Answer:
Hydrogen (H) Sulfur(S) and Oxygen(O)
Explanation:
The concept used here is the Law of Conservation of Mass. Technically, it's more specifically included in the Law of Definite Proportions. According to Dalton's atomic theory, when substances react together, they form a compound that has the same fixed ratio of the individual elements. That is the main reason why we balance, because stoichiometric coefficients are essential to obey the Law.
For the reaction a + b ⇒ ab, this is a combination reaction. For every 1 mole of a and 1 mole of b, 1 mole of product ab is formed. This is the fixed ratio we have to follow: 1:1:1. Now, the next thing to note is the limiting and excess reactant. If initially, there are 2 moles of A and 3 moles of B, the limiting reactant is A and the excess is B. Since the ratio between reactants is 1:1, 3 moles of B requires 3 moles of A. But since only 2 moles are available, reactant A is limited. In this problem, we assume that B is provided in excess. So, we just focus on the amount of the limiting reactant a.
If there are 5,000 molecules of a, we can determine the molecules of ab using the fixed ratio, 1 part a is to 1 part ab. Then, that means that 5,000 molecules of a would yield also 5,000 molecules of ab.
Answer:
The answers are explained below
Explanation:
a)
Given: concentration of salt/base = 0.031
concentration of acid = 0.050
we have
PH = PK a + log[salt]/[acid] = 1.8 + log(0.031/0.050) = 1.59
b)
we have HSO₃⁻ + OH⁻ ------> SO₃²⁻ + H₂O
Moles i............0.05...................0.01.................0.031.....................0
Moles r...........-0.01.................-0.01................0.01........................0.01
moles f...........0.04....................0....................0.041.....................0.01
c)
we will use the first equation but substituting concentration of base as 0.031 + 10ml = 0.031 + 0.010 = 0.041
Hence, we have
PH = PK a + log[salt]/[acid] = 1.8 + log(0.041/0.050) = 1.71
d)
pOH = -log (0.01/0.510) = 1.71
pH = 14 - 1.71 = 12.29
e)
Because the buffer solution (NaHSO3-Na2SO3) can regulate pH changes. when a buffer is added to water, the first change that occurs is that the water pH becomes constant. Thus, acids or bases (alkali = bases) Additional may not have any effect on the water, as this always will stabilize immediately.
Answer:
4g/cm^3
Explanation:
The formula for density is mass divided by volume. 180g is your mass, 45cm^3 is your volume, divide the two to get your answer.
Answer:
The correct option is: (A) Ketones cannot be oxidized further
Explanation:
Oxidation refers to the gain of oxygen or the formation of carbon-oxygen bond (C-O bond).
Alcohols are organic compounds containing at least one hydroxyl group. These compounds can be further classified into <em>primary (R-CH₂-OH), secondary (R¹R²CH-OH) and tertiary alcohols (R¹R²R³C-OH).</em>
- The <u>partial oxidation of </u><u>primary alcohol</u><u> gives </u><u>aldehyde</u> (R-CHO). Further <u>oxidation of aldehyds gives </u><u>carboxylic acid </u>(R-COOH). <u>Primary alcohol can also be oxidized directly to carboxylic acid.</u> <em>However, the carboxylic acid can not be further oxidized.</em>
![R-CH_{2}-OH \overset{[O]}{\rightarrow} R-CHO \overset{[O]}{\rightarrow} R-COOH \\R-CH_{2}-OH \overset{[O]}{\rightarrow} R-COOH](https://tex.z-dn.net/?f=R-CH_%7B2%7D-OH%20%5Coverset%7B%5BO%5D%7D%7B%5Crightarrow%7D%20R-CHO%20%5Coverset%7B%5BO%5D%7D%7B%5Crightarrow%7D%20R-COOH%20%5C%5CR-CH_%7B2%7D-OH%20%5Coverset%7B%5BO%5D%7D%7B%5Crightarrow%7D%20R-COOH)
- <u>Oxidation of </u><u>secondary alcohols</u><u> gives </u><u>ketones</u> (R¹R²C=O), <em>which can not be oxidized further.</em>
![R^{1}R^{2}CH-OH \overset{[O]}{\rightarrow} R^{1}R^{2}C=O](https://tex.z-dn.net/?f=R%5E%7B1%7DR%5E%7B2%7DCH-OH%20%5Coverset%7B%5BO%5D%7D%7B%5Crightarrow%7D%20R%5E%7B1%7DR%5E%7B2%7DC%3DO)
- Whereas, tertiary alcohols are resistant to oxidation.
<u>Therefore, the correct statement regarding oxidation is </u><u>(A)</u><u> </u><em><u>Ketones cannot be oxidