The compound diborane can be used as a rocket propellant. What is the percent composition of boron in diborane (B2H6)?
Α.22%
Β.25%
C.39%
D.75%
E.78%

Answers

Answer 1

Answer:

Explanation:

atomic weight of B= 10.8

atomic weight of h = 1.0

B2H6 = 2*10.81+6*1= 27.62

out of total 27.62 the Boron contrubtion was = 2*10.81= 21.62

For 27.62 B2H6  ---- 21.62 is B

for 100 B2H6  = 21.62*100 /27.62

                       = ...........%

can you calcualte and the coorect answer ?

Answer 2

Answer:

The correct answer is E. 78%.

[Image Below]

The Compound Diborane Can Be Used As A Rocket Propellant. What Is The Percent Composition Of Boron In

Related Questions

Duncan knows that it takes 36400 cal of energy to heat a pint of water from room temperature to boiling. However, Duncan has prepared ramen noodles so many times he does not need to measure the water carefully. If he happens to heat 0.800 pint of room-temperature water, how many kilojoules of heat energy will have been absorbed by the water at the moment it begins to boil?

Answers

Answer:

[tex]\large \boxed{\text{122 000 J}}[/tex]

Explanation:

1. Calculate the energy needed

[tex]\text{Energy} = \text{0.800 pt} \times \dfrac{\text{36 400 cal}}{\text{1 pt}} = \text{ 29 120 cal}[/tex]

2. Convert calories to joules

[tex]\text{Energy} = \text{29 120 cal} \times \dfrac{\text{4.184 J}}{\text{1 cal}} = \textbf{122 000 J}\\\\\text{The water will have absorbed $\large \boxed{\textbf{122 000 J}}$}[/tex]

If one contraction cycle in muscle requires 55 kJ55 kJ , and the energy from the combustion of glucose is converted with an efficiency of 35%35% 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:

18 moles

Explanation:

Here the combustion of one mole of glucose ----> carbon dioxide + water, releases 2870 kilojoules / moles.

_______________________________________________________

With one contraction cycle requiring 55 kilojoules,

2870 / 55 ≈ 52.18

And with the efficiency being 35 percent,

52.1818..... * 0.35 = ( About ) 18 moles

Hope that helps!

Harvey kept a balloon with a volume of 348 milliliters at 25.0˚C inside a freezer for a night. When he took it out, its new volume was 322 milliliters, but its pressure was the same. If the final temperature of the balloon is the same as the freezer’s, what is the temperature of the freezer?

Answers

Answer:

[tex]T2=276K[/tex]

Explanation:

Given:

Initial volume of the balloon V1 = 348 mL

Initial temperature of the balloon T1 = 255C

Final volume of the balloon V2 = 322 mL

Final temperature of the balloon T2 =

To calculate T1 in kelvin

T1= 25+273=298K

Based on Charles law, which states that the volume of a given mass of a ideal gas is directly proportional to the temperature provided that the pressure is constant. It can be applied using the below formula

[tex](V1/T1)=(V2/T2)[/tex]

T2=( V2*T1)/V1

T2=(322*298)/348

[tex]T2=276K[/tex]

Hence, the temperature of the freezer is 276 K

Answer: 276 kelvins

Explanation:

Why need to add NaAlF6 to Al2O3?

Answers

The electrolysis of pure alumina faces some difficulties. Pure alumina is a bad conductor of electricity. The fusion temperature of pure alumina is about 2000 C and at this temperature when electrolysis is carried out on the fused mass, the metal formed vaporizes, as the boiling point of aluminium is 1800 ∘C. These difficulties are overcome by using a mixture containing alumina, cryolite (Na 3AlF 6), and fluorspar (CaF 2).


So in the electrolytic reduction of alumina, cryolite is added along with fluorspar to–
– decrease melting point of alumina
– decrease viscosity of electrolyte (CaF
2is used) – increase conductivity
Hope this helps

Choose all that apply. According to the theory of matter proposed by John Dalton: all atoms of the same element have the same mass combinations of atoms create chemical change all matter is made up of atoms all atoms of the same element have the same size

Answers

Answer: all matter is made up of atoms

all atoms of the same element have the same mass combinations of atoms create chemical change

all atoms of the same element have the same size

Explanation:

Dalton's Atomic theory suggests that all matter are made up of atoms. All atoms are made up of same elements. The atoms of the same elements will have similar physical and chemical properties. The atoms will have same size, mass, and will show similar chemical changes. The atoms in the elements are indestructible blocks and indivisible.

