Answer:
![K_2=\frac{[NOBr]^4_{eq}}{[NO]^4_{eq}[Br]^2_{eq}}](https://tex.z-dn.net/?f=K_2%3D%5Cfrac%7B%5BNOBr%5D%5E4_%7Beq%7D%7D%7B%5BNO%5D%5E4_%7Beq%7D%5BBr%5D%5E2_%7Beq%7D%7D)
Explanation:
Hello,
In this case, for the equilibrium condition, the equilibrium constant is defined via the law of mass action, which states that the division between the concentrations of the products over the concentration of the reactants at equilibrium equals the equilibrium constant, for the given reaction:

The suitable equilibrium constant turns out:
![K_2=\frac{[NOBr]^4_{eq}}{[NO]^4_{eq}[Br]^2_{eq}}](https://tex.z-dn.net/?f=K_2%3D%5Cfrac%7B%5BNOBr%5D%5E4_%7Beq%7D%7D%7B%5BNO%5D%5E4_%7Beq%7D%5BBr%5D%5E2_%7Beq%7D%7D)
Or in terms of the initial equilibrium constant:

Since the second reaction is a doubled version of the first one.
Best regards.
<span>C7H8
First, lookup the atomic weight of all involved elements
Atomic weight of carbon = 12.0107
Atomic weight of hydrogen = 1.00794
Atomic weight of oxygen = 15.999
Then calculate the molar masses of CO2 and H2O
Molar mass CO2 = 12.0107 + 2 * 15.999 = 44.0087 g/mol
Molar mass H2O = 2 * 1.00794 + 15.999 = 18.01488 g/mol
Now calculate the number of moles of each product obtained
Note: Not interested in the absolute number of moles, just the relative ratios. So not going to get pedantic about the masses involved being mg and converting them to grams. As long as I'm using the same magnitude units in the same places for the calculations, I'm OK.
moles CO2 = 3.52 / 44.0087 = 0.079984
moles H2O = 0.822 / 18.01488 = 0.045629
Since each CO2 molecule has 1 carbon atom, I can use the same number for the relative moles of carbon. However, since each H2O molecule has 2 hydrogen atoms, I need to double that number to get the relative number of moles for hydrogen.
moles C = 0.079984
moles H = 0.045629 * 2 = 0.091258
So we have a ratio of 0.079984 : 0.091258 for carbon and hydrogen. We need to convert that to a ratio of small integers. First divide both numbers by 0.079984 (selected since it's the smallest), getting
1: 1.140953
The 1 for carbon looks good. But the 1.140953 for hydrogen isn't close to an integer. So let's multiply the ratio by 1, 2, 3, 4, ..., etc and see what each new ratio looks like (Effectively seeing what 1, 2, 3, 4, etc carbons look like)
1 ( 1 : 1.140953) = 1 : 1.140953
2 ( 1 : 1.140953) = 2 : 2.281906
3 ( 1 : 1.140953) = 3 : 3.422859
4 ( 1 : 1.140953) = 4 : 4.563812
5 ( 1 : 1.140953) = 5 : 5.704765
6 ( 1 : 1.140953) = 6 : 6.845718
7 ( 1 : 1.140953) = 7 : 7.986671
8 ( 1 : 1.140953) = 8 : 9.127624
That 7.986671 in row 7 looks extremely close to 8. I doubt I'd get much closer unless I go to extremely high integers. So it looks like the empirical formula for toluene is C7H8</span>
Answer:
Group 4A (or IVA) of the periodic table includes the nonmetal carbon (C), the metalloids silicon (Si) and germanium (Ge), the metals tin (Sn) and lead (Pb), and the yet-unnamed artificially-produced element ununquadium (Uuq).
The Group 4A elements have four valence electrons in their highest-energy orbitals (ns2np2). Carbon and silicon can form ionic compounds by gaining four electrons, forming the carbide anion (C4-) and silicide anion (Si4-), but they more frequently form compounds through covalent bonding. Tin and lead can lose either their outermost p electrons to form 2+ charges (Sn2+, the stannous ion, and Pb2+, the plumbous ion) or their outermost s and p electrons to form 4+ charges (Sn4+, the stannic ion, and Pb4+, the plumbic ion).
Carbon (C, Z=6).
Carbon is most familiar as a black solid is graphite, coal, and charcoal, or as the hard, crystalline diamond form. The name is derived from the Latin word for charcoal, carbo. It is found in the Earth's crust at a concentration of 480 ppm, making it the 15th most abundant element. It is found in form of calcium carbonate, CaCO3, in minerals such as limestone, marble, and dolomite (a mixture of calcium and
Explanation:
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<u>E</u><u>N</u><u>J</u><u>O</u><u>Y</u><u> </u><u>THE</u><em><u> </u></em><em><u>A</u></em><em><u>N</u></em><em><u>S</u></em><em><u>W</u></em><em><u>E</u></em><em><u>R</u></em>
Answer:
Sulfuric acid is a strong acid in any state
Explanation:
You may be thinking if the definition that a weak acid produces relatively few ions in aqueous solution,
If you have 100 % sulfuric acid, there is no water, so the definition does not apply.
Commercial sulfuric acid contains about 2 % water. The sulfuric acid in that small amount of water consists of about 99 % HSO₄⁻ and 1 % SO₄²⁻. The acid is almost completely ionized.
Sulfuric acid is a strong acid.