Some students have said that a buffer is like a proton (H ) sponge. Evaluate this statement and explain both how a buffer is like a sponge and how it is not like a sponge. For your explanation, include chemical equations, data, and create and explain molecular-level drawings. (Question is taken from Orgill and Sutherland, 2008.) (Saloni)

Answer:

0.01804 M KMnO4 Solution.

Explanation:

3 Percent(w/w) H2O2 Are Sold In Drug Stores As A Disinfectant. ... H2O2 In A Bottle Of Peroxide Solution Was Done By Diluting 1.50 ML To 100 ML With Water, Acidifying With Dilute H2SO4 And Titrating With A 0.01804 M KMnO4 Solution.

Answer: The percentage of potassium iodide in the sample is 2.63 %.

Explanation:

The chemical equation for the reaction of iodide ions with bromine gas follows:

[tex]I^-+3Br_2+3H_2O\rightarrow 6Br^-+IO_3^-+6H^+[/tex]                (i)

The chemical equation for the reaction of iodate ions with barium ions follows:

[tex]Ba^{2+}+2IO_3^-\rightarrow Ba(IO_3)_2[/tex]                 ......(ii)

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

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

Given mass of barium iodate = 0.0596 g

Molar mass of barium iodate = 487.13 g/mol

Using equation 1:

[tex]\text{Moles of barium iodate}=\frac{0.0596 g}{487.13 g/mol}\\\\\text{Moles of barium iodate}=1.22\times 10^{-4} moles[/tex]

By Stoichiometry of the reaction (ii):

1 mole of barium iodate is produced by 2 moles of iodate ions

So, [tex]1.22\times 10^{-4} moles[/tex] of barium iodate will be produced by [tex]\frac{2}{1}\times 1.22\times 10^{-4} =2.44\times 10^{-4}moles[/tex] of iodate ions

By the stoichiometry of the reaction (i):

1 mole of iodate ions are produced by 1 moles of iodine ions

So, [tex]2.44\times 10^{-4} moles[/tex] of iodate ions will be produced by [tex]\frac{1}{1}\times 2.44\times 10^{-4} =2.44\times 10^{-4}moles[/tex] of iodine ions

Moles of potassium iodide = Moles of iodide ions = [tex]2.44\times 10^{-4}[/tex]

Since, the molar mass of potassium iodide = 166 g/mol

Using equation 1:

[tex]\text{Mass of potassium iodide}=2.44\times 10^{-4}mol\times 166 g/mol\\\\\text{Mass of potassium iodide}=0.0405 g[/tex]

To calculate the percentage by mass of a substance, the equation used is:

[tex]\text{Percent by mass}=\frac{\text{Mass of a substance}}{\text{Mass of solution}}\times 100[/tex]

Mass of a solution = 1.54 g

Mass of potassium iodide = 0.0405 g

Using above equation:

[tex]\text{Percent potassium iodide}=\frac{0.0405 g}{1.54g}\times 100\\\\\text{Percent potassium iodide}=2.63\%[/tex]

Hence, the percentage of potassium iodide in the sample is 2.63 %.

Expressing the results of potassium iodide in percentage = 2.63%

The chemical reaction of iodine ions with Bromine gas can be expressed as :

I⁻  + 3Br₂  + 3H₂O  -- > 6Br⁻ + IO₃ + 6H⁺ ----- ( 1 )

Chemical reaction between iodate ions with barium ions  can be expressed as :  Ba²⁺  + 2IO⁻₃ ------> Ba ( IO₃ )₂   --------- ( 2 )

Step 1 : Calculate the number of Barium iodate moles

mass of Barium iodate = 0.0596 g

molar mass of Barium iodate = 487.13 g/mol

from equation ( 1 )

moles of Barium iodate = ( 0.0596 ) / ( 487.13 ) = 1.22 * 10⁻⁴ moles

also from equation ( 1 ) the moles of potassium iodide = moles of iodide ions

= 2.44 * 10⁻⁴

molar mass of potassium iodide = 166 g/mol

Next step : Determine the mass of potassium iodide

moles of potassium * molar mass

= 2.44 * 10⁻⁴  * 166 g/mol  = 0.0405 g

Final step : Determine the percentage of potassium iodide in the solution

Percentage = ( mass of potassium iodide / mass of solution ) * 100

                   = ( 0.0405 / 1.54 ) * 100

                   = 2.63%

Hence we can conclude that potassium iodide in percentage = 2.63%

Learn more : https://brainly.com/question/3388761

Answer:

nacl with water

they are capable of conducting electricity

Answer:

reactivity

Explanation:

for example atoms in group 7 react by gaining 1 electron to become stable but they do not have the same number of energy levels

Answer:

Al,Ga,B

Explanation:

Now since i helped you can you help me with this plz

Matteo took 5 math quizzes. The mean of the 5 quizzes was 8.2. Here are four of his quiz scores 7, 7, 8, 10. What is the 5th quiz score? Show work.

