what mass of lead sulfate is formed in a lead-acid storage battery when 1.21 g of pb undergoes oxidation?

Points A, D, and H are noncoplanar with points B, C, and F.

To determine which points are noncoplanar with points B, C, F, and G, we can follow these steps:

Identify the plane formed by points B, C, and F using the equation of a plane.

Substitute the coordinates of point G into the equation of the plane.

If the equation is satisfied, then point G is coplanar with points B, C, and F.

Otherwise, it is noncoplanar.

Repeat steps 2 and 3 for each of the other points (A, D, E, and H) to determine their coplanarity with B, C, and F.

Points A, D, E, and H that do not satisfy the equation of the plane are noncoplanar with points B, C, and F.

Based on the given options, points A, D, and H are noncoplanar with points B, C, and F.

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The correct answers are Potential and Kinetic energy.

An object can store energy as a result of its position. For example, the heavy ball of a demolition machine is storing energy when it is held at an elevated position. This stored energy of position is referred to as potential energy. Similarly, a drawn bow is able to store energy as a result of its position.

Kinetic energy is a form of energy that an object or a particle has by reason of its motion. If work, which transfers energy, is done on an object by applying a net force, the object speeds up and thereby gains kinetic energy.

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a) Free-radical polymerization of 2-chloro-1,3-butadiene would produce a polymer consisting of repeating units of 2-chloro-1,3-butadiene. The general structure of the polymer would be:

  Cl

  |

  CH2 - CH = CH - CH2 - Cl

  |        |

  Cl      Cl

This polymer is known as poly(2-chlorobutadiene).

b)

The retrosynthesis of compound S is as follows:

      Br          Br

       |           |

       |           |

H2C = CH - CH2 - CH2 - C - CH = CH

       |           |

       |           |

      Cl          Cl

Starting material: 1,4-dibromo-2-butene

Step 1: Dehalogenation of 1,4-dibromo-2-butene using zinc dust and acetic acid to give 1,4-butadiene.

      Br          Br

       |           |

       |           |

H2C = CH - CH = CH - CH = CH2

       |           |

       |           |

      Br          Br

Step 2: Reaction of 1,4-butadiene with hydrogen chloride in the presence of benzoyl peroxide as a free radical initiator to give compound S.

H2C = CH - CH2 - CH2 - C - CH = CH

       |           |

       |           |

      Cl          Cl

Note: The reaction of 1,4-butadiene with HCl in the presence of a free radical initiator is an example of free-radical addition reaction.

c) The synthesis of the product on the right from all the starting materials on the left can be accomplished through the following steps:

Step 1: Bromination of ethylbenzene using N-bromosuccinimide (NBS) in the presence of light or heat to give 1-bromo-2-phenylethane.

         H

         |

         CH3

          |

          C6H5

          |

    Br --- CH2 --- CH3

Step 2: Conversion of 1-bromo-2-phenylethane to 1-phenylethanol using lithium aluminum hydride (LiAlH4) reduction.

         H

         |

         CH3

          |

          C6H5

          |

    OH --- CH2 --- CH3

Step 3: Reaction of 1-phenylethanol with thionyl chloride (SOCl2) to give 1-chloro-1-phenylethane.

        H

        |

        CH3

         |

         C6H5

         |

   Cl --- CH2 --- CH3

Step 4: Reaction of 1-chloro-1-phenylethane with sodium iodide (NaI) in acetone to give 1-iodo-1-phenylethane through the Finkelstein reaction.

        H

        |

        CH3

         |

         C6H5

         |

   I --- CH2 --- CH3

Step 5: Conversion of 1-iodo-1-phenylethane to phenylethene (styrene) using potassium tert-butoxide (KOtBu) in dimethylformamide (DMF) through the dehydrohalogenation reaction.

        H

        |

        CH3

         |

         C6H5

         |

   CH = CH2

Overall, the reaction sequence can be represented as:

      H

      |

      CH3                          H

       |                           |

       C6H5                       CH

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There are approximately 4.31 × 10²² atoms of nitrogen in 1g of ammonia (NH₃).  There are approximately 8.62 × 10²² atoms of nitrogen in 1g of hydrazine (N₂H₄).