State the effect of anion hydrolysis on the pH of water​

Answers

Answer:

Depending on the anions and cations present within a hydrolysis reaction, the solution can be more... ... This lesson will explain how this occurs. ... that could react with water and create products that affect the characteristics of the solution.

Answer:

Salts of weak bases and strong acids do hydrolyze, which gives it a pH less than 7. This is due to the fact that the anion will become a spectator ion and fail to attract the H+, while the cation from the weak base will donate a proton to the water forming a hydronium ion.

Explanation:

I hope this is the answer your looking for

Under certain conditions, the substances zinc oxide and water combine to form zinc hydroxide. If 30.1 grams of zinc oxide and 6.7 grams of water combine to form zinc hydroxide, how many grams of zinc hydroxide must form

Answers

Answer:

36.8g of Zinc hydroxide

Explanation:

Based on the reaction:

ZnO + H₂O → Zn(OH)₂

Where 1 mole of zinc oxide reacts with 1 mole of water to produce 1 mole of zinc hydroxide.

Moles of 30.1g of ZnO (FW = 81.38g/mol) are:

30.1g ZnO ₓ (1mol / 81.38g) = 0.370 moles of ZnO

And moles of 6.7g of H₂O (FW = 18.01g/mol) are:

6.7g H₂O ₓ (1mol / 18.01g) = 0.372 moles of H₂O

As 1 mole of ZnO reacts per mole of H₂O, limiting reactant is ZnO because has a less number of moles than water.

Thus, moles of Zn(OH)₂ produced are 0.370 moles.

As Molar mass of Zinc hydroxide is 99.424 g/mol, there are formed:

0.370 moles Zn(OH)₂ ₓ (99.424g / mol) =

36.8g of Zinc hydroxide

Why does a chemical change occur when copper is heated?

Answers

Answer:

When copper is heated, it decomposes to form copper oxide and carbon dioxide. It is an endothermic reaction, which means that it absorbs heat. When heated, copper is easily bent or molded into shapes.

Explanation:

The cytochromes are heme‑containing proteins that function as electron carriers in the mitochondria. Calculate the difference in the reduction potential (ΔE∘′) and the change in the standard free energy (ΔG∘′) when the electron flow is from the carrier with the lower reduction potential to the higher. cytochrome c1 (Fe3+)+e−↽−−⇀cytochrome c1 (Fe2+)E∘′=0.22 V cytochrome c (Fe3+)+e−↽−−⇀cytochrome c (Fe2+)E∘′=0.254 V Calculate ΔE∘′ and ΔG∘′ .

Answers

Complete Question

The complete question is shown on the first uploaded image

Answer:

The change in reduction potential is  [tex]\Delta E^o=E^o_{cell} = 0.034 V[/tex]

The change in standard free energy is  [tex]\Delta G^o = -3.2805 \ KJ/mol[/tex]

Explanation:

From the question we are told that

At the anode

      [tex]cytochrome \ c_1 \ (Fe^{3+}) + e^-[/tex]⇔[tex]cytochrome \ c_1 \ (Fe^{2+}) \ \ E^o = 0.22 \ V[/tex]

At the cathode

      [tex]cytochrome \ c \ (Fe^{3+}) + e^-[/tex]⇔[tex]cytochrome \ c \ (Fe^{2+}) \ \ E^o = 0.254 \ V[/tex]

The difference in the reduction potential is mathematically represented as

     [tex]\Delta E^o = E^o_{cathode} - E^o_{anode}[/tex]

substituting values

      [tex]\Delta E^o = 0.254 - 0.220[/tex]

     [tex]\Delta E^o=E^o_{cell} = 0.034 V[/tex]

The change in the standard free energy is mathematically represented as

      [tex]\Delta G^o = -n * F * E^o_{cell}[/tex]

Where  F is the Faraday constant with value  F = 96485 C

and  n i the number of the number of electron = 1

   So

       [tex]\Delta G^o = -(1) * 96485 * 0.034[/tex]

       [tex]\Delta G^o = -3.2805 \ KJ/mol[/tex]

During the winter months, many locations experience snow and ice storms. It is a common practice to treat roadways and sidewalks with salt, such as NaCl . If a 11.3 kg bag of NaCl is used to treat the sidewalk, how many moles of NaCl does this bag contain

