The Raoult's law for liquid phase is:

[tex]p_A=x_A\times p^o_A[/tex]     .............(1)

where,

[tex]p_A[/tex] = partial vapor pressure of A

[tex]p^o_A[/tex] = vapor pressure of pure substance A

[tex]x_A[/tex] = mole fraction of A

The Raoult's law for vapor phase is:

[tex]p_A=y_A\times p_T[/tex]      .............(2)

where,

[tex]p_A[/tex] = partial vapor pressure of A

[tex]p_T[/tex] = total pressure of the mixture

[tex]y_A[/tex] = mole fraction of A

Now comparing equation 1 and 2, we get:

[tex]x_A\times p^o_A=y_A\times p_T[/tex]

[tex]p_T=\frac{x_A\times p^o_A}{y_A}[/tex]    ............(3)

First we have to calculate the total pressure of the mixture.

Given:

[tex]x_A=0.4[/tex] and [tex]x_B=1-x_A=1-0.4=0.6[/tex]

[tex]y_A=0.7[/tex] and [tex]y_B=1-y_A=1-0.7=0.3[/tex]

[tex]p^o_A=70kPa[/tex]

Now put all the given values in equation 3, we get:

[tex]p_T=\frac{0.4\times 70kPa}{0.7}=40kPa[/tex]

Now we have to calculate the vapor pressure of B.

Formula used :

[tex]x_B\times p^o_B=y_B\times p_T[/tex]

[tex]p^o_B=\frac{y_B\times p_T}{x_B}[/tex]

Now put all the given values in this formula, we get:

[tex]p^o_B=\frac{0.3\times 40kPa}{0.6}=20kPa[/tex]

Therefore, the vapor pressure of B is 20 kPa.

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