To calculate the number of atoms of nitrogen in a given amount of a compound, we need to use the molar mass of nitrogen and the formula of the compound.

The molar mass of nitrogen (N) is approximately 14.01 g/mol.

Ammonia (NH₃):

The formula for ammonia is NH₃. It contains one atom of nitrogen.

To calculate the number of moles of nitrogen element in 1g of ammonia, we divide the given mass by the molar mass of nitrogen:

Number of moles of nitrogen = 1g / 14.01 g/mol ≈ 0.0714 mol

Since 1 mole of a substance contains Avogadro's number of particles (6.022 × 10²³), the number of atoms of nitrogen in 0.0714 mol is:

Number of atoms of nitrogen = 0.0714 mol × 6.022 × 10²³ atoms/mol ≈ 4.31 × 10²² atoms

Therefore, there are approximately 4.31 × 10²² atoms of nitrogen in 1g of ammonia (NH₃).

Hydrazine (N₂H₄):

The formula for hydrazine is N₂H₄. It contains two atoms of nitrogen.

To calculate the number of moles of nitrogen in 1g of hydrazine, we again divide the given mass by the molar mass of nitrogen:

Number of moles of nitrogen = 1g / 14.01 g/mol ≈ 0.0714 mol

Since there are two atoms of nitrogen in each molecule of hydrazine, the number of atoms of nitrogen in 0.0714 mol is:

Number of atoms of nitrogen = 0.0714 mol × 2 × 6.022 × 10²³ atoms/mol ≈ 8.62 × 10²² atoms

Therefore, there are approximately 8.62 × 10²² atoms of nitrogen in 1g of hydrazine (N₂H₄).

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--The  question is incomplete, the given complete question is:

"What number of atoms of nitrogen are present in 1g of each of the compounds that is  ammonia (NH₃) and hydrazine (N₂H₄)?"--

The chemical equation for the reaction between iron(III) oxide and carbon monoxide is given by:

2Fe₂O₃ + 3CO → 4Fe + 3CO₂

The reaction between iron(III) oxide and carbon monoxide is a redox reaction in which carbon monoxide is the reducing agent while iron(III) oxide is the oxidizing agent. In this reaction, Fe₂O₃ is reduced to Fe, and CO is oxidized to CO₂.

The balanced equation shows that for every 2 moles of iron(III) oxide that reacts, we get 4 moles of iron and 3 moles of carbon dioxide as the products. This reaction is also called the thermite reaction and is an exothermic reaction that produces a lot of heat, making it useful in welding and metallurgical processes. The reaction takes place at high temperatures and is self-sustaining, meaning that it does not require external energy input.

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Answer:

D. all of the above

Explanation:

Ice made from water is a solid because the atoms or molecules

- vibrate in place

- are unable to move around

- are closely packed and touching

In a solid, the atoms or molecules are unable to move around so they just vibrate in fixed positions and they are closely packed and touching.

The hydrolysis of the given acetal, H₂C(CH₃)(OCH₃)CH₃, in aqueous HCl results in the formation of two organic products: 2-methoxy-2-methylpropane and methanol.

The hydrolysis of an acetal involves the cleavage of the acetal bond and the formation of two separate organic compounds. In this case, the given acetal H₂C(CH₃)(OCH₃)CH₃ will undergo hydrolysis in aqueous HCl.

The hydrolysis of the acetal bond leads to the formation of 2-methoxy-2-methylpropane and methanol as the organic products. The structural formula for 2-methoxy-2-methylpropane can be represented as CH₃CH(OCH₃)CH₃, where the oxygen atom of the acetal is now bonded to the central carbon atom. The structural formula for methanol remains as CH₃OH.

To indicate stereochemistry, wedge/hash bond tools can be used. For example, if there is a chiral center in the acetal, the stereochemistry of the resulting products can be shown by using wedges and hashes to depict the spatial arrangement of the substituents around that chiral center.

In summary, the hydrolysis of the given acetal in aqueous HCl yields 2-methoxy-2-methylpropane and methanol as the organic products. The structural formulas for these compounds can be represented, considering the stereochemistry where applicable, using wedge/hash bond tools.

