Do you have the picture of the data?
Answer:
The answer is
Explanation:
To find the number of moles in a substance given it's number of entities we use the formula
where n is the number of moles
N is the number of entities
L is the Avogadro's constant which is
6.02 × 10²³ entities
From the question we have
We have the final answer as
Hope this helps you
The answer is D, Magnesium.
For an anion or cation(negatively or positively charged ion) to replace a similarly-charged ion, it needs to be more active. This means that only a cation that is above Zinc on the reactivity scale can replace Zinc.
Since Aluminum was displaced in the reaction, that means the unknown metal is more reactive.
This means that our options are:
Lithium, Potassium, Calcium, Sodium, and Magnesium.
But because the metal did not displace the Sodium, it must be less reactive than it.
This gives us a new list of metals that are less-reactive than Sodium:
Magnesium, Aluminum, Zinc, Iron, Nickel, etc.
The only metal on both lists is Magnesium, and that means that your answer is D, Magnesium.
The first step in balancing any redox reaction is determining whether or not it is even an oxidation-reduction reaction, which requires that species exhibits changing oxidation statesduring the reaction. To maintain charge neutrality in the sample, the redox reaction will entail both a reduction component and an oxidation components and is often separated into independent two hypothetical <span>half-reactions </span>to aid in understanding the reaction. This requires identifying which element is oxidized and which element is reduced. For example, consider this reaction:
<span><span><span>Cu(s)+2A<span>g+</span>(aq)→C<span>u<span>2+</span></span>(aq)+2Ag(s)</span>(1)</span><span>(1)<span>Cu(s)+2A<span>g+</span>(aq)→C<span>u<span>2+</span></span>(aq)+2Ag(s)</span></span></span>
The first step in determining whether the reaction is a redox reaction is to splitting the equation into two hypothetical half-reactions. Let's start with the half-reaction involving the copper atoms:
<span><span><span>Cu(s)→C<span>u<span>2+</span></span>(aq)</span>(2a)</span><span>(2a)<span>Cu(s)→C<span>u<span>2+</span></span>(aq)</span></span></span>
The oxidation state of copper on the left side is 0 because it is an element on its own. The oxidation state of copper on the right hand side of the equation is +2. The copper in this half-reaction is oxidized as the oxidation states increases from 0 in Cu to +2 in Cu2+. Now consider the silver atoms
<span><span><span>2A<span>g+</span>(aq)→2Ag(s)</span>(2b)</span><span>(2b)<span>2A<span>g+</span>(aq)→2Ag(s)</span></span></span>
In this half-reaction, the oxidation state of silver on the left side is a +1. The oxidation state of silver on the right is 0 because it is an element on its own. Because the oxidation state of silver decreases from +1 to 0, this is the reduction half-reaction.
Consequently, this reaction is a redox reaction as both reduction and oxidation half-reactions occur (via the transfer of electrons, that are not explicitly shown in equations 2). Once confirmed, it often necessary to balance the reaction (the reaction in equation 1 is balanced already though), which can be accomplished in two ways because the reaction could take place in neutral, acidic or basic conditions.
To make food for the plant
