Answer:
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Explanation:
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The bonds of a glucose molecule store chemical energy
Answer:
In an acid-base equilibrium, acid becomes a conjugate base and base becomes a conjugate acid.
Explanation:
Let's remember the Bronsted-Lowry theory to answer this specific question. According to the theory, acid is a proton donor, while a base is a proton acceptor.
Consider an acid in a form HA (aq) and base in a form of B (aq). Since acid is a proton donor, it will donate its hydrogen ion to the base, B. The resultant products would be (aq) and (aq).
Remember that an acid-base reaction is an equilibrium reaction. This means we may also look at this proton transfer reaction from the product side towards the reactants. Summarizing what has been said, we may write the equilibrium as:
⇄
Now acid, HA, donates a proton to become a conjugate base. The conjugate base, if we look from the reverse equation side, is actually a base, since it can accept a proton to become HA. Similarly, B accepts a proton to become a conjugate acid. Looking from the reverse reaction, it can now donate a proton, so in reality we can consider it a base.
To summarize, your logic is correct.
Answer:
Explanation:
Group one elements are alkali metals. All alkali metal have one valance electron. They loses their one valance electron and from cation with charge of +1.
Charges on group one.
Hydrogen = +1
Lithium = +1
Sodium = +1
Potassium = +1
Rubidium = +1
Cesium = +1
Francium = +1
Group two elements are alkaline earth metals. All alkaline earth metal have two valance electron. They loses their two valance electron and from cation with charge of +2.
Charges on group two.
Beryllium = +2
Magnesium = +2
Calcium = +2
Strontium = +2
Barium= +2
Radium = +2
Group 13 elements are boron family. All elements have three valance electrons. They loses their three valance electron and from cation with charge of +3.
Charges on group 13.
Boron = +3
Aluminium = +3
Gallium = +3
Indium = +3
Thallium= +3
Group 13 elements are also shows +1 charge by losing one valance electron.
First find the oxidation states of the various atoms:
<span>in Cr2O2 2- Cr @ +1; In NH3 N @ +3; in CrO3 Cr @ +3, N2 N @ 0 </span>
<span>Note that N gained electrons, ie, was reduced; Cr was oxidized </span>
<span>Now there is a problem, because B has NH4+ which the problem did not, and is not balanced, showing e- in/out </span>
<span>B.NH4+ → N2 </span>
<span>Which of the following is an oxidation half-reaction? </span>
<span>A.Sn 2+ →Sn 4+ + 2e- </span>
<span>Sn lost electrons so it got oxidized</span>