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3 years ago

How many grams of zinc would be required to produce 9.65g of zinc hydroxide

Chemistry

1 answer:

Effectus [21]3 years ago

6 0

The question is incomplete, here is the complete question:

How many grams of zinc would be required to produce 9.65g of zinc hydroxide

Zn+2MnO₂+H₂O→Zn(OH)₂+Mn₂O₃

Answer: The mass of zinc required is 6.35 grams

Explanation:

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of zinc hydroxide = 9.65 g

Molar mass of zinc hydroxide = 99.4 g/mol

Putting values in equation 1, we get:

$\text{Moles of zinc hydroxide}=\frac{9.65g}{99.4g/mol}=0.0971mol$

The given chemical equation follows:

$Zn+2MnO_2+H_2O\rightarrow Zn(OH)_2+Mn_2O_3$

By Stoichiometry of the reaction:

1 mole of zinc hydroxide is produced from 1 mole of zinc

So, 0.0971 moles of zinc hydroxide will be produced from = $\frac{1}{1}\times 0.0971=0.0971mol$ of zinc

Now, calculating the mass of zinc from equation 1, we get:

Molar mass of zinc = 65.4 g/mol

Moles of zinc = 0.0971 moles

Putting values in equation 1, we get:

$0.0971mol=\frac{\text{Mass of zinc}}{65.4g/mol}\\\\\text{Mass of zinc}=(0.0971mol\times 65.4g/mol)=6.35g$

Hence, the mass of zinc required is 6.35 grams

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Carbons from acetyl CoA are transferred to the citric acid cycle. Which is the first round of the citric acid cycle that could p

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Answer:

In the third step of the citric acid cycle, the oxidation of isocitrate takes place and one molecule of carbon dioxide is released.

Explanation:

In the first step of citric acid cycle, acetylCoA combines with a four-carbon molecule, oxaloacetate, forming a six-carbon molecule, citrate.

In the second step, the citrate in the presence of enzyme anicotase is converted into isocitrate.

In the third step, the oxidation of isocitrate takes place and one molecule of carbon dioxide is released leaving behind one five-carbon molecule called as α-ketoglutarate. During this step, NAD⁺ is reduced to form NADH.

This is first round of the citric acid cycle that could possibly release a carbon atom originating from this acetyl CoA.

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3 years ago

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Gather and respond to information

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2 years ago

A galvanic (voltaic) cell consists of an electrode composed of zinc in a 1.0 M zinc ion solution and another electrode composed

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Answer:

The E°cell for the galvanic cell is 1.56 V.

Explanation:

A galvanic cell is a device that uses redox reactions to convert chemical energy into electrical energy. The chemical reaction used is always spontaneous.

Oxide-reduction reactions, also called redox, involve the transfer or transfer of electrons between two or more chemical species. In these reactions two substances interact: the reducing agent and the oxidizing agent.

The gain of electrons is called reduction and the loss of electrons oxidation. That is to say, there is oxidation whenever an atom or group of atoms loses electrons (or increases its positive charges) and in the reduction an atom or group of atoms gains electrons, increasing its negative charges or decreasing the positive ones.

The species that supplies electrons is the reducing agent (that is, it is that species that oxidizes, yielding electrons and increasing its positive charge, or decreasing the negative one causing the reduction of the other species) and the one that gains them is the oxidizing agent ( that is, it is that species that is reduced, capturing electrons and increasing its negative charge, or decreasing its positive charge, causing oxidation of the other species).

The galvanic cell works as follows: In the anodic half-cell oxidations occur, while in the cathodic half-cell reductions occur. The anode electrode, conducts the electrons that are released in the oxidation reaction, to the metallic conductors. These electrical conductors conduct the electrons and carry them to the cathode electrode; the electrons thus enter the cathode half-cell and the reduction takes place in it.

To determine the oxidizing and reducing agent you must first know the reduction potentials. For this you consult the list of standard reduction potentials. In this list you can see that the semi-reactions that occur with their corresponding potentials are:

Ag⁺ + e⁻ ⇒ Ag E°= 0.80 V

Zn²⁺ + 2 e⁻ ⇒ Zn E° -0.76 V

The species that has the greatest potential for reduction will be the species that will be reduced, that is, it will be the oxidizing agent. In this case, it will be the experience corresponding to silver (Ag). Therefore, to obtain the redox reaction, the half-reaction corresponding to zinc (Zn) must be reversed to be an oxidation, keeping its E ° value constant. Then:

Reduction: Ag⁺ + e⁻ ⇒ Ag E°= 0.80 V

Oxidation: Zn ⇒ Zn²⁺ + 2 e⁻ E° -0.76 V

So: E°cell=Ereduction - Eoxidation

Or what is the same E°cell=Ecathode - Eanode because the reduction always occurs in the cathode and oxidation in the anode.

E°cell=0.80 V - (-0.76) V

E°cell= 1.56 V

Then the E°cell for the galvanic cell is 1.56 V.

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3 years ago

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Answer: d) -705.55 kJ

Explanation:

Heat of reaction is the change of enthalpy during a chemical reaction with all substances in their standard states.

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Reversing the reaction, changes the sign of $\Delta H$

$C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)$

$\Delta H=-1411.1kJ$

On multiplying the reaction by $\frac{1}{2}$ , enthalpy gets half:

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Thus the enthalpy change for the given reaction is -705.55kJ

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3 years ago

. how many lone pairs of electrons are present in the lewis structure of calcium sulfide?

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Answer:
Four

Explanation:
Calcium Sulfide is an ionic compound made up of Ca²⁺ and S²⁻.
Ca²⁺ is formed as,

Ca → Ca²⁺ + 2 e⁻

These two electrons are accepte by Sulfur as,

S + 2 e⁻ → S²⁻

So, before accepting 2 electrons S was having six valence electrons, after accepting two electrons from Ca it has 8 electrons which are present in four pairs as shown below,

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4 years ago