A gas originally at 27 °c and 1.00 atm pressure in a 2.6 l flask is cooled at constant pressure until the temperature is 11 °c. the new volume of the gas is

Answers

Answer 1
Final answer:

According to Charles's law, the final volume of a gas that is cooled from 27 °c to 11 °c at constant pressure, and initially in a 2.6 L flask, is 2.46 L.

Explanation:

This question pertains to Charles's law, which states that the volume of a gas is directly proportional to its temperature if kept at a constant pressure. In this case, the initial temperature T₁ is 27°C (which converts to 300 K) and the final temperature T₂ is 11°C (or 284 K). The initial volume V₁ is 2.6 L. Remembering to convert degrees Celsius to Kelvin (by adding 273), we can solve for V₂, the unknown final volume, by setting up the equation V₁/T₁ = V₂/T₂.

Solving for V₂ yields V₂ = V₁ × (T₂/T₁) = 2.6 L × (284K/300K) = 2.46 L. So, if a gas originally at 27 °c and 1.00 atm pressure in a 2.6 L flask is cooled at constant pressure until the temperature is 11 °c, the new volume of the gas is 2.46 L.

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Related Questions

13) Given the redox reaction: Cr3+ + Al Cr + Al3+ As the reaction takes place, there is a transfer of A) electrons from Al to Cr3+ B) electrons from Cr3+ to Al C) protons from Al to Cr3+ D) protons from Cr3+ to Al

Answers

Cr³⁺+ Al⁰---> Cr ⁰+ Al³⁺
Cr³⁺+ 3e⁻---> Cr ⁰ 
Cr³⁺ takes electrons to become neutral atom.
 Al⁰---> Al³⁺ + 3e⁻
Al⁰ loses electrons to become ion.

Electrons transferred from  Al to Cr³⁺.
Correct answer is  
A) electrons from Al to Cr³⁺.

The main interactions between molecules of hydrogen chloride are examples of

Answers

Answer:
            The main interactions between molecules of hydrogen chloride are examples of Dipole-Dipole Interactions.

Explanation:
                   The bond between Hydrogen and Chlorine is Polar Covalent Bond as the electronegativity difference between these two elements is 0.96 which is greater than 0.4. Chlorine being more electronegative attracts the electrons from Hydrogen making the Hydrogen partial positive and itself partial negative. The two poles on HCl makes it a dipole. Now, one HCl (a dipole) interacts with another HCl (another Dipole) through their opposite charges and creates Dipole-Dipole Interaction. The picture is as below, the green dashed lines are interactions,

Final answer:

The main interactions between molecules of hydrogen chloride are examples of dipole-dipole interactions, specifically hydrogen bonding.

Explanation:

Hydrogen chloride (HCl) is a polar molecule, which means it has a partially positive end (hydrogen) and a partially negative end (chlorine). The main interactions between molecules of hydrogen chloride are examples of dipole-dipole interactions, specifically hydrogen bonding. These interactions occur when the partially positive hydrogen atom in one molecule is attracted to the partially negative end of another molecule, creating a bond between them.

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A 3.00-l flask is filled with gaseous ammonia, nh3. the gas pressure measured at 15.0 ∘c is 2.15 atm . assuming ideal gas behavior, how many grams of ammonia are in the flask? express your answer to three significant figures and include the appropriate units.

Answers

we can find the mass of ammonia using the ideal gas law equation,
PV = nRT
where 
P - pressure - 2.15 atm x 101 325 = 2.18 x 10⁵
V - volume - 3.00 x 10⁻³ m³
n - number of moles 
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in kelvin - 15.0 °C + 273 = 288 K
substituting these values in the equation 
2.18 x 10⁵ Pa x 3.00 x 10⁻³ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 288 K
n = 0.273 mol 
number of moles of NH₃ is 0.273 mol 
molar mass of NH₃ - 17.0 g/mol 
mass pf ammonia present - 0.273 mol x 17.0 g/mol = 4.64 g
mass of NH₃ present is 4.64 g

Which part of an atom is most directly involved in chemical bonding?

Answers

The outer or valence electrons

Chemical bonding is defined as the attraction between elements, ions or molecules which results in the formation of compounds. The bonding takes place due to sharing of electrons as in covalent bond or by the electrostatic force of attraction between positive charge ion (cation) and negative charge ion (anion) as in ionic bond.

