Answer: C) Roman numeral following the name.
Explanation: If we want to name an ionic compound like NaCl then we can easily write its name. Na is sodium and Cl is chlorine. First we write the name of the metal ion and then the name of the anion it has. Here we have mono atomic ion and for these we use the suffix -ide. So, the name will be sodium chloride.
Now, if we have something like 
and we write the name as Iron chloride, then it will not be correct since Fe is a transition metal and it shows +2 and +3 oxidation states. So, to overcome this difficulty, a roman numeral is used for the oxidation state of the metal ion. The name of above compound will be Iron(III)chloride. Similarly, if we have 
then its name will be written as Iron(II)chloride.
In both the above names, the roman numerals (III) and (II) are indicating numerical value of the charge of the metal ion. So, the correct choice is C) Roman numerals following the name.
when naming a transition metal ion that can have more than one common ionic charge, the numerical value of the charge can be indicated by a Roman numeral following the name.
It is important to test only one variable, because 2 or more may cause changes in the original variable that weren't supposed to occur
Answer:
a
Explanation:
The pollution will kill the local pond life, and amphibians are very delicate to change.
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To balance any redox equation, first divide the reaction into two half reactions, and balance the atom whose oxidation number is changing:
<span>NO2 --> NH3 </span>
<span>H2 --> H2O </span>
<span>Next, balance oxygens by adding H2O where needed, and then balance hydrogen by adding H+: </span>
<span>NO2 --> NH3 + 2 H2O </span>
<span>H2O + H2 --> H2O </span>
<span>7 H+ + NO2 --> NH3 + 2 H2O </span>
<span>H2O + H2 --> H2O + 2 H+ </span>
<span>Next, balance charges by adding electrons (e-): </span>
<span>7 H+ + NO2 + 7 e- --> NH3 + 2 H2O </span>
<span>H2O + H2 --> H2O + 2 H+ + 2 e- </span>
<span>Now, multiply one or both equations by small numbers to make the number of electrons the same in the two half reactions. In this case, multiply the top equation by 2 and the bottom one by 7 to give you 14 electrons in each half reaction: </span>
<span>14 H+ + 2 NO2 + 14 e- --> 2 NH3 + 4 H2O </span>
<span>7 H2O + 7 H2 --> 7 H2O + 14 H+ + 14 e- </span>
<span>Finally, add the two half reactions and simplify by cancelling anything that is on both sides. This will give you the final balanced reaction: </span>
<span>2 NO2 + 7 H2 --> 2 NH3 + 4 H2O </span>
<span>So, the answer to the original question is (A) 7
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