Answers

Answer:The moles of NaCL in the 11.3kg bag  is   193.36moles

Explanation:

Given that  a bag of NaCl = 11.3kg

1kg = 1000g

therefore 11.3 kg = 11,300g

Remember that

No of moles = mass of subatance/ molar mass of substance

The molar mass of NaCl = Na + Cl= 22.989769  + 35.453 =58.442769≈ 58.44g/mol

No of moles = mass of subatance/ molar mass of substance

                    = 11300g/ 58.44g/mol  =  193.36 moles

The moles of NaCL in the 11.3kg bag  is   193.36 moles.

describe how would you use chromatography to show whether blue ink contains a single purple dye or a mixture of dyes​

Answers

Explanation:

if the solution placed on the chromatography is pure there will be formation of one spot from the baseline and will go farthest to the front line unlike the impure one

With ink chromatography, a small amount of ink is added to the paper, one end is submerged in water, and the different colors of the ink are revealed as the water moves up the paper. All of this is made possible by the water base and variety of salabilities or densities that make up ink.

What is chromatography ?

Separating mixture's constituent parts by chromatography is a method. The mixture is dissolved in a material known as the mobile phase to start the process, which then transports it through a material known as the stationary phase.

A little dot of the ink to be separated is placed at one end of a strip of filter paper to perform ink chromatography. The paper strip's opposite end is submerged in a solvent. The solvent moves up the paper strip, dissolving the chemical combination as it goes and pulling it up the paper.

Throughout the experiment, the dyes are pulled along by the mobile phase (water) as it gently advances up the stationary phase (paper).

Thus, With ink chromatography, a small amount of ink is added to the paper, one end is submerged in water.

To learn more about chromatography, follow the link;

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When ethanol, C2H5OH (a component in some gasoline mixtures) is burned in air, one molecule of ethanol combines with three oxygen molecules to form two CO2 molecules and three H2O molecules.


A) Write the balanced chemical equation for the reaction described.


B) How many molecules of CO2 and H2O would be produced when 2 molecules ethanol are consumed? Equation?


C) How many H2O molecules are formed, then 9 O2 molecules are consumed? What conversion factor did you use? Explain!


D) If 15 ethanol molecules react, how many molecules O2 must also react? What conversion factor did you use? Explain!

Answers

Answer:

1) C2H5OH(l)+3O2(g)⟶2CO2(g)+3H2O(l)

2) four molecules of CO2 will be produced and six molecules of water

3)9 molecules of water are formed when 9 molecules of oxygen are consumed.

4) 45 molecules of oxygen

Explanation:

The balanced chemical reaction equation is shown here and must guide our work. When ethanol is burned in air, it reacts as shown;

C2H5OH(l)+3O2(g)⟶2CO2(g)+3H2O(l)

Hence, if we use 2 molecules of ethanol, the balanced reaction equation will look like this;

2C2H5OH(l)+6O2(g)⟶4CO2(g)+6H2O(l)

Hence four molecules of CO2 are formed and six molecules of water are formed

From the balanced stoichiometric equation;

3 molecules of oxygen yields 3 molecules of water

Therefore, 9 molecules of oxygen will yield 9 × 3/3 = 9 molecules of water

Therefore, 9 molecules of water are formed when 9 molecules of oxygen are consumed.

From the reaction equation;

1 molecule of ethanol reacts with 3 molecules of oxygen

Therefore 15 molecules of ethanol will react with 15 × 3/1 = 45 molecules of oxygen

The reaction of 15 moles carbon with 30 moles O2 will
result in a theoretical yield of __ moles CO2.

Answers

Answer:

15 moles.

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

[tex]C+O_2\rightarrow CO_2[/tex]

Clearly, since carbon and oxygen are in a 1:1 molar ratio, 15 moles of carbon will completely react with 15 moles of oxygen, therefore 15 moles of oxygen remain as leftovers. In such a way, since carbon and carbon dioxide are also in a 1:1 molar ratio, the theoretical yield of carbon dioxide is 15 moles based on the stoichiometry:

[tex]n_{CO_2}=15molC*\frac{1molCO_2}{1molC} \\\\n_{CO_2}=15molCO_2[/tex]

Best regards.

An experimenter studying the oxidation of fatty acids in extracts of liver found that when palmitate (16:0) was provided as substrate, it was completely oxidized to CO2. However, when undecanoic acid (11:0) was added as substrate, incomplete oxidation occurred unless he bubbled CO2 through the reaction mixture. The addition of the protein avidin, which binds tightly to biotin, prevented the complete oxidation of undecanoic acid even in the presence of CO2, although it had no effect on palmitate oxidation. Explain these observations in light of what you know of fatty acid oxidation reactions.