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Answer:

Reaction types are :

Answer:

Synthesis and Combustion

Explanation:

For edge

The identity of the gas is hydrogen (H2).The average kinetic energy of gas molecules is directly proportional to the temperature. Using the formula for kinetic energy (KE = (1/2)mv^2), we can equate the kinetic energy of the unknown gas (Z) at 250K to that of hydrogen (H2) at 500K.

Since the average velocity is proportional to the square root of the kinetic energy, we can set up the following equation:

(1/2)mz(vz^2) = (1/2)mh(vh^2)

Given that mz is the molar mass of gas Z and mh is the molar mass of hydrogen (2 g/mol), and vz is the velocity of gas Z at 250K, and vh is the velocity of hydrogen at 500K.We can simplify the equation by canceling out the common factors:

mz(vz^2) = mh(vh^2)

The gas with an average velocity equal to hydrogen at twice the temperature is hydrogen gas itself (H2).

Since the average velocity of gas Z is equal to the average velocity of hydrogen at twice the temperature, the molar mass of gas Z must be half of the molar mass of hydrogen.

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                 Answer:

                     ⬇️

There are 1.9926 * 10 ^24 atoms.


M ( Mg ) = 80.45 g

We can find the molar mass of magnesium in the Periodic table:M ( Mg ) =              24.31 g / molen = 80.45 g :

                                     24.31 g/mole = 3.31 molesN a ( Avogadro`s constant ) = 6.02 * 10 ^23N = Na * n = 6.02 * 10 ^24 * 3.31 = 19.926 * 10 ^24

Answer: True

Explanation:

The second sample produced 30 g of element a and 40 g of element b, indicating the composition of the compound.

Based on the given information, we have two samples of a compound containing elements a and b. In the first sample, 15 g of element a and 35 g of element b were produced upon decomposition.

For part a, we are given that the second sample produced 30 g of element a. To find the mass of element b in the second sample, we can use the concept of conservation of mass. Since the total mass of the compound remains constant, we can subtract the mass of element a from the total mass of the second sample to determine the mass of element b.

The total mass of the second sample is 70 g (30 g of element a + mass of element b). Subtracting the mass of element a (30 g) from the total mass of the second sample (70 g), we find that the mass of element b in the second sample is 40 g (70 g - 30 g = 40 g).

To summarize, the second sample produced 30 g of element a and 40 g of element b.

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Explanation:

2Al(OH)3 + 3H2SO4 --> Al2 (SO4)3 + 6H2O

Answer:

The first ionization energy is the energy it takes to remove an electron from a neutral atom.

hope it is helpful :)

Answer:

a. 69.5% and 30.5%

Explanation:

we know there are two naturally occurring isotopes of Mg, Mg-24  and Mg-25.

First of all we will set the fraction for both isotopes

X for the isotopes having mass 25

1-x for isotopes having mass 24

The average atomic mass of Mg is 24.305 amu.

we will use the following equation,

25x + 24 (1-x) =  24.305

25x + 24 - 24x =  24.305

25x- 24x = 24.305 - 24

1x = 0.305

x= 0.305 /1

x= 0.305

0.305 × 100 = 30.5 %

30.5 % is abundance of Mg-25 because we solve the fraction x.

now we will calculate the abundance of Mg-24.

(1-x)

1-0.305 =0.695

0.695  × 100 = 69.5%

69.5%   for Mg-24.

Answer:

Kinetic Energy

Explanation:

When we run we are using chemical energy in our bodies to produce movement ( kinetic energy ) .

Answer:

The correct option of the first option

Explanation:

Dissolution of a solute in a solution is the process in which the solute interacts with the solvent leading to the disappearance of the solute to form a solution. Rate of dissolution is primarily affected by three factors namely

Stirring of the solution: Stirring the solution causes the solute to interact more with the solvent, thus increasing the rate of dissolution.

Surface area of the solute: The smaller the surface area of the solute, the more easier it is for it (the solute) to interact with the solvent. Hence, the smaller the surface area, the faster the rate of dissolution.

Temperature of the solvent: An increase in temperature of the solvent causes the generally causes more things to dissolve in this solvent due to increased energy of molecules in the solvent, which causes the molecules of the solute and solvent to interact more.