Boron ( B )

Aluminium ( Al )

Gallium ( Ga )

Answer:

liquid to gas

Explanation:

when boiling water when evaporating heat is given out

Answer:

2.04x10⁻⁴ mol

Explanation:

First we convert 3.06x10⁻³ grams of glycine into moles of glycine, using its molar mass:

In order to calculate the number of hydrogen moles, we multiply the number of glycine moles by 5, as there are 5 hydrogen moles per glycine mol:

Answer: The percent yield of the reaction is 83.5 %

Explanation:

The given balanced equation is

[tex]SiO_2+3C\rightarrow SiC+2CO[/tex]

[tex]SiO_2[/tex] is the limiting reagent as it limits the formation of product and [tex]C[/tex] is the excess reagent.

According to stoichiometry :

60.08 g of [tex]SiO_2[/tex] produce = 40.11 of [tex]SiC[/tex]

Thus 50.0 of [tex]SiO_2[/tex] will produce=[tex]\frac{40.11}{60.08}\times 50.0=33.4[/tex]  of [tex]SiC[/tex]

 Experimental yield of SiC = 27.9 g

Percent yield = [tex]\frac{\text {Experimental yield}}{\text {theoretical yield}}\times 100=\frac{27.9g}{33.4g}\times 100=83.5\%[/tex]

Thus  percent yield of the reaction is 83.5 %

The percent yield of 83.5 % of 50.0 g of silicon dioxide is heated with an excess of carbon, and 27.9 g of silicon carbide is produced in the reaction.

The chemical equations are balanced when the reactants react to form products. The reactants and products react in proper ratios and if they are not in ratio then we balance them by adding the required quantity in the reactants and the products.

The given balanced equation is

[tex]\rm SiO_2+3C---- > SiC+2CO[/tex]

[tex]SiO_2[/tex] is the limiting reagent as it limits the formation of product and is the excess reagent.

According to stoichiometry

60.08 g [tex]SiO_2[/tex] of  produce = 40.11 of [tex]SiC[/tex]

Thus 50.0 of  [tex]SiO_2[/tex] will produce=  [tex]\dfrac{40.11}{60.08} \times 50=33.4\ SiC[/tex]

The experimental yield of SiC = 27.9 g

The percentage yield will be calculated as

[tex]\rm Percentage \ Yield = \frac{Experimental\ yield}{Theoretical \ yield }\times 100[/tex]

[tex]\rm Percentage \ yield =\dfrac{27.9}{33.49} \times 100=83.5[/tex]

Thus the percent yield of 83.5 % of 50.0 g of silicon dioxide is heated with an excess of carbon, and 27.9 g of silicon carbide is produced in the reaction.

To know more about balanced chemical equations follow

https://brainly.com/question/26227625

The redox equation C6H12O6 + 6O2 -> 6H2O + 6CO2 represents the process of cellular respiration. Option C

Cellular respiration is a biochemical process that occurs in living organisms, including plants and animals, to convert organic compounds, such as glucose (C6H12O6), into usable energy in the form of adenosine triphosphate (ATP). It is the primary process by which cells derive energy to carry out their functions.

In the given equation, glucose (C6H12O6) is being oxidized, losing electrons, and releasing carbon dioxide (CO2) as a byproduct. This oxidation process results in the production of energy-rich molecules, such as ATP.

Additionally, oxygen (O2) is being reduced, accepting the electrons from glucose and combining with hydrogen (H) to form water (H2O). This reduction process allows for the transfer of electrons and the generation of energy.

The process of cellular respiration is essential for the survival and functioning of organisms, as it provides the necessary energy for various metabolic activities, growth, and maintenance of cellular processes.

It is a fundamental metabolic pathway found in both plants and animals, enabling them to extract energy from organic molecules through the oxidation of glucose or other fuel sources.

Therefore, option C, cellular respiration, is the correct answer that represents the process described by the given redox equation.

