Part D Questions and Problems

Chemistry

1 answer:

jonny [76]3 years ago

4 0

Answer:

There are 120 possible ways.To calculate the number of permutations here, where order is important and repetition is not allowed, we use the following formula:Number of permutations = n! / (n - r)! = 5!/0! = 12345 / 1 (note: "!" means "factorial" and 0! equals 1) = 120/1 = 120.For a complete list see below. Let a,b,c,d & e represent the 5 players and their order determine their position:{a,b,c,d,e}

Explanation:

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Does anyone have the student exploration sheet answers for the Drug Dosage (Forensic Science) Gizmos lab?

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3 years ago

Based on the chemical equation, use the drop-down menu to choose the coefficients that will balance the chemical equation: ()zn

kompoz [17]

Based on the chemical equation, to balance the given equation we should use 2 as a coefficients of HCl on reacatnt side of the reaction.

<h3>What is balanced equation?</h3>

Balanced chemical equations are those equation in which each entities are present in same amount on reactant side as well as on the product side of the chemical reaction.

Given chemical reaction is:

Zn + HCl → ZnCl₂ + H₂

In the above reaction equation is not balanced as number of chlorine and hydrogen atoms are not equal on both sides, so balanced equation will be:

Zn + 2HCl → ZnCl₂ + H₂

Hence we add 2 as a coefficient of HCl to balance the equation.

To know more about balance equation, visit the below link:

brainly.com/question/15355912

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2 years ago

Determine the reducing agent in the following reaction. Explain your answer. 2 Li(s) + Fe(C2H3O2)2(aq) → 2 LiC2H3O2(aq) + Fe(s)

Brilliant_brown [7]

The reducing agent in the reaction 2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) is lithium (Li).

The general reaction is:

2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) (1)

We can write the above reaction in two reactions, one for oxidation and the other for reduction:

Oxidation reaction

Li⁰(s) → Li⁺(aq) + e⁻ (2)

Reduction reaction

Fe²⁺(aq) + 2e⁻ → Fe⁰(s) (3)

We can see that Li⁰ is oxidizing to Li⁺ (by losing one electron) in the lithium acetate (reaction 2) and that Fe²⁺ in iron(II) acetate is reducing to Fe⁰ (by gaining two electrons) (reaction 3).

We must remember that the reducing agent is the one that will be oxidized by reducing another element and that the oxidizing agent is the one that will be reduced by oxidizing another species.

In reaction (1), the reducing agent is Li (it is oxidizing to Li⁺), and the oxidizing agent is Fe(CH₃COO)₂ (it is reducing to Fe⁰).

Therefore, the reducing agent in reaction (1) is lithium (Li).

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