Answers

Answer:

Even-number fatty acids such as palmitate undergoes complete β-oxidation in the liver motochondria to CO₂ because the product, acetyl-CoA can enter the TCA cycle.

Oxidation of odd-number fatty acids such as undecanoic acid yields acetyl-CoA + propionyl-CoA in their last pass. Propionyl-CoA requires additional reactions including carboxylation in order to be able to enter the TCA cycle.

The reaction CO2 + propionyl-CoA ----> methylmalonyl-CoA is catalyzed by propionyl-CoA carboxylase, a biotin-containing enzyme, which is inhibited by avidin.  Palmitate oxidation however, does not involve carboxylation.

Explanation:

Even-number fatty acids such as palmitate undergoes complete β-oxidation in the liver motochondria to CO₂ because their oxidation product, acetyl-CoA, can enter the TCA cycle where it is oxidized to CO₂.

Undecanoic acid is an odd-number fatty acid having 11 carbon atoms. Oxidation of odd-number fatty acids such as undecanoic acid yields a five -carbon fatty acyl substrate for their last pass through β-oxidation which is oxidized and cleaved into acetyl-CoA + propionyl-CoA. Propionyl-CoA requires additional reactions including carboxylation in order to be able to enter the TCA cycle. Since oxidation is occuring in a liver extract, CO₂ has to be externally sourced in order for the carboxylation of propionyl-CoA to proceed and thus resulting in comlete oxidation of undecanoic acid.

The reaction CO2 + propionyl-CoA ----> methylmalonyl-CoA is catalyzed by propionyl-CoA carboxylase, a biotin-containing enzyme.  The role of biotin is to activate the CO₂ before its tranfer to the propionate moiety. The addition of the protein avidin prevents the complete oxidation of undecanoic acid by  binding tightly to biotin, hence inhibiting the activation and transfer of CO₂ to propionate.

Palmitate oxidation however, does not involve carboxylation, hence addition of avidin has no effect on its oxidation.

A chemist prepares a solution of barium chlorate by measuring out of barium chlorate into a volumetric flask and filling the flask to the mark with water. Calculate the concentration in of the chemist's barium chlorate solution. Be sure your answer has the correct number of significant digits.

Answers

Complete Question

A chemist prepares a solution of barium chlorate BaClO32 by measuring out 42.g of barium chlorate into a 500.mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in /molL of the chemist's barium chlorate solution. Be sure your answer has the correct number of significant digits.

Answer:

The concentration is [tex]C = 0.28 \ mol/L[/tex]

Explanation:

   From the question we told that

     The mass of [tex]Ba(ClO_{3})_2[/tex] is  [tex]m_b = 42 \ g[/tex]

     The volume of the solution  [tex]V_s = 500 mL = 500*10^{-3} L[/tex]

Now the number f moles of  [tex]Ba(ClO_{3})_2[/tex] in the solution is mathematically represented as

        [tex]n = \frac{m_b}{Z_b}[/tex]

Where  [tex]Z_b[/tex] is the molar mass of [tex]Ba(ClO_{3})_2[/tex] which a constant with a value

             [tex]Z_b = 304.23 \ g/mol[/tex]

Thus

       [tex]n = \frac{42}{304.23}[/tex]

      [tex]n = 0.14 \ mol[/tex]

The concentration of [tex]Ba(ClO_{3})_2[/tex]  in the solution is mathematically evaluated as

       [tex]C = \frac{n}{V_2}[/tex]

substituting  values  

      [tex]C = \frac{0.14}{500*10^{-3}}[/tex]

      [tex]C = 0.28 \ mol/L[/tex]

how many grams are there in 9.4x10^25 molecules of H2

Answers

Answer:

You start with 9.4 x 1025 molecules of H2.

You know that an Avogadro's number of molecules of H2 has a mass of 2.0 g.

To solve, 9.4 x 1025 molecules H2 x (2.0 g H2 / 6.023 x 1023 molecules H2) = 312. g H2

Explanation:

The iceman known as Otzi was discovered on a mountain on the Austrian-Italian border. Samples of his hair and bones had carbon-14 activity that was about 12.5% of that present in new hair or bone. How long ago did Otzi live if the half-life for C-14 is 5730 years

Answers

Answer:

1432.5  years

Explanation:

The rate of decay of a radioactive isotope is the characteristics of the isotope and it is usually expressed in terms of its half-life.