For more such questions on redox equation visit:

https://brainly.com/question/21851295

#SPJ8

Answer:

1.75 g HCl

Explanation:

First we convert 2.00 g of PCl₅ into moles, using its molar mass:

Then we convert PCl₅ moles into HCl moles, using the stoichiometric coefficients of the equation:

Finally we convert HCl moles into grams, using its molar mass:

Answer:

True

Explanation:

I learned about it a few years ago

Answer:

A. mushroom

Explanation:

Answer:

Explanation:

b

The rate of the reaction determines the product formation and depends on various factors. The increase in pressure will change the concentration of reactants. Thus, option B is correct.

The rate of the reaction is the speed at which the chemical reaction moves in a direction to yield products. The reaction rate is influenced by the surface area of the substance, temperature, and concentration.

The pressure directly affects the concentration which in turn affects the reaction rate. The increased pressure raises the concentration and more collisions will occur between the particles resulting in a fast rate of reaction.

Therefore, the increase in pressure increases the concentration and reaction rate.

Learn more about reaction rate here:

https://brainly.com/question/12866245

#SPJ5

Answer:

The answers to your questions are given below.

Explanation:

__ NaBr + CaF₂ —> __ NaF + CaBr₂

The above equation can be balance as follow:

NaBr + CaF₂ —> NaF + CaBr₂

There are 2 atoms of F on the left side and 1 atom on the right side. It can be balance by writing 2 before NaF as shown below:

NaBr + CaF₂ —> 2NaF + CaBr₂

There are 2 atoms of Na on the right side and 1 atom on the left side. It can be balance by writing 2 before NaBr as shown below:

2NaBr + CaF₂ —> 2NaF + CaBr₂

Now, the equation is balanced.

Atom >>>> Reactant >>>> Product

Na >>>>>> 2 >>>>>>>>>>> 2

Br >>>>>>> 2 >>>>>>>>>>> 2

Ca >>>>>> 1 >>>>>>>>>>>> 1

F >>>>>>> 2 >>>>>>>>>>>> 2

Answer:

False

Explanation:

A liquid has a definite shape and takes on the volume of its container. A gas has both the shape and the volume of its container. The particles in a gas cannot be compressed into a smaller volume. Liquids tend to contract when heated.

Hope this helps

Answer:

(a) The normal freezing point of water (J·K−1·mol−1) is [tex]-22Jmole^-^1k^-^1[/tex]

(b) The normal boiling point of water (J·K−1·mol−1) is [tex]-109Jmole^-^1K^-^1[/tex]

(c) the chemical potential of water supercooled to −5.0°C exceed that of ice at that temperature is  109J/mole

Explanation:

Lets calculate

(a) - General equation -

      [tex](\frac{d\mu(\beta )}{dt})p-(\frac{d\mu(\alpha) }{dt})_p[/tex] = [tex]-5_m(\beta )+5_m(\alpha )[/tex] =  [tex]-\frac{\Delta H}{T}[/tex]

 [tex]\alpha ,\beta[/tex] → phases

ΔH → enthalpy of transition

T → temperature transition

 [tex](\frac{d\mu(l)}{dT})_p -(\frac{d\mu(s)}{dT})_p[/tex] =[tex]= -\frac{\Delta_fH}{T_f}[/tex]

            = [tex]\frac{-6.008kJ/mole}{273.15K}[/tex] ( [tex]\Delta_fH[/tex] is the enthalpy of fusion of water)

           = [tex]-22Jmole^-^1k^-^1[/tex]

(b) [tex](\frac{d\mu(g)}{dT})_p-(\frac{d\mu(l)}{dT})_p= -\frac{\Delta_v_a_p_o_u_rH}{T_v_a_p_o_u_r}[/tex]

                                  = [tex]\frac{40.656kJ/mole}{373.15K}[/tex] ([tex]\Delta_v_a_p_o_u_rH[/tex] is the enthalpy of vaporization)

                               = [tex]-109Jmole^-^1K^-^1[/tex]

(c) [tex]\Delta\mu =\Delta\mu(l)-\Delta\mu(s)[/tex] =[tex]-S_m\DeltaT[/tex]

[tex][\mu(l-5[/tex]°[tex]C)-\mu(l,0[/tex]°[tex]C)][/tex] =  [tex][\mu(s-5[/tex]°[tex]C)-\mu(s,0[/tex]°[tex]C)][/tex][tex]=-S_m[/tex]ΔT

[tex]\mu(l,-5[/tex]°[tex]C)-\mu(s,-5[/tex]°[tex]C)=-Sm\DeltaT [\mu(l,0[/tex]

[tex]\Delta\mu=(21.995Jmole^-^1K^-^1)\times (-5K)[/tex]

     = 109J/mole

Answer:

The correct answer is - 1 and 3 are true.