A decrease in the amount of solute will reduce the amount of interaction the solute (since it is in lesser quantity) have with the solvent hence the rate of dissolution will also be low (even though a solution will still be formed).

There are different factors that influences chemical equations.  The factor that causes a decrease in the rate of dissolution is decreasing the surface area of the solute.

Surface area is known to affect the rate of dissolution.  Note that When the total surface area of the solute particles is increased, the solute will then dissolves more quickly than ever.

The breaking of a solute into smaller pieces in turn do increases its surface area and increases its rate of solution and the act of stirring, do allows the solute to dissolve faster.

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Answer:

Backbone help us to be straight ,walk ,sleep etc

Explanation:

Backbone is the part of human body which is located back of our body.

It effort helps us to be straight do various work

The spine or the backbone is the central structure of the vertebrate body and it serves a few imperative capacities:

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Answer:

kenetic

Explanation:

<3

The major products from the reaction of methyl butanoate with diisobutylaluminum hydride at -78 degrees Celsius, followed by acidic work-up, are 2-methylbutanol and isobutyl acetate.

1. Reaction with diisobutylaluminum hydride: Diisobutylaluminum hydride (DIBAL-H) is a strong reducing agent that can convert esters into alcohols. In this case, methyl butanoate undergoes reduction to form 2-methylbutanol.

2. Acidic work-up: After the reduction step, the reaction mixture is treated with an acidic solution. This step helps in the hydrolysis of any remaining DIBAL-H and in the conversion of the intermediate alkoxyaluminum species to the corresponding alcohol and aluminum hydroxide.

Overall reaction:

Methyl butanoate + Diisobutylaluminum hydride → 2-Methylbutanol + Aluminum hydroxide

Additional product: Isobutyl acetate may also be formed as a minor product, resulting from the reaction of diisobutylaluminum hydride with the carbonyl group of the ester.

It is important to note that the reaction conditions, such as temperature and reagent concentrations, can influence the selectivity and yield of the products. The specific reaction conditions used in the experimental setup can provide more detailed information about the major products obtained.

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Answer:

2:8

Explanation:

Answer:

2:8

Explanation:

got it right on edge

When a diprotic acid is titrated with a strong base, and the Ka1 and Ka2 are significantly different, then the pH vs. volume plot of the titration will have: two distinct equivalence points. The answer is C.

There are two distinct steps in the titration curve, the first equivalence point is the point at which the base has reacted with all of the H+ ions from the first acidic hydrogen, while the second equivalence point is the point at which the base has reacted with all of the H+ ions from the second acidic hydrogen.

The pH at the first equivalence point will be less than 7, and the pH at the second equivalence point will be greater than 7, indicating that the solution is acidic for the first equivalence point and basis for the second equivalence point.

The Ka1 and Ka2 values for diprotic acids are typically different because the first hydrogen ion is more strongly bound to the molecule than the second hydrogen ion, resulting in different dissociation constants for each hydrogen ion.

Therefore, the pH vs. volume plot of the titration of a diprotic acid with a strong base will have two distinct equivalence points if Ka1 and Ka2 are significantly different.

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Answer:

Explanation:

The systematic name of the coordination compound [Ir(NH3)4Br2]Br is tetraamminedibromoiridium (III) bromide

Answer:

C air resistance is the answer

Answer: Im not sure but i think its B

Explanation: Its hard to explain so im

not gonna do it

Taking into account the definition of strong acid, pH and pOH, it is obtained that the correct option is option A) [H₃O⁺]= 5.5×10⁻² M, [OH⁻]= 1.82×10⁻¹³ M, pH= 1.26

It is called strong acid, that acid that dissociates completely in solution at constant temperature and pressure. Under these conditions, the concentration of a strong acid is equal to the concentration of hydrogen ions (Hydronium or H₃O⁺). In other words, a strong acid is an acid that completely dissociates into hydrogen ions and anions in solution.  