Now, electrons are the negative charged particles of an atom and it is found in clouds which is surrounded by the nucleus of an atom. Electrons play a vital role in chemical bonding. In both type of bonding i.e. ionic bonding in which electrons are transferred from on atom to other atom and covalent bonding which results due to sharing of electrons between two atoms.

Thus, electron is a part of an atom is most directly involved in chemical bonding.





Which would be most useful if you were lost in the Sahara desert? A. Landsat satellite B. OSTM/Jason-2 satellite C. Global Positioning System D. topographic map of Africa

Answers

A Global Positioning System would be most useful if you were lost in the Sahara desert.
It would most likely be a GPS system because it has different options and the other wont even help you, there just random satellites and land form maps, hope this helps

how many grams of salt are in 5 L of a solution with a concentration of 3 g/L

Answers

Three grams per liter times five liters tells us that there are fifteen grams of salt in this five liter solution.

When hydrochloric acid is added to sodium hydrogen carbonate?

Answers

The reaction of hydrochloric acid (HCl) and sodium carbonate (NaCO3) yields a product of water (H20), carbon dioxide gas (CO2), and table salt (NaCl). The intermediate that is formed in situ is carbonic acid (H2CO3), but due to its unstable nature, it breaks down into water, carbon dioxide, and a salt.

Hope this helps!

Describe what changes occur during electron capture. describe what changes occur during electron capture. the mass number and atomic number increases. the mass number and atomic number decreases. the mass number is unchanged and the atomic number decreases. the mass number and atomic number do not change. the mass number is unchanged and the atomic number increases.

Answers

The mass number is unchanged and the atomic number decreases.
Final answer:

Electron capture is a nuclear process where an inner shell electron combines with a proton, creating a neutron and resulting in a decrease in atomic number while keeping the mass number unchanged. The process generates energy often in the form of an X-ray. An example of this is seen in potassium-40, wherein electron capture transforms it into a different nuclide.

Explanation:

Electron capture is a process during which an inner shell electron combines with a proton in the nucleus and transforms into a neutron. This results in the creation of a vacancy within the atom, which is then filled by an electron from an outer shell. This electron, as it falls into the vacancy, releases energy, often in the form of an X-ray.

One significant outcome of electron capture is that the atomic number of the atom decreases by one, while the mass number remains unchanged. This happens because the process effectively converts a proton into a neutron within the atomic nucleus. Thus, the atom moves closer to the band of stability, improving the neutron to proton (n:p) ratio.

For instance, in the case of potassium-40, the atom undergoes electron capture, transforming it into a different nuclide with an atomic number one less than the original, and an unchanged mass number.

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Compared to a solution with a pH value of 7 a solution with a thousand times greater hydronium ion concentration has a pH value of what

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Answer is: solution has a pH value of 4.
pH₁ = 7.
pH = -log[H₃O⁺].
[H₃O⁺]₁ = 10∧(-pH).
[H₃O⁺]₁ = 10⁻⁷ M.
[H₃O⁺]₂ = 10⁻⁷ M ·10³.
[H₃O⁺]₂ = 10⁻⁴ M.
pH₂ = -log[H₃O⁺]₂.
pH₂ = -log(10⁻⁴ M).
pH₂ = 4.
pH (potential of hydrogen) is a numeric scale used to specify the acidity or basicity an aqueous solution.
When pH is less than seven (pH<7), solution is acidic..
When is equal seven (pH = 7), solution is neutral.
When pH is greater than seven (pH > 7), solution is basic.

The pH of the new solution has been 4. Thus, the solution has been more acidic than the precious solution.

The pH has been defined as the negative log of hydronium ion concentration in the solution. The pH has been expressed as:

[tex]\rm pH=-\;log\;[H_3O^+][/tex]

Computation for the pH of the solution

The pH of the given solution has been 7. The hydronium ion concentration has been given as:

[tex]\rm pH=\;-log\;[H_3O^+]\\\\ 7=\;-\;log[H_3O^+]\\\\ H_3O^+=10^-^7\;M[/tex]

The concentration of the new solution has been 1000 times greater than the previous solution.

