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amm1812
3 years ago
15

A child is upset because his ice cream is melting. He thinks he now has less dessert. Which of these explanations correctly stat

es why the child is incorrect?
Chemistry
1 answer:
11Alexandr11 [23.1K]3 years ago
4 0
The ice cream melts because it is gaining heat and energy from an endothermic reaction but the mass of the ice cream never leaves the cone. This is because the Law of Conservation of Mass says that matter is not created nor destroyed. Where can the mass go exactly?
Nowhere because it stays in the cone and is the same amount of ice cream as before.
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What does the oxidizing agent do in a redox reaction apex?
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Same as balancing a regular chemical reaction! Please see the related question to the bottom of this answer for how to balance a normal chemical reaction. This is for oxidation-reduction, or redox reactions ONLY! These instructions are for how to balance a reduction-oxidation, or redox reaction in aqueous solution, for both acidic and basic solution. Just follow these steps! I will illustrate each step with an example. The example will be the dissolution of copper(II) sulfide in aqueous nitric acid, shown in the following unbalanced reaction: CuS (s) + NO 3 - (aq) ---> Cu 2+ (aq) + SO 4 2- (aq) + NO (g) Step 1: Write two unbalanced half-reactions, one for the species that is being oxidized and its product, and one for the species that is reduced and its product. Here is the unbalanced half-reaction involving CuS: CuS (s) ---> Cu 2+ (aq) + SO 4 2- (aq) And the unbalanced half-reaction for NO 3 - is: NO 3 - (aq) --> NO (g) Step 2: Insert coefficients to make the numbers of atoms of all elements except oxygen and hydrogen equal on the two sides of each half-reaction. In this case, copper, sulfur, and nitrogen are already balanced in the two half-reaction, so this step is already done here. Step 3: Balance oxygen by adding H 2 O to one side of each half-reaction. CuS + 4 H 2 O ---> Cu 2+ + SO 4 2- NO 3 - --> NO + 2 H 2 O Step 4: Balance hydrogen atoms. This is done differently for acidic versus basic solutions. . For acidic solutions: Add H 3 O + to each side of each half-reaction that is "deficient" in hydrogen (the side that has fewer H's) and add an equal amount of H 2 O to the other side. For basic solutions: add H 2 O to the side of the half-reaction that is "deficient" in hydrogen and add an equal amount of OH - to the other side. Note that this step does not disrupt the oxygen balance from Step 3. In the example here, it is in acidic solution, and so we have: CuS + 12 H 2 O ---> Cu 2+ + SO 4 2- + 8 H 3 O + . NO 3 - + 4 H 3 O + --> NO + 6 H 2 O Step 5: Balance charge by inserting e - (electrons) as a reactant or product in each half-reaction. Oxidation: CuS + 12 H 2 O ---> Cu 2+ + SO 4 2- + 8 H 3 O + + 8 e - . Reduction: NO 3 - + 4 H 3 O + + 3 e - --> NO + 6 H 2 O . Step 6: Multiply the two half-reactions by numbers chosen to make the number of electrons given off by the oxidation step equal to the number taken up by the reduction step. Then add the two half-reactions. If done correctly, the electrons should cancel out (equal numbers on the reactant and product sides of the overall reaction). If H 3 O + , H 2 O, or OH - appears on both sides of the final equation, cancel out the duplication also. Here the oxidation half-reaction must be multiplied by 3 (so that 24 electrons are produced) and the reduction half-reaction must by multiplied by 8 (so that the same 24 electrons are consumed). 3 CuS + 36 H 2 O ---> 3 Cu 2+ + 3 SO 4 2- + 24 H 3 O + + 24 e - 8 NO 3 - + 32 H 3 O + + 24 e - ---> 8 NO + 48 H 2 O Adding these two together gives the following equation: 3 CuS + 36 H 2 O + 8 NO 3 - + 8 H 3 O + ---> 3 Cu 2+ + 3 SO 4 2- + 8 NO + 48 H 2 O Step 7: Finally balancing both sides for excess of H 2 O (On each side -36) This gives you the following overall balanced equation at last: 3 CuS (s) + 8 NO 3 - (aq) + 8 H 3 O + (aq) ---> 3 Cu 2+ (aq) + 3 SO 4 2- (aq) + 8 NO (g) + 12 H 2 O (l)


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Answer:

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Explanation:

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hodyreva [135]

The greater amount is the solvent and the lesser amount is the solute.

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Answer:

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Valency of Helium 2 0

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Valency of Nitrogen 7 3

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Valency of Fluorine 9 1

Valency of Neon 10 0

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Valency of Magnesium (Mg) 12 2

Valency of Aluminium 13 3

Valency of Silicon 14 4

Valency of Phosphorus 15 3

Valency of Sulphur 16 2

Valency of Chlorine 17 1

Valency of Argon 18 0

Valency of Potassium (K) 19 1

Valency of Calcium 20 2

Valency of Scandium 21 3

Valency of Titanium 22 4

Valency of Vanadium 23 5,4

Valency of Chromium 24 2

Valency of Manganese 25 7, 4, 2

Valency of Iron (Fe) 26 2, 3

Valency of Cobalt 27 3, 2

Valency of Nickel 28 2

Valency of Copper (Cu) 29 2, 1

Valency of Zinc 30 2

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2 years ago
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