Answer:
15 grams of water
Explanation:
15 grams of water of water would lose heat the faster compared to higher masses of water.
Water generally is a poor conductor heat.
- To heat up a unit of water, significant amount of energy must be added to the body of water.
- With time, the body continues to increase in temperature.
- A 500g mass of water will take more time to lose heat.
Answer:
You should start with 63.54 grams of copper.
Explanation:
The chemical reactions are processes in which the nature of the substances changes, that is, from some initial substances called reactants, totally different ones called products are obtained.
In the chemical reaction, the formulas of reagents and products appear preceded by numbers (the stoichiometric coefficients) that indicate the proportions according to which the transformation occurs. So you can say that stoichiometry establishes relationships between the molecules or elements that make up the reactants of a chemical equation with the products of said reaction. The relationships that are established are MOLAR relationships between the compounds or elements that make up the chemical equation: always in MOLES.
The stoichiometric coefficients of a chemical equation are due to the fact that the atoms present before the reaction must be the same after the reaction, although they will have been rearranged to produce new substances.
If you want 2 moles of silver (Ag), for stoichiometry of the reaction you need a moles of copper Cu. Being the molar mass of copper Cu 63.54 g / mole, then:
1 mole*63.54 g/mole= 63.54 g
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You should start with 63.54 grams of copper.</em></u>
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The best substance to heat up the fastest would be blue fire
Answer:
E° = 1.24 V
Explanation:
Let's consider the following galvanic cell: Fe(s) | Fe²⁺(aq) || Ag⁺(aq) | Ag(s)
According to this notation, Fe is in the anode (where oxidation occurs) and Ag is in the cathode (where reduction occurs). The corresponding half-reactions are:
Anode: Fe(s) ⇒ Fe²⁺(aq) + 2 e⁻
Cathode: Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an
E° = 0.80 V - (-0.44 V) = 1.24 V