The previous solution has hydronium ion concentration of [tex]\rm 10^-^7\;M[/tex]. The concentration of new solution will be:

[tex]\rm New\;solution=10^-^7\;\times\;1000\;M\\ New\;solution=10^-^4\;M[/tex]

The concentration of the new solution has been [tex]\rm 10^-^4\;M[/tex]. The pH of the solution has been given as:

[tex]\rm pH=-log\;[H_3O^+]\\ pH=-log\;[10^-^4]\\ pH=4[/tex]

The pH of the new solution has been 4. Thus, the solution has been more acidic than the precious solution.

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When preparing the diazonium salt, the solution is tested with potassium iodide-starch paper. a positive test is the immediate formation of a blue color. what is the ki-starch paper testing for ? explain?

Answers

KI-starch paper allows the detection of strong oxidizers such as nitrite. It is used here to control diazotization of 4-nitroaniline. Nitrite oxidizes potassium iodide in order to form elemental iodine which reacts with starch to a blue-violet complex. With KI-starch paper, enough sodium nitrite is added to produce nitrous acid, which then will react with 4-nitroaniline to form a diazonium salt.

Final answer:

Potassium iodide-starch paper is used to detect the presence of nitrogen dioxide in the preparation of diazonium salts by revealing a blue-black coloration upon reacting with iodine produced from the reaction between NO2 and potassium iodide.

Explanation:

The potassium iodide-starch paper is used to test for the presence of nitrogen dioxide (NO2), which is a byproduct of the reaction used to prepare diazonium salts. During the diazo coupling reaction, excess nitrous acid can decompose and produce nitrogen dioxide. This gas can then react with the potassium iodide (KI) in the starch paper to produce iodine (I2), which subsequently forms a blue-black complex with starch. The immediate formation of the blue color on the potassium iodide-starch paper is a positive test indicating the presence of nitrogen dioxide. It's important to monitor this because the presence of NO2 suggests that the diazonium salt solution might be unsafe due to the potential release of toxic gases.

The iodine-starch test is a well-known reaction in which iodine (I2), produced by the oxidation of iodide ions by NO2, interacts with starch to produce a characteristic blue-black color. This test provides a quick and sensitive method for detecting the presence of iodine, which, in this context, indirectly indicates the generation of nitrogen dioxide in the reaction mixture.

Calculate the volume in liters of a ×4.3910−5/moll mercury(ii) iodide solution that contains 500.mg of mercury(ii) iodide hgi2 . round your answer to 3 significant digits.

Answers

the molarity of the solution is 4.39 x 10⁻⁵ mol/L
molarity is the number of moles of solute in 1 L of solution 
in this instance there are 4.39 x 10⁻⁵ mol of HgI₂ in 1 L solution 
the mass of HgI₂ in 1 L - 4.39 x 10⁻⁵ mol x 454.4 g/mol = 19.9 mg
if 19.9 mg are in 1 L 
then 500.0 mg are in - 500.0 mg / 19.9 mg/L = 25.1 L
therefore volume is 25.1 L

Final answer:

To calculate the volume of the mercury(II) iodide solution in liters, convert the mass of the solute to moles and use the Molarity equation. The volume is approximately 25.1 L.

Explanation:

To calculate the volume in liters of the mercury(II) iodide solution, we need to convert the mass of the solute (HgI2) to moles, using the molar mass of HgI2. Then, we can use the equation Molarity = moles of solute / volume of solution in liters to find the volume.

First, calculate the moles of HgI2:

Moles = mass / molar mass = 500 mg / (454.39 g/mol) = 1.101 x 10^-3 mol

Next, rearrange the equation to solve for volume:

Volume = moles of solute / Molarity = 1.101 x 10^-3 mol / (4.3910^-5 M) = 25.07 L

Rounding to three significant digits, the volume of the solution is approximately 25.1 L.

How many times greater is the rate of effusion of molecular fluorine than that of molecular bromine at the same temperature and pressure

Answers

According to Graham's Law of Diffusion,"the rates of diffusion of two gases are inversely proportional to the square root of their Molar masses or Densities at the same pressure and temperature".

                                           r₁ / r₂  =  [tex] \sqrt{M2 / M1} [/tex]
Where,
            r₁  =  Rate of Fluorine

            r₂  =  Rate of Bromine

            M₂  =  Molar mass of Bromine  =  159.8 g/mol

            M₁  =  Molar mass of Fluorine  =  37.98 g/mol

Putting values,

                                           r₁ / r₂  =  [tex] \sqrt{159.8 / 37.98} [/tex]

                                           r₁ / r₂  =  [tex] \sqrt{4.20} [/tex]

                                           r₁ / r₂  =  2.04

Result:
          Fluorine effuses 2 times faster than Bromine gas.

