Which of the following statements is false?

Which best describes an oxidizing agent?

stellarik [79]

Answer:B) a reactant that undergoes reduction

Explanation:

Oxidation reaction is defined as the reaction in which a substance looses electrons. The oxidation state of the substance increases during oxidation.

$M\rightarrow M^++e^-$

Reduction reaction is defined as the reaction in which a substance gains electrons. The oxidation state of the substance gets reduced during reduction.

$N^++e^-\rightarrow N$

Overall reaction: $M+N^+\rightarrow M^++N$

The substance M which itself gets oxidized, reduces other and is called as reducing agent. The substance N which itself gets reduced, oxidizes other and is called as oxidizing agent.

6 0

3 years ago

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An isotope of lead can be represented by the following symbol ²⁰⁷ 82 PB. From this symbol, it can be deduced that one atom of th

9966 [12]

Answer:

B) 82 protons

Explanation:

The number of protons for an element never change. 82 Is the atomic number for Pb which is how many protons it has. An isotope means it has more neutrons.

6 0

3 years ago

please help!!!!!! What is the mass of 1 mole of magnesium (Mg)? Express your answer to four significant figures. The mass of 1 m

Semmy [17]

I believe you just look at your periodic table for this value. I don't think there is any math involved.
Therefore one mole of Mg = 24.305g.

5 0

2 years ago

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9. Using the balanced equation from Question #8, how many grams of lead will be produced if 2.54 grams of PbS is burned with 1.8

MissTica

Answer: 2.24 grams of Pb

Explanation:

Step 1

Balanced chemical reaction;

2PbS + 3O2 → 2Pb + 2SO3

Step 2

Moles of both PbS and O2

Moles = mass / molar mass

Moles of PbS = 2.54 g / 239.3 g/mol = 0.0108 moles

Moles of O2 = 1.88 / 32 g/mol = 0.0588 moles

Step 3

Finding the limiting reactant.

Limiting reactant, is that reactant which is completely used in the reaction;

If we assume that PbS is the limiting reactant;

We have 0.0588 moles of O2. This needs ( 0.0588 * 2) / 3 = 0.0392 moles of PbS to fully react. But we have only 0.0108 moles of PbS available. That means that the PbS will be completely consumed hence the limiting reactant

If we assume O2 is the limiting reactant;

We have 0.0108 moles of PbS. That needs ( 0.0108 * 3) / 2 = 0.0162 moles of O2. But we have 0.0588 moles of O2 which is in excess further confirming that PbS is the limiting reactant since it will be depleted in the reaction.

Step 4

Moles of lead

For this step we apply the mole ratios with the limiting reactant;

Mole ratio of PbS : Pb = 2 : 2 = 1 : 1

Therefore;

Moles of Pb = (0.0108 moles * 1 ) 1

Moles of Pb =0.0108 moles

Step 5

Mass of Pb

Mass = moles * molar mass

Mass of Pb =0.0108 moles * 207.2 g/mol

Mass of Pb = 2.24 grams

5 0

2 years ago

A sample of 3.62 moles of diphosphorous trioxide is

Sindrei [870]

$H_3PO_3$ $2H_3PO_3$ $2H_3PO_3$ $P_2O_3$ Answer:

B

Explanation:

This question is about stoichiometry. From the balanced equation $P_2O_3 + 3H_2O$ ⇒ $2H_3PO_3$ , we see that 3 moles of water is needed to react with 1 mole of $P_2O_3$ .

This means that, to fully react 3.62 moles of $P_2O_3$ , we would need 3*3.62 or 10.86 moles of water. However, we only have 6.31 moles, so water is the limiting reactant.

Since 3 moles of water react with 1 mole of $P_2O_3$ , 6.31 moles of water can fully react with 6.31÷3 or 2.1033 moles of $P_2O_3$ .

From the balanced equation, we see that every mole of $P_2O_3$ reacted gets you 2 moles of $2H_3PO_3$ . Therefore, 2.1033 moles of $P_2O_3$ would give you approximately 4.21 moles of $H_3PO_3$ .

5 0

2 years ago