Answer:
The answer to your question is letter B, 2-methylhexane.
Explanation:
Remember that for naming organic compounds first, we need to look for the largest chain of carbons.
In your example, the largest chain is horizontal and has 6 carbons.
Later, we need to circle all the branches, in your example there is only one branch located close to the left side
After that, we number the carbons of the main chain, starting in the corner with more branches, in your example we start from the first carbon on the left.
Finally, start naming the number of the carbon branch, later hte name of the branch and finally the name of the main chain.
Answer:
2Fe + 3H2SO4 + Fe2(SO4)3+ 3H2
Explanation:
1. Fe (SO4) 3 is an incorrectly written formula because iron is trivalent as we can see by this three ahead of SO4. SO4 is divalent always.
2. since (SO4) is 3, this three shows us that there must be 3 in the reactants as well.
so now there is 3H2SO4
3. Since we have added 3 to one hydrogen we must add another. So now it's 3H2
4. and finally iron. In Fe2 (SO4) 3 we see this 2 in front of Fe which means it goes 2Fe.
Answer:
b) 2H+(aq) + 2C1-(aq) + Zn(s) → H2(g) + Zn2+(aq) + 2Cl-(aq)
Explanation:
The equation is given as;
2HCl(aq) + Zn(s) + H2(g) + ZnCl2(aq)
In writing an ionic equation, only the aqueous compounds dissociates into ions. This means HCl and ZnCl2 would dissociate to form ions.
This is given as;
2H+ + 2Cl- + Zn(s) --> H2(g) + Zn2+ + 2Cl-
The correct option is;
b) 2H+(aq) + 2C1-(aq) + Zn(s) → H2(g) + Zn2+(aq) + 2Cl-(aq)
Answer:
Described by a redox reaction below
Explanation:
Iron(III) oxide is an ionic compound, since it consists of a metal, iron, and a nonmetal, oxygen.
Ionic compounds are formed when metals lose their valence electrons in order to have an octet in their previous shell and donate them to nonmetal atoms, so that nonmetals fill their outer shell to have an octet.
As a result, positive ions (cations) and negative ions (anions) are formed. When iron reacts with oxygen, the following reaction takes place:
This is a redox (oxidation–reduction) reaction, since we have electron loss and gain. Four iron atoms lose a total of 12 electrons to obtain a +3 charge in the final compound, while 3 oxygen molecules gain these 12 electrons to become 6 oxide anions with a -2 charge.
Answer:
%N = 25.94%
%O = 74.06%
Explanation:
Step 1: Calculate the mass of nitrogen in 1 mole of N₂O₅
We will multiply the molar mass of N by the number of N atoms in the formula of N₂O₅.
m(N): 2 × 14.01 g = 28.02 g
Step 2: Calculate the mass of oxygen in 1 mole of N₂O₅
We will multiply the molar mass of O by the number of O atoms in the formula of N₂O₅.
m(O): 5 × 16.00 g = 80.00 g
Step 3: Calculate the mass of 1 mole of N₂O₅
We will sum the masses of N and O.
m(N₂O₅) = m(N) + m(O) = 28.02 g + 80.00 g = 108.02 g
Step 4: Calculate the percent composition of N₂O₅
We will use the following expression.
%Element = m(Element)/m(Compound) × 100%
%N = m(N)/m(N₂O₅) × 100% = 28.02 g/108.02 g × 100% = 25.94%
%O = m(O)/m(N₂O₅) × 100% = 80.00 g/108.02 g × 100% = 74.06%