If you add 500 kj of heat to 500 g of water at 50.0°c, how much water is left in the container? the latent heat of vaporization of water is 2.26 ×103 j/g and its specific heat is is 4.190 j/(g∙k)

Answers

First, we will get the heat required to raise the 500 g of water from 50°C to 100°C = m*C*ΔT 

when m is the mass = 500 g 

and C is the specific heat capacity of water = 4.19

ΔT change in temperature = 50 °C

by substitution:

q = 0.5 Kg * 4.19 * 50°C

   =104.75 KJ

∴ heat left to boil the water= 500KJ - 104.75KJ = 395 KJ

the heat required to boil water from 100°C to steam = mass *latent heat of vaporization

  395KJ     = M * 2.26 x 10^3

 Mass = 0.17Kg = 170 g

∴ water remain= 500 g - 170 = 330 g


The coordinate plane is separated into four quadrants as shown. let p: x < 0 let q: y < 0 what is represented by p ∨ q?
quadrant 1 because both x and y are positive coordinates

quadrant 3 because both x and y are negative coordinates

quadrants 1, 2, and 4 because in these quadrants x, y, or both are positive coordinates

quadrants 2, 3, and 4 because in these quadrants x, y, or both are negative coordinates

Answers

Answer:

The correct option is C which is it is the third coordinate because p and q both are negative.

Explanation:

If we look at the four quadrants of coordinate plane:

Quadrant 1: If x and y both are positive.Quadrant 2: If x is negative and y is positive.Quadrant 3: If both x and y are negative.Quadrant 4: If x is positive and y is negative.

So according to given data, x is less than 0 means xis negative. y is also negative because y is less than 0.

So both x and y are negative and it is in quadrant 3.

Answer:

Quadrant II, because the slope is positive and the y intercept is negative

Explanation:

A solution has [oh−] = 2.3×10−2 m . use the ion product constant of water kw=[h3o+][oh−] to find the [h3o+] of the solution.

Answers

Answer:

The concentration of hydronium ions in the solution is [tex]4.35\times 10^{-13} M[/tex].

Explanation:

Concentration of hydroxide ions = [tex][OH^-]=2.3\times 10^{-2}[/tex]

Concentration of hydroxide ions = [tex][H_3O^+]=?[/tex]

[tex]H_2O+H_2O\rightleftharpoons H_3O^++OH^-[/tex]

The ionic product of water is given as:

[tex]K_w=[H_3O^+][OH^-][/tex]

The value of ionic product of water, [tex]K_w=1\times 10^{-14}[/tex]

[tex]K_w=1\times 10^{-14}=[H_3O^+][OH^-][/tex]

[tex]1\times 10^{-14}=[H_3O^+]\times 2.3\times 10^{-2}M[/tex]

[tex][H_3O^+]=\frac{1\times 10^{-14}}{2.3\times 10^{-2}}=4.35\times 10^{-13} M[/tex]

The concentration of hydronium ions in the solution is [tex]4.35\times 10^{-13} M[/tex].

The concentration of hydronium, ion [H₃O⁺] in the solution containing 2.3×10⁻² M concentration of hydroxide ion, [OH⁻] is 4.35×10⁻¹³ M

How to determine the concentration of hydronium, ion [H₃O⁺]?

The following data were obtained from the question given above:

Concentration of hydroxide ion, [OH⁻] = 2.3×10⁻² MConcentration of hydronium, ion [H₃O⁺] = ?

[H₃O⁺] × [OH⁻] = 10¯¹⁴

[H₃O⁺] × 2.3×10⁻² = 10¯¹⁴

Divide both side by 2.3×10⁻²

[H₃O⁺] = 10¯¹⁴ / 2.3×10⁻²

= 4.35×10⁻¹³ M

Thus, we can conclude that the concentration of hydronium ion, [H₃O⁺] in the solution is 4.35×10⁻¹³ M

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What solution has the lowest pH?

1.0 M HNO2

1.0 M CH3COOH

1.0 M HCOOH

1.0 M HNO3

1.0 M HPO4-

Answers

Letter D. I hope this helps.