Hydrobromic acid HBr is a strong acid, so a concentration of 0.055 M of the acid generates a H₃O⁺ concentration of the same value. This is:

[HBr]= [H₃O⁺]= 0.055 M= 5.5×10⁻² M

On the other hand, pH is a measure of acidity or alkalinity that indicates the amount of hydrogen ions present in a solution or substance. Mathematically it is calculated as the negative logarithm in base 10 of the activity of hydrogen ions:

pH= -log [H₃O⁺]

Being [H₃O⁺]= 5.5×10⁻² M, the pH in this case is:

pH= -log (5.5×10⁻² M)

Solving:

pH= 1.26

Similarly, pOH is defined as the negative base 10 logarithm of the activity of the OH⁻ ions:

pOH= - log [OH⁻]

The following relationship can be established between pH and pOH:

pH + pOH= 14

Then, being pH = 1.26, the pOH is calculated by:

1.26 + pOH= 14

pOH= 14 - 1.26

pOH= 12.74

Replacing in the definition of pOH:

12.74= - log [OH⁻]

and solving you get:

[OH⁻]= 1.82×10⁻¹³ M

Finally, the correct option is option A) [H₃O⁺]= 5.5×10⁻² M, [OH⁻]= 1.82×10⁻¹³ M, pH= 1.26

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Answer:

H₂CO₃

Explanation:

Water + Carbon di oxide

H20+ CO2

= H2CO3

The limiting reactant is potassium aluminum sulfate.

To determine the limiting reactant, we need to compare the amount of each reactant to the stoichiometry of the balanced chemical equation. The balanced equation for the reaction between ammonium hydroxide (NH4OH) and potassium aluminum sulfate (KAl(SO4)2) is:

2 NH4OH + KAl(SO4)2 -> Al(OH)3 + (NH4)2SO4 + K2SO4

From the equation, we can see that the stoichiometric ratio between ammonium hydroxide and potassium aluminum sulfate is 2:1. Therefore, we need twice as many moles of ammonium hydroxide as potassium aluminum sulfate.

To calculate the moles of each reactant, we use the formula:

moles = concentration (M) × volume (L)

For the ammonium hydroxide:

moles of NH4OH = 3.0 M × 0.100 L = 0.300 mol

For the potassium aluminum sulfate:

moles of KAl(SO4)2 = mass (g) / molar mass (g/mol)

moles of KAl(SO4)2 = 30.0 g / (39.1 g/mol + 26.98 g/mol + 2(32.1 g/mol) + 4(16.0 g/mol)) = 0.083 mol

Since the stoichiometric ratio is 2:1, the moles of ammonium hydroxide are in excess.

To determine the theoretical yield of aluminum hydroxide (Al(OH)3), we need to convert the moles of the limiting reactant (potassium aluminum sulfate) to moles of the product using the stoichiometry of the balanced equation. From the balanced equation, we can see that the stoichiometric ratio between potassium aluminum sulfate and aluminum hydroxide is 1:1.The moles of aluminum hydroxide produced will be the same as the moles of potassium aluminum sulfate used, which is 0.083 mol.

To calculate the theoretical yield in grams, we use the formula:

mass = moles × molar mass

The molar mass of aluminum hydroxide (Al(OH)3) is (26.98 g/mol + 3(16.0 g/mol)) = 78.0 g/mol.

The theoretical yield of aluminum hydroxide is:

mass = 0.083 mol × 78.0 g/mol = 6.474 g

Therefore, the theoretical yield of aluminum hydroxide is 6.474 grams.

Aluminum hydroxide is a precipitate, which means it forms solid particles when the reaction occurs. It can be collected on a filter paper using a filtration process. Filtration is a common method to separate solids from liquids. The reaction mixture can be poured through a filter paper funnel, and the solid aluminum hydroxide particles will be trapped on the filter paper while the liquid and soluble salts (such as ammonium sulfate and potassium sulfate) pass through.However, it's important to note that the success of the filtration process depends on the particle size and the nature of the solid precipitate. If the particles of aluminum hydroxide are too fine or colloidal in nature, they may pass through the filter paper and affect the efficiency of the filtration. In such cases, additional techniques like centrifugation or using a finer filter may be required to achieve better separation.

Overall, collecting aluminum hydroxide on a filter paper is a feasible method in this scenario, provided the precipitate is of the appropriate size and nature for filtration.

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