Answer:
V ∝ abc
Explanation:
This task is a joint variation task involving only direct proportionality:
Direct variation is one in which two variables are in direct proportionality to each other. This means that as one increases, the other variable also increases and vice - versa.
Joint variation is one in which one variable is dependent on two or more variables and varies directly as each of them.
In this exercise:
If a ∝ b and a ∝ c, then a ∝ bc
Taking the above three proportionalities,
V ∝ a ∝ b ∝ c
V ∝ a ∝ bc
V ∝ abc
Answer:
Atoms of metal elements give away electrons in their reactions to form positive ions. The ions formed have a full outer electron shell, so are very stable.
Explanation:
To become stable, the metal atom wants to get rid of one or more electrons in its outer shell. ... Alternately, an atom that loses electrons becomes a positively charged ion (aka cations). The particles in an ionic compound are held together because there are oppositely charged particles that are attracted to one another.
Answer:
See explanation
Explanation:
At the same temperature, the average kinetic energy of gases depends on the molecular mass of the gas. Hence, given a particular temperature, lighter molecules travel faster on the average than heavier molecules.
Considering ethanol and octane, The molar mass of ethanol is 46.07 g/mol while the molar mass of octane is 114.23 g/mol. Hence, ethanol has a greater average kinetic energy than octane at the same temperature.
Answer:
The answer is "Option B".
Explanation:
![\to C \ CH_3COONa = \frac{(0.01\ mol + 5 \ E-4\ mol)}{(0.105\ L )}\\\\\to C \ CH_3COONa = 0.1 \ M\\\\\therefore Ka = ([H_3O^{+}]\times \frac{(0.1 + [H_3O^+]))}{(0.0905 - [H_3O^+])} = 1.75\ E-5\\\\\to 0.1[H_3O^+] + [H_3O^+]^2 = (1.75 E-5)\times (0.0905 - [H_3O^+])\\\\](https://tex.z-dn.net/?f=%5Cto%20C%20%5C%20CH_3COONa%20%3D%20%20%5Cfrac%7B%280.01%5C%20%20mol%20%2B%205%20%5C%20E-4%5C%20%20mol%29%7D%7B%280.105%5C%20L%20%29%7D%5C%5C%5C%5C%5Cto%20C%20%5C%20CH_3COONa%20%3D%200.1%20%5C%20M%5C%5C%5C%5C%5Ctherefore%20Ka%20%3D%20%28%5BH_3O%5E%7B%2B%7D%5D%5Ctimes%20%5Cfrac%7B%280.1%20%2B%20%5BH_3O%5E%2B%5D%29%29%7D%7B%280.0905%20-%20%5BH_3O%5E%2B%5D%29%7D%20%3D%201.75%5C%20E-5%5C%5C%5C%5C%5Cto%200.1%5BH_3O%5E%2B%5D%20%2B%20%5BH_3O%5E%2B%5D%5E2%20%3D%20%281.75%20E-5%29%5Ctimes%20%280.0905%20-%20%5BH_3O%5E%2B%5D%29%5C%5C%5C%5C)
![\to [H_3O^+]^2 \ 0.1[H_3O^+] = 1.584\ E-6 - 1.75\ E-5[H_3O^+]\\\\\to [H_3O^+]^2 + 0.1000175[H_3O^+] - 1.584 \ E-6 = 0\\\\\to [H_3O^+] = 1.5835\ E-5 \ M\\\\\therefore pH = - \log [H_3O^+]\\\\\to pH = - \log (1.5835 \ E-5)\\\\ \to pH = 4.8004 > 4.7](https://tex.z-dn.net/?f=%5Cto%20%5BH_3O%5E%2B%5D%5E2%20%5C%200.1%5BH_3O%5E%2B%5D%20%3D%201.584%5C%20%20E-6%20-%201.75%5C%20%20E-5%5BH_3O%5E%2B%5D%5C%5C%5C%5C%5Cto%20%5BH_3O%5E%2B%5D%5E2%20%2B%200.1000175%5BH_3O%5E%2B%5D%20-%201.584%20%5C%20E-6%20%3D%200%5C%5C%5C%5C%5Cto%20%20%5BH_3O%5E%2B%5D%20%3D%201.5835%5C%20%20E-5%20%5C%20M%5C%5C%5C%5C%5Ctherefore%20pH%20%3D%20-%20%5Clog%20%5BH_3O%5E%2B%5D%5C%5C%5C%5C%5Cto%20%20pH%20%3D%20-%20%5Clog%20%281.5835%20%5C%20E-5%29%5C%5C%5C%5C%20%5Cto%20pH%20%3D%204.8004%20%3E%204.7)
Answer:
What reaction is balanced