What volume (in liters) does 3.91 moles of nitrogen gas at 5.35 atm and 323 K occupy

Answers

Answer:

19.4

Explanation:

Answer:

19.4

Explanation:

Determine the molecular mass ratio of two gases whose rates of effusion have a ratio of 16 : 1.

Answers

Answer is: the molecular mass ratio of two gases is 1 : 256.
rate of effusion of gas1 : rate of effusion of gas = 16 : 1.
rate of effusion of gas1 = 1/√M(gas1).
rate of effusion of gas2 = 1/√M(gas2).
rate of effusion of gas1 = rate of effusion of gas2 · 16.
1/√M(gas1) = 1/√M(gas2) · 16 /².
1/M(gas1) = 1/M(gas2) · 256.
M(gas1) · 256 = M(gas2).

Wolff-kishner reduction (hydrazine, koh, ethylene glycol, 130°c) of the compound shown gave compound
a. treatment of compound a with m−chloroperoxybenzoic acid (mcpba) gave compound b, which on reduction with lithium aluminum hydride gave compound
c. oxidation of compound c with chromic acid gave compound d (c9h14o). identify compounds a through d in this sequence, and do not indicate stereochemistry.

Answers

The scheme is shown below, the steps involved are as follow,

Step one: Reduction:
               The carbonyl group of given compound on reduction using Wolf Kishner reagent converts the carbonyl group into -CH₂- group.

Step two: Epoxidation:
              The double bond present in starting compound when treated with m-CPBA (meta-Chloroperoxybenzoic acid) gives corrsponding epoxide.

Step three: Reduction:
                The epoxide is reduced to alcohol on treatment with Lithium Aluminium Hydride (LiAlH₄) followed by hydrolysis.

Step four: Oxidation:
               The hydroxyl group (alcohol) is oxidized to carbonyl (ketonic group) using oxidizing agent Chromic acid (H₂CrO₄).

The ksp of pbbr2 is 6.60× 10–6. what is the molar solubility of pbbr2 in 0.500 m kbr solution

Answers

The solubility of PbBr₂(s) with the presence of 0.500 M of KBr is 2.64 x 10⁻⁵ M.

Explanation:

The given equation is as follows.

       [tex]PbBr_2 \rightleftharpoons Pb^{2+} + 2Br^{-}[/tex]

                          s       2s

It is given that,

       [tex]K_{sp} = [Pb^{2+}][Br^{-}]^{2} = 6.60 \times 10^{-6}[/tex]

Let the solubility of given ions be "s".

Since, KBr on dissociation will given bromine ions.

Hence,  [tex]K_{sp} = [Pb^{2+}] \times ([Br^{-}])^{2}[/tex]

        [tex]6.60 \times 10^{-6}[/tex]  = [tex]s \times (2s)^{2}[/tex]

                        = [tex]1.18 \times 10^{-2}[/tex] M

Therefore, solubility of [tex][PbBr_{2}][/tex] is [tex]1.18 \times 10^{-2}[/tex] M  in KBr.

Now, we will calculate the molar solubility of [tex]PbBr_{2}[/tex] in 0.5 M KBr solution as follows.

           [tex]K_{sp} = (s) \times (2s + 0.5)^{2}[/tex]

  [tex]6.60 \times 10^{-6}[/tex]  = [tex]4s^{3} + 0.25s + 2s^{2}[/tex]

                         s = [tex]2.63 \times 10^{-5}[/tex]

Thus, we can conclude that molar solubility of [tex][PbBr_{2}][/tex] in 0.500 m KBr solution is [tex]2.63 \times 10^{-5}[/tex].

An industrial chemist introduces 8.1 atm h2 and 8.1 atm co2 into a 1.00-l container at 25.0°c and then raises the temperature to 700.0°c, at which keq = 0.534: h2(g) + co2(g) ⇔ h2o(g) + co(g) how many grams of h2 are present after equilibrium is established?

Answers

i reccomend searching it up

Identify each of the following mixtures as either homogeneous or heterogeneous and as a solution, a suspension, or a colloid.
Blood
homogeneous heterogeneous solution colloid suspension
Salad dressing
homogeneous heterogeneous solution colloid suspension
PLEASE HELP!!!!

Answers

With reference to the solutions, following things may be noted
1) Homogeneous solutions comprises of single phase.
2) Heterogeneous solution comprises of more than one phase.
3) in colloidal solution, solute is evenly distributed in solvent. in such system, it is difficult to separate solute and solvent.
4) in suspension, there is an uneven distribution of solute in solvent. In such system, solute and solvent gets separated on standing long.  

