what combination will result in a acidic solution? a reaction of strong acid and strong base b reaction of strong acid and weak base c a reaction of weak acid and strong base d reaction of weak acid and weak base

Answer:

1.8g

Explanation:

Initial volume = 43.5ml

Final volume = 49.4ml

Mass = 10.88g

Density = ?

Volume = Final volume - initial volume

= 49.4 - 43.5

= 5.9ml

Density = Mass/volume

Density = 10.88/5.9

= 1.8g/ml

Answer:

True

Explanation:

The penetrating ability of electrons in the orbitals is in the order s > p > d > f

An electron in a 3s orbital is closer to the nucleus than the one in a 3p orbital and as a result, there will be lesser shielding effect on it. This low shielding effect experienced by the 3s electron gives it a high penetration ability and hence will be able to easily penetrate regions occupied by core electrons. Conversely, the 3p orbital is farther away from the nucleus, electrons revolving around it are highly shielded which limits their ability to penetrate regions of core electrons.

Note that the maximum electrons that the s orbital can accommodate is 2 while p orbital can accommodate a maximum of 8.

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]

Answer:

How are bees able to fly?

Explanation:

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) =

Answer:

-226.0kJ = ΔH°f COCl₂(g)

Explanation:

Using Hess' law, it is possible to obtain the enthalpy of formation of a substance from the enthalpy change of a reaction and the other enthalpies of formation involved in the reaction.

For the reaction:

CO(g) + Cl₂(g) → COCl₂(g)

Hess's law is:

ΔHr = -115.5kJ = ΔH°f COCl₂(g) - (ΔH°f CO(g) + ΔH°f Cl₂(g))

ΔH°f CO(g) is -110.5kJ/mol

ΔH°f Cl₂(g) is 0 kJ/mol

Replacing in Hess's law:

-115.5kJ = ΔH°f COCl₂(g) - (-110.5kJ/mol + 0kJ/mol)

-115.5kJ = ΔH°f COCl₂(g) + 110.5kJ

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]

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]

Answer:

Yes.

Explanation:

The nuclear equation {226/88 Ra → 222/26 Rn + 4/2 He} is balanced. As we know that an alpha particle is identical to a helium atom. This implies that if an alpha particle is eliminated from an atom's nucleus, an atomic number of 2 and a mass number of 4 is lost.

Therefore, the equation will be reduced to:

226 - 4 = 222

88 - 2 = 86

Hence, the equation is balanced.

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

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]

Answer: 323.61 g of [tex]Ag[/tex] will be produced

Explanation:

The given balanced chemical reaction is :

[tex]Cu+2AgNO_3\rightarrow Cu(NO_3)_2+2Ag[/tex]

According to stoichiometry :

2 moles of [tex]AgNO_3[/tex] require 1 mole of [tex]Cu[/tex]

Thus 3.00 moles of  [tex]AgNO_3[/tex] will require=[tex]\frac{1}{2}\times 3.00=1.50moles[/tex]  of [tex]Cu[/tex]

Thus [tex]AgNO_3[/tex] is the limiting reagent as it limits the formation of product.

As 2 moles of [tex]AgNO_3[/tex] give =  2 moles of [tex]Ag[/tex]

Thus 3.00 moles of [tex]AgNO_3[/tex] give =[tex]\frac{2}{2}\times 3.00=3.00moles[/tex]  of [tex]Ag[/tex]

Mass of [tex]Ag=moles\times {\text {Molar mass}}=3.00moles\times 107.87g/mol=323.61g[/tex]

Thus 323.61 g of [tex]Ag[/tex] will be produced from the given moles of both reactants.

Answer:

B). 2KNO3 Is your answer

Explanation:

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.

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.

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]

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.

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.

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]

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.

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.

Answer:

A

Explanation:

Answer:

(aq) meaning aqueous solution

Explanation:

hope it helps .