The half-life of a radioactive element is the time taken for  half of the total number of atoms in a given sample of the element to decay or the time taken for the intensity of radiation to fall to half of its original value.

From the given question.

Since the same of his bones had a carbon-14 activity that was about 12.5% of that present in new hair or bone.

Thus; the time taken to reduce the amount of the sample to one-quarter  of its amount(12.5%) = the half life for C-14 (5730 years)

The time taken for how long Otzi live = 5730/4  = 1432.5  years

a solid x when heated gives up a brown gas. If x is soluble in excess sodium hydroxide solution but insoluble in excess ammonium hydroxide solition. What is X?

Answers

Answer:

Lead (ii) nitrate

Explanation:

It is soluble in sodium hydroxide but insoluble in aqueous ammonia

When heated it produces nitrogen (IV) oxide that isbrown in colour

A common laboratory reaction is the neutralization of an acid with a base. When 31.8 mL of 0.500 M HCl at 25.0°C is added to 68.9 mL of 0.500 M NaOH at 25.0°C in a coffee cup calorimeter (with a negligible heat capacity), the temperature of the mixture rises to 28.2°C. What is the heat of reaction per mole of NaCl (in kJ/mol)? Assume the mixture has a specific heat capacity of 4.18 J/(g·K) and that the densities of the reactant solutions are both 1.07 g/mL. Enter your answer to three significant figures in units of kJ/mol.

Answers

Answer:

The correct answer to the following question will be "90.6 kJ/mol".

Explanation:

The total reactant solution will be:

[tex](31.8 \ mL+68.9 \ mL)\times 1.07\ g/mL = 107.74 \ g[/tex]

The produced energy will be:

[tex]=4.18 \ J/(gK)\times 107.74 \ g\times (28.2-25.0)K[/tex]

[tex]=450.35\times 3.2[/tex]

[tex]=1441.12 \ J[/tex]

The reaction will be:

⇒  [tex]HCl+NaOH \rightarrow NaCl+H_{2}O[/tex]

Going to look at just the amounts of reactions with the same concentrations, we notice that they're really comparable.  

Therefore, the moles generated by NaCl will indeed be:

=  [tex](\frac{31.8}{1000} \ L)\times (0.500 \ M \ HCl/L)\times \frac{1 \ mol \ NaCl}{1 \ mol \ HCl}[/tex]

=  [tex]0.0318\times 0.500[/tex]

=  [tex]0.0159 \ mole \ of \ NaCl[/tex]

Now,

=  [tex]\frac{1441.12 \ J}{0.0159 \ moles \ NaCl}[/tex]

=  [tex]906364.7[/tex]

=  [tex]90.6 \ KJ/mol \ NaCl[/tex]

What are the concentrations of Cu2+, NH3, and Cu(NH3)42+ at equilibrium when 18.8 g of Cu(NO3)2 is added to 1.0 L of a 0.800 M solution of aqueous ammonia? Assume that there is no volume change upon the addition of the solid, and that the reaction goes to completion and forms Cu(NH3)42+.

Answers

Answer:

Explanation:

Cu(NO₃)₂ + 4NH₃ = Cu(NH₃)₄²⁺  + 2 NO₃⁻

187.5 gm      4M           1 M

187.5 gm reacts with 4 M ammonia

18.8 g     reacts with  .4 M ammonia

ammonia remaining left after reaction

= .8 M - .4 M = .4 M .

187.5 gm reacts with 4 M ammonia   to form 1 M Cu(NH₃)₄²⁺

18.8 g reacts with .4 M ammonia  to form 0.1 M Cu(NH₃)₄²⁺  

At equilibrium , the concentration of Cu²⁺ will be zero .

concentration of ammonia will be .4 M

concentration of  Cu(NH₃)₄²⁺ formed will be 0.1 M

The bromine test (part d) is often used as an indication of unsaturation(double and triple bonds). Explain why your result for trichloroethylene and toluene were different than for the simple alkene produc

Answers

Answer:

Toluene is an aromatic compound not an alkene

Bromine test is used to determine the presence of unsaturation in the given compound. The trichloroethylene does not have any unsaturation while toluene have double bonds of benzene ring. Therefore, the Bromine test can differentiate between trichloroethylene and toluene.