Blood is a homogeneous and colloidal solution,
Salad dressing is a heterogeneous and suspension solution.   

blood: heterogeneous and suspension

salad dressing: heterogeneous and suspension

Determine the empirical formula for a compound that is 70.79% carbon, 8.91% hydrogen, 4.59% nitrogen, and 15.72% oxygen. determine the empirical formula for a compound that is 70.79 carbon, 8.91 hydrogen, 4.59 nitrogen, and 15.72 oxygen. c18h27no2 c18h27no3 c17h27no3 c17h26no3

Answers

Hello!

We use the amount in grams (mass ratio) based on the composition of the elements, see: (in 100g solution)

C: 70.79% = 70,79 g 
H: 8.91% = 8.91 g
N: 4.59% = 4.59 g
O: 15.72% = 15.72 g

Let us use the above mentioned data (in g) and values will be ​​converted to amount of substance (number of moles) by dividing by molecular mass (g / mol) each of the values, lets see:

[tex]C: \dfrac{70.79\:\diagup\!\!\!\!\!g}{12\:\diagup\!\!\!\!\!g/mol} \approx 5.89\:mol[/tex]

[tex]H: \dfrac{8.91\:\diagup\!\!\!\!\!g}{1\:\diagup\!\!\!\!\!g/mol} = 8.91\:mol[/tex]

[tex]N: \dfrac{4.59\:\diagup\!\!\!\!\!g}{14\:\diagup\!\!\!\!\!g/mol} \approx 0.328\:mol[/tex]

[tex]O: \dfrac{15.72\:\diagup\!\!\!\!\!g}{16\:\diagup\!\!\!\!\!g/mol} = 0.9825\:mol[/tex]

We note that the values ​​found above are not integers, so let's divide these values ​​by the smallest of them, so that the proportion is not changed, let's see:

[tex]C: \dfrac{5.89}{0.328}\to\:\:\boxed{C\approx 18}[/tex]

[tex]H: \dfrac{8.91}{0.328}\to\:\:\boxed{H\approx 27}[/tex]

[tex]N: \dfrac{0.328}{0.328}\to\:\:\boxed{N\approx 1}[/tex]

[tex]O: \dfrac{0.9825}{0.328}\to\:\:\boxed{O\approx 3}[/tex]

Thus, the minimum or empirical formula found for the compound will be:

[tex]\boxed{\boxed{C_{18}H_{27}N_1O_3\:\:or\:\:C_{18}H_{27}NO_3}}\end{array}}\qquad\checkmark[/tex]

I hope this helps. =)

The empirical formula of the compound is C₁₈H₂₇NO₃

From the question given above, the following data were obtained:

Carbon (C) = 70.79%

Hydrogen (H) = 8.91%

Nitrogen (N) = 4.59%

Oxygen (O) = 15.72%

Empirical formula =?

The empirical formula of the compound can be obtained as follow:

C = 70.79%

H = 8.91%

N = 4.59%

O = 15.72%

Divide by their molar mass

C = 70.79 / 12 = 5.899

H = 8.91 / 1 = 8.91

N = 4.59 / 14 = 0.328

O = 15.72 / 16 = 0.9825

Divide by the smallest

C = 5.899 / 0.328 = 18

H = 8.91 / 0.328 = 27

N = 0.328 / 0.328 = 1

O = 0.9825 / 0.328 = 3

Therefore, the empirical formula of the compound is C₁₈H₂₇NO₃

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Consider the electrolysis of molten barium chloride (bacl2). write the half-reactions. include the states of each species.

Answers

Molten barium chloride is separeted into two species :
BaCl₂(l) → Ba(l) + Cl₂(g),
but first ionic bonds in this salt are separeted because of heat:
BaCl₂(l) → Ba²⁺(l) + 2Cl⁻(l).

Reaction of reduction at cathode(-): Ba²⁺(l) + 2e⁻ → Ba(l).

Reaction of oxidation at anode(+): 2Cl⁻(l) → Cl₂(g) + 2e⁻.

The anode is positive and the cathode is negative.

Hclo is a weak acid (ka = 4.0 × 10–8) and so the salt naclo acts as a weak base. what is the ph of a solution that is 0.030 m in naclo at 25 °c?