The question is incomplete; the complete question is;

Before running a titration, you calculate the expected endpoint. However, when performing the experiment, you pass the expected endpoint with no visible color change. What is the most likely problem with the titration set- up? Select one

a) There was an air bubble in the burette tip.

b) There is not enough indicator in the analyte.

c) The burette tip is leaking titrant into the analyte.

d) The analyte solution is being stirred too quickly

Answer:

a) There was an air bubble in the burette tip.

Explanation:

Titration involves the determination of the concentration of a solution by measuring the volumes of reactants used in the reaction. The concentration of one of the species must be known while the concentration of the other specie is to be determined by the volumetric analysis.

However, if there are air bubbles at the tip of the burette, this will cause less volume of titrant to be delivered from the burette than expected. Hence, the analyst may think that a certain volume of titrant has been delivered while in reality, a lesser volume was actually delivered due to the air bubbles present. Hence, the analyst may pass the expected endpoint without any colour change because of this problem.

From the available options to the question:

a) There was an air bubble in the burette tip.

b) There is not enough indicator in the analyte.

c) The burette tip is leaking titrant into the analyte.

d) The analyte solution is being stirred too quickly

The most likely problem with the titration setup that could make one to pass the expected endpoint with no visible color change would be if there is not enough indicator in the analyte. The correct option would be B.

A suitable quantity (in drops) of the indicator should be added to the analyte in the conical flask before carrying out a titration. The color of indicators changes quickly near their pKa.

If too few drops of the indicator is used, the color change will be too faint to be obvious and the endpoint will be exceeded. If too many drops of the indicator is used, the final pH of the reaction would be affected and the titer value will be inaccurate.

In this case, the expected endpoint has been exceeded without any color change. The most likely problem would, therefore, be that there is not enough indicator in the analyte.

More on indicators can be found here: https://brainly.com/question/4050911

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

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.

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:

https://brainly.com/question/13404978

#SPJ2

Answer:

Answers are in the explanation.

Explanation:

A buffer is defined as the aqueous mixture of a weak acid and its conjugate base or vice versa. Having this in mind:

0.31 M ammonium bromide + 0.39 M ammonia . Is a good buffer system because ammonia is a weak base and its conjugate base, ammonium ion is in the solution.

0.31 M nitrous acid + 0.25 M potassium nitrite . Is a good buffer system because nitrous acid is the weak acid and nitrite ion its conjugate base.

0.21 M perchloric acid + 0.21 M potassium perchlorate . Perchloric acid is a strong acid. Thus, Is not a good buffer system.

0.16 M potassium cyanide + 0.21 M hydrocyanic acid . Hydrocyanic acid is a weak acid and cyanide ion is its conjugate base. Is a good buffer system.

0.14 M hypochlorous acid + 0.21 M sodium hypochlorite . Hypochlorous acid is a weak acid and hypochlorite ion its conjugate base. Is a good buffer system.

0.13 M nitrous acid + 0.12 M potassium nitrite . Is a good buffer system as I explained yet.

0.15 M potassium hydroxide + 0.22 M potassium bromide . Potassium hydroxide is a strong base. Is not a good buffer system.

0.23 M hydrobromic acid + 0.20 M potassium bromide . HBr is a strong acid. Is not a good buffer system.

0.34 M calcium iodide + 0.29 M potassium iodide . CaI and KI are both salts, Is not a good buffer system.

0.33 M ammonia + 0.30 M sodium hydroxide . Ammonia is a weak base but its conjugate base ammonium ion is not in solution. Is not a good buffer system.

0.20 M nitrous acid + 0.18 M potassium nitrite . Is a good buffer system.

0.30 M ammonia + 0.34 M ammonium bromide . Ammonia and ammonium in solution, Good buffer system.

0.29 M hydrobromic acid + 0.22 M sodium bromide . HBr is a strong acid, is not a good buffer system.

0.17 M calcium hydroxide + 0.28 M calcium bromide . CaOH is a strong base, is not a good buffer system.

0.34 M potassium iodide + 0.27 M potassium bromide. KI and KBr are both salts, is not a good buffer system.

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]

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]