What is degree of unsaturation?

The degree of unsaturation of an organic compounds can be categorised two types: saturated and unsaturated. Saturated compounds are those that have only single bonds. An unsaturated compound are those that has a double bond, triple bond, and/or ring(s).

The alkanes with only single bonds are classified as saturated whereas the alkenes and alkynes with double and triple bonds are classified as unsaturated hydrocarbons.

The degree of unsaturation formula helps in finding whether a compound is saturated or unsaturated.

In the Bromine test when the bromine solution will be added into the compound if the brown color of the solution will disappear it means the unsaturation is present in the given compound.

Therefore, the we can distinguish between trichloroethylene and toluene with bromine test.

Learn more about degree of unsaturation, here:

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Be sure to answer all parts. Three 8−L flasks, fixed with pressure gauges and small valves, each contain 4 g of gas at 276 K. Flask A contains He, flask B contains CH4, and flask C contains H2. Rank the flask contents in terms of:

Answers

Here is the complete question.

Be sure to answer all parts. Three 8−L flasks, fixed with pressure gauges and small valves, each contain 4 g of gas at 276 K. Flask A contains He, flask B contains CH4, and flask C contains H2. Rank the flask contents in terms of:  the following properties. (Use the notation >, <, or =, for example B=C>A.)

(a) pressure

(b) average molecular kinetic energy

(c) diffusion rate after the valve is opened

(d) total kinetic energy of the molecules

Answer:

Explanation:

Given that:

Three flask A,B, C:

contains a volume of 8-L

mass m = 4g    &;

Temperature = 276 K

Flask A = He

Flask B = H₂

Flask C = CH₄

a) From the ideal gas equation:

PV = nRT

where;

n = number of moles = mass (m)/molar mass (mm)

Then:

PV = m/mm RT

If  T ,m and V are constant for the three flasks ; then

P ∝ 1/mm

As such ; the smaller the molar mass the larger the pressure.

Now; since the molecular weight of CH₄ is greater than He and H₂ and also between He and H₂,  He has an higher molecular weight .

Then the order of pressure in the flask is :

[tex]\mathbf{P_B >P_A>P_C}[/tex]

where :

[tex]P_A[/tex] = pressure in the flask A

[tex]P_B[/tex] = pressure in the flask B

[tex]P_C[/tex]= Pressure in the flask C

b)

average molecular kinetic energy

We all know that  the average molecular kinetic energy  varies directly proportional to the temperature.

Thus; the given temperature = 276 K

∴ The order of the average molecular kinetic energy is [tex]\mathbf{K.E_A =K.E_B =K.E_C}[/tex]

c)

The rate of diffusion of gas is inversely proportional to the square root of it density . Here the density is given in relation to their molar mass.

So;

rate of diffusion ∝ [tex]\dfrac{1}{\sqrt{mm} }[/tex]

where;

[tex]D_A[/tex] = rate of diffusion in flask A

[tex]D_B[/tex] = rate of diffusion in flask B

[tex]D_C[/tex] = rate of diffusion in flask C

Thus; the order of the rate of diffusion = [tex]D_B[/tex]  > [tex]D_A[/tex] > [tex]D_C[/tex]

d)  total kinetic energy of the molecules .

The kinetic energy deals with how the speed of particles of a  substance determines how fast the substances will diffuse in a given set of condition.

The the order of the total kinetic energy depends on the molecular speed

Thus; the order of the total kinetic energy  for the three flask is as follows:

[tex]\mathbf{ K.E_B>K.E_A>K.E_C}[/tex]

Air is compressed from an inlet condition of 100 kPa, 300 K to an exit pressure of 1000 kPa by an internally reversible compressor. Determine the compressor power per unit mass flow rate if the device is (a) isentropic, (b) polytropic with n =1.3, (c) isothermal

Answers

Answer:

(a) [tex]W_{isoentropic}=8.125\frac{kJ}{mol}[/tex]

(b) [tex]W_{polytropic}=7.579\frac{kJ}{mol}[/tex]

(c) [tex]W_{isothermal}=5.743\frac{kJ}{mol}[/tex]

Explanation:

Hello,

(a) In this case, since entropy remains unchanged, the constant [tex]k[/tex] should be computed for air as an ideal gas by:

[tex]\frac{R}{Cp_{air}}=1-\frac{1}{k} \\\\\frac{8.314}{29.11} =1-\frac{1}{k}\\[/tex]

