Answer:
N2 (g) + 3 H2 (g) -> 2 NH3 (g)
Explanation:
First, write the reactants on the left and the products on the right side. Also, add the states of matter after the reactants and products. Note, nitrogen gas is written as N2 and hydrogen gas is written as H2, not N or H.
N2 (g) + H2 (g) -> NH3 (g)
Next, balance the equation! Since there are two nitrogens on the left, add a two to the product NH3. Now there are 6 hydrogens on the right side. To balance that, add a 3 to the H2.
N2 (g) + 3 H2 (g) -> 2 NH3 (g)
Double checking, there are 2 nitrogens and 6 hydrogens on each side.
Success!
What is the theoretical yield of aluminum oxide if 2.40 mol of aluminum metal is exposed to 2.10 mol of oxygen?<span><span>Ask for details </span><span>Follow </span>Report</span><span> by <span>Sapperd2248 </span><span>1 minute ago
1879</span></span>
Ra, would have the lowest ionization energy. Remember ionization energy increases going up and to the right.
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Here we have to choose the specific iron ion preset in the compound FeO among the given ions.
The iron ion present in FeO is iron (II) ion.
The compound FeO is a neutral compound. Where the anion is oxide ion (O²⁻) thus the cation should also contain two unit positive charge. Thus Fe²⁺. Hence , the iron ion is Fe (II) ion.
The ferride ion means Fe²⁻ ion which cannot be present in FeO as the anion is already present which is O²⁻.
The iron (I) ion means iron present in +1 state, then to neutralize the compound 2 iron (I) is needed but in compound FeO there is only one iron present. Thus it cannot be iron (I) ion.
The ferride (I) ion means Fe¹⁻, which cannot be the state of the iron in FeO as there already present an anion which is O²⁻ ion.