Answers

Chemical reaction 1: NaClO(aq) → Na⁺(aq) + ClO⁻(aq).
Chemical reaction 2: ClO⁻(aq) + H₂O(l) ⇄ HClO(aq)+ OH⁻(aq).
c(NaClO) = 0,030 M.
[ClO⁻] = 0,03 M - x.
Ka(HClO) = 4·10⁻⁸.
Ka · Kb = 10⁻¹⁴.
Kb(ClO⁻) = 2,5·10⁻⁷.
Kb(ClO⁻) = [OH⁻] · [HClO] / [ClO⁻].
[OH⁻] = [HClO] = x.
2,5·10⁻⁷ = x² / (0,03 M -x).
Solve quadratic equation: x = [OH⁻] = 0.0000893 M.
pOH = -log(0.0000893 M) = 4.05.
pH = 14 - 4.05.
pH = 9.95.

Balance this equation. If a coefficient of "1" is required, choose "blank" for that box. CH3OH + O2 → CH2O + H2O

Answers

If the number of different atoms of elements on the reactant side is equal to that present on the product side than the reaction is said to be chemically balanced. 

In above case, following is the balanced reaction.

2 CH3OH + O2 → 2 CH2O + 2 H2O

Answer:

look at the screenshot below

Explanation:

elaborate on the difference in natural occurrences between nuclear fission and nuclear fusion reactions. A) Neither nuclear fission nor nuclear fusion reactions naturally occur on earth or outer space. B) Nuclear fission reactions rarely occur naturally while nuclear fusion reactions occur in the stars. C) Both nuclear fission and nuclear fusion reactions readily naturally occur in the pressure of the earth's core. D) Nuclear fusion reactions do not naturally occur while nuclear fission reactions occur in the volcanoes.

Answers

B) Nuclear fission reactions rarely occur naturally while nuclear fusion reactions occur in the stars.

Option B: Nuclear fission reactions rarely occur naturally while nuclear fusion reactions occur in the stars.

Nuclear fission reaction is defined as formation of two or more atoms by splitting of large atoms. On the other hand, nuclear fusion reaction is formation of large atom from small atoms.

The splitting of large atoms into small atoms (nuclear fission) generally does not occur naturally because it requires high speed neutrons and critical mass of the substance undergoing fission.

Nuclear fusion reaction occurs in stars such as sun because it requires extremely high energy to bring two more proton closer to each other by overcoming electronic repulsion.

Therefore, nuclear fission reactions rarely occur naturally while nuclear fusion reactions occur in the stars.

A certain half-reaction has a standard reduction potential . an engineer proposes using this half-reaction at the anode of a galvanic cell that must provide at least of electrical power. the cell will operate under standard conditions. note for advanced students: assume the engineer requires this half-reaction to happen at the anode of the cell.

Answers

An electrochemical cell are of two types 1) Chemical 2) concentration

For the present application, since the engineer requires this half-reaction to happen at the anode of the cell, chemical cell must be preferred. Further since, the engineer required, the emf of cell equal to half cell reaction, the counter electrode/electrolyte that can be used is Pt/H2/H+. It must be noted that std reduction potential of this electrode is 0v. It must also be noted that, engineer requires this half-reaction to happen at the anode of the cell. Oxidation occurs at anode compartment. Hence, the remaining half cell should comprise of chemical, whose std. oxidation potential is higher than that of Pt/H2/H+ or std.  reduction potential is less that that of Pt/H2/H+. For, this purpose, those chemicals must be chosen, which are placed below H2/H+ in electrochemical series. If this material is of the form X/X+, the cell representation of the overall electrochemical cell is
X/X+//H+/H2/Pt

Notice that " s o 4 " appears in two different places in this chemical equation. s o 2− 4 is a polyatomic ion called "sulfate." what number should be placed in front of cas o 4 to give the same total number of sulfate ions on each side of the equation? ?cas o 4 +alc l 3 →cac l 2 +a l 2 (s o 4 ) 3 express your answer numerically as an integer.

Answers

a "3" should but put in front of
"cas o 4 "

A compound consisting of atoms of large atomic mass is more likely to require:

Answers

Answer:

A) a higher temperature to liquefy

BRAINLIST?

Explanation:

Answer:

The answer is

A: a higher temperature to liquefy

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