[tex]0.2856=1-\frac{1}{k}\\\\k=1.4[/tex]

Next, we compute the final temperature:

[tex]T_2=T_1(\frac{p_2}{p_1} )^{1-1/k}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.4}=579.21K[/tex]

Thus, the work is computed by:

[tex]W_{isoentropic}=\frac{kR(T_2-T_1)}{k-1} =\frac{1.4*8.314\frac{J}{mol*K}(579.21K-300K)}{1.4-1}\\\\W_{isoentropic}=8.125\frac{kJ}{mol}[/tex]

(b) In this case, since [tex]n[/tex] is given, we compute the final temperature as well:

[tex]T_2=T_1(\frac{p_2}{p_1} )^{1-1/n}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.3}=510.38K[/tex]

And the isentropic work:

[tex]W_{polytropic}=\frac{nR(T_2-T_1)}{n-1} =\frac{1.3*8.314\frac{J}{mol*K}(510.38-300K)}{1.3-1}\\\\W_{polytropic}=7.579\frac{kJ}{mol}[/tex]

(c) Finally, for isothermal, final temperature is not required as it could be computed as:

[tex]W_{isothermal}=RTln(\frac{p_2}{p_1} )=8.314\frac{J}{mol*K}*300K*ln(\frac{1000kPa}{100kPa} ) \\\\W_{isothermal}=5.743\frac{kJ}{mol}[/tex]

Regards.

Calculate the amount of heat (kcal) released when 50.0g of steam at 100 degrees celsius hits the skin, condenses and cools to a body temperature of 37 degrees celsius

Answers

Answer:

[tex]Q=-126.1kJ[/tex]

Explanation:

Hello,

In this case, by means of the released heat, we need to consider the cooling of water in two steps:

1. Condensation of steam at 100 °C.

2. Cooling of water from 100 °C to 37 °C.

Therefore, we need the enthalpy of condensation of water that is 40.65 2258.33 J/g and the specific heat that is 4.18 J/g°C for the same amount of cooled water to obtain:

[tex]Q=50.0g*[-2258.33\frac{J}{g}+4.18\frac{J}{g\°C}(37-100)\°C]\\\\Q=-126.1kJ[/tex]

Best regards.

Consider the insoluble compound zinc carbonate , ZnCO3 . The zinc ion also forms a complex with hydroxide ions . Write a balanced net ionic equation to show why the solubility of ZnCO3 (s) increases in the presence of hydroxide ions and calculate the equilibrium constant for this reaction. For Zn(OH)42- , Kf = 2.9×1015 . Use the pull-down boxes to specify states such as (aq) or (s).

Answers

Answer:

The net ionic equation is  [tex]ZnCO_3 _{(s)} + 4 OH^{-}_{(aq)} \to [Zn(OH)_4]^{2-} _{(aq)} + CO_3^{2-} _{(aq)}[/tex]

The equilibrium constant is  [tex]K = 4.06 *10^{4}[/tex]

Explanation:

From the question we are that

      The  [tex]K_f = 2.9 *10^{15 }[/tex]

The ionic equation is chemical represented as

    Step 1

         [tex]ZnCO_3 _{(s)}[/tex]  ⇔   [tex]Zn^{2+} _{aq} + CO_3^{2-} _{aq}[/tex]   The  solubility product constant for stage is     [tex]K_{sp} = 1.4*10^{-11}[/tex]

 Step 2

        [tex]Zn^{2+} _{(aq)} + 4 0H^{-} _{(aq)}[/tex]    ⇔  [tex][Zn(OH_4)]^{2-} _{(aq)}[/tex]  The formation constant for this step is given as [tex]K_f = 2.9 *10^{15 }[/tex]

 The net reaction is  

           [tex]ZnCO_3 _{(s)} + 4 OH^{-}_{(aq)} \to [Zn(OH)_4]^{2-} _{(aq)} + CO_3^{2-} _{(aq)}[/tex]

The equilibrium constant is mathematically evaluated as

         [tex]K = K_{sp} * K_f[/tex]

substituting values

         [tex]K = 1.4*10^{-11} * 2.9 *10^{15}[/tex]

        [tex]K = 4.06 *10^{4}[/tex]

How many moles of each product form when the given amount of each reactant completely reacts. C3H8(g)+5O2yields 3CO2(g)+4H2O(g). 4.6 moles of C3H8