Explanation:

Hydrogen bonds require a lot of energy to be broken as these bonds are relatively strong intermolecular forces which makes the boiling point of liquid like water very high, as it requires a lot of heat to break the intermolecular bonds holding the water molecules together.

An increase in the surface area of the liquid also increases the rate of evaporation as it takes place among the molecules of the liquid on the surface.

Answer:

238000 N

Explanation:

Use F=ma

Mass= 17000 kg

Accel.= 14 m/s²

Answer:

Cl₂ is the limiting reactant.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

N₂ + 3Cl₂ —> 2NCl₃

From the balanced equation above,

1 L of N₂ reacted with 3 L of Cl₂.

Finally, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

1 L of N₂ reacted with 3 L of Cl₂.

Therefore, 134 L of N₂ will react with = 134 × 3 = 402 L of Cl₂.

From the calculation made above, we can see that a higher volume (i.e 402 L) of Cl₂ than what was given (i.e 99 L) is needed to react completely with 134 L of N₂.

Therefore, Cl₂ is the limiting reactant and N₂ is the excess reactant.

Answer:

The final units are mol

Explanation:

The units of molar mass are grams per mole, g/mol.

This means that when dividing mass by molar mass, the units of grams cancel each other, leaving only units of moles:

Another way of writing that might help is as follows:

Thus, the final units are moles.

Answer:

Nitrates compared to sulphates are much more soluble therefore won't form precipitates easily.

Answer:

35.8g of Cl₂ is the yield

Explanation:

Based on the reaction:

MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂

1 mole of MnO₂ and 4 moles of HCl react producing 1 mole of Cl₂

To solve this question we must find limiting reactant. with limiting reactant we can find the theoretical yield of Cl₂. As the actual yield is the 62.7% we can find actual yield of Cl₂ in grams:

Moles MnO₂ -Molar mass: 86.9368g/mol-:

70.0g * (1mol / 86.9368g) = 0.805 moles

Moles HCl -Molar mass: 36.46g/mol-:

128.0g * (1mol / 36.46g) = 3.51 moles

For a complete reaction of 3.51 moles of HCl are required:

3.51 moles HCl * (1mol MnO₂ / 4mol HCl) = 0.878 moles MnO₂.

As there are just 0.805 moles of MnO₂, MnO₂ is limiting reactant.

1 mole of MnO₂ produce 1 mole of Cl₂. The theoretical moles of Cl₂ produced are 0.805 moles.

As the yield is of 62.7%, the yield of Cl₂ is:

0.805 moles * (62.7 / 100) = 0.505 moles Cl₂. In grams:

0.505 moles Cl₂ * (70.906g / mol) =

Answer:

221.22K or -51°C

Explanation:

We will be using the Ideal Gas Law to calculate the temperature of the gas. It is a mathematical relationship that describes the behavior of ideal gas ample for any combo of varying pressure, volume, temperature, and # of moles (n). It is derived by combing Boyle's Law, Charles' Law, Gay-Lussac's & Avogadro's Law.

Note: As always, remember that temperature must be in Kelvin not Celsius when using this equation.

Ideal Gas Law: [tex]PV = nRT[/tex], where P = pressure, V = volume (in Liters), n = # of moles, R = the ideal gas constant, and T = temperature (in Kelvin).

Based on the problem, we are given the pressure, volume, and # of moles. We are asked to find the temperature. What about R you ask? Well, R is a constant that is the value of 1 mole of gas at STP. R has various values depending on the pressure units. In this case, our pressure is in atm so the R value = 0.0821.

Onto the math - all that needs to be done now is to plug and chug. Plug in the given values to find the temperature:

Set up: [tex](7.6 atm)(12L) = (5 mol)(0.0821 L*atm/(mol*K))(T)[/tex]

==> [tex]T = \frac{(7.6 atm)(12L)}{(5 mol)(0.0821 L*atm/(mol*K))}[/tex]

==> T = 221.17K

The answer is 221.17K. To convert into Celsius, subtract by 273.15 to get -50.99 or -51°C.

Answer:

1.5x1024

Explanation:

Answer:

A,C,D

Explanation:

Variations effect the likelyhood of survival for a organism, it can give off an advantage or disadvantage.

Brainliest? pls