Answer:
1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)
Explanation:
Main reaction: 2Ag⁺(aq) + Mn(s) ⇄ 2Ag(s) + Mn²⁺(aq)
In the oxidation half reaction, the oxidation number increases:
Mn changes from 0, in the ground state to Mn²⁺.
The reduction half reaction occurs where the element decrease the oxidation number, because it is gaining electrons.
Silver changes from Ag⁺ to Ag.
1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)
To balance the hole reaction, we need to multiply by 2, the second half reaction:
Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
(Ag⁺(aq) + 1e⁻ ⇄ Ag(s)) . 2
2Ag⁺(aq) + 2e⁻ ⇄ 2Ag(s)
Now we sum, and we can cancel the electrons:
2Ag⁺(aq) + Mn(s) + 2e⁻ ⇄ 2Ag(s) + Mn²⁺(aq) + 2e⁻
It is C.ti and o and you don’t have to make me brainlest
Fusion is a type of <u>n</u><u>u</u><u>c</u><u>l</u><u>e</u><u>ar</u><u> </u><u>reaction</u><u> </u>....
Answer:
The answer is 6.25g.
Explanation:
First create your balanced equation. This will give you the stoich ratios needed to answer the question:
2C8H18 + 25O2 → 16CO2 + 18H2O
Remember, we need to work in terms of NUMBERS, but the question gives us MASS. Therefore the next step is to convert the mass of O2 into moles of O2 by dividing by the molar mass:
7.72 g / 16 g/mol = 0.482 mol
Now we can use the stoich ratio from the equation to determine how many moles of H2O are produced:
x mol H2O / 0.482 mol O2 = 18 H2O / 25 O2
x = 0.347 mol H2O
The question wants the mass of water, so convert moles back into mass by multiplying by the molar mass of water:
0.347 mol x 18 g/mol = 6.25g
Answer:
- <u><em>It is positive when the bonds of the product store more energy than those of the reactants.</em></u>
Explanation:
The <em>standard enthalpy of formation</em>, <em>ΔHf</em>, is defined as the energy required to form 1 mole of a substance from its contituent elements under standard conditions of pressure and temperature.
Then, per defintion, when the elements are already at their standard states, there is not energy involved to form them from that very state; this is, the standard enthalpy of formation of the elements in their standard states is zero.
It is not zero for the compounds in its standard state, because energy should be released or absorbed to form the compounds from their consituent elements. Thus, the first choice is false.
When the bonds of the products store more energy than the those of the reactants, the difference is:
- ΔHf = ΔHf products - ΔHf reactants > 0, meaning that ΔHf is positive. Hence, the second statement is true.
Third is false because forming the compounds may require to use (absorb) or release (produce) energy, which means that ΔHf could be positive or negative.
Fourth statement is false, because the standard state of many elements is not liquid. For example, it is required to supply energy to iron to make it liquid. Thus, the enthalpy of formation of iron in liquid state is not zero.