Answers

Answer:  13.8 moles of [tex]CO_2[/tex] and 18.4 moles of [tex]H_2O[/tex] will be produced

Explanation:

The given balanced reaction is;

[tex]C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)[/tex]

Given : 4.6 moles of [tex]C_3H_8[/tex]

According to stoichiometry :

1 mole of [tex]C_3H_8[/tex] give = 3 moles of [tex]CO_2[/tex]

Thus 4.6 moles of  [tex]C_3H_8[/tex] will give =[tex]\frac{3}{1}\times 4.6=13.8moles[/tex]  of [tex]CO_2[/tex]

1 mole of [tex]C_3H_8[/tex]  give =  4 moles of [tex]H_2O[/tex]

Thus 4.6 moles of [tex]C_3H_8[/tex] give =[tex]\frac{4}{1}\times 4.6=18.4moles[/tex]  of [tex]H_2O[/tex]

Thus 13.8 moles of [tex]CO_2[/tex] and 18.4 moles of [tex]H_2O[/tex] will be produced from the given moles of reactant [tex]C_3H_8[/tex]

Draw structural formulas for all the alkene(s) formed by treatment of each haloalkane or halocycloalkane with sodium ethoxide in ethanol. Assume that elimination occurs by an E2 mechanism.

Answers

Answer:

Explanation:

Kindly note that I have attached the complete question as an attachment.

Here, we are told that elimination occurs by an E2 mechanism. What this means is that the hydrogen and the halogen must be above and below for the reaction to proceed.

The possible products are as follows;

Please check attachment for complete equations and diagrams of compounds too.

Look at the picture and observations below.
Observations: The bee's wings are moving very fast.
The bee's wings are much smaller than its body.
what’s the answer ?

Answers

First of all what is it asking for and second of all there is no picture of the observations, sorry can’t help with this question.

Answer:

How are bees able to fly?

Explanation:

How does the number of valence electrons in an atom relate to the element's
placement on the periodic table?
A. Elements in the same group have the same number of valence
electrons.
B. The number of valence electrons increases as the atomic number
increases.
O C. The number of valence electrons is the same for all elements on
the periodic table.
D. Elements in the same period have the same number of valence
electrons.

Answers

The believe the answer is A

An ethylene glycol solution contains 21.4 g of ethylene glycol (C2H6O2) in 97.6 mL of water. (Assume a density of 1.00 g/mL for water.) Determine the freezing point and boiling point of the solution. (Assume a density of 1.00 g/ mL for water.)

Answers

Answer: The freezing point and boiling point of the solution are [tex]-6.6^0C[/tex] and [tex]101.8^0C[/tex] respectively.

Explanation:

Depression in freezing point:

[tex]T_f^0-T^f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}[/tex]

where,

[tex]T_f[/tex] = freezing point of solution = ?

[tex]T^o_f[/tex] = freezing point of water = [tex]0^0C[/tex]

[tex]k_f[/tex] = freezing point constant of water = [tex]1.86^0C/m[/tex]

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

m = molality

[tex]w_2[/tex] = mass of solute (ethylene glycol) = 21.4 g

[tex]w_1[/tex]= mass of solvent (water) = [tex]density\times volume=1.00g/ml\times 97.6ml=97.6g[/tex]

[tex]M_2[/tex] = molar mass of solute (ethylene glycol) = 62g/mol

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

[tex](0-T_f)^0C=1\times (1.86^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}[/tex]

[tex]T_f=-6.6^0C[/tex]

Therefore,the freezing point of the solution is [tex]-6.6^0C[/tex]

Elevation in boiling point :

[tex]T_b-T^b^0=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}[/tex]

where,

[tex]T_b[/tex] = boiling point of solution = ?

[tex]T^o_b[/tex] = boiling point of water = [tex]100^0C[/tex]

[tex]k_b[/tex] = boiling point constant of water = [tex]0.52^0C/m[/tex]

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

m = molality

[tex]w_2[/tex] = mass of solute (ethylene glycol) = 21.4 g

[tex]w_1[/tex]= mass of solvent (water) = [tex]density\times volume=1.00g/ml\times 97.6ml=97.6g[/tex]

[tex]M_2[/tex] = molar mass of solute (ethylene glycol) = 62g/mol

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

[tex](T_b-100)^0C=1\times (0.52^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}[/tex]

[tex]T_b=101.8^0C[/tex]

Thus the boiling point of the solution is [tex]101.8^0C[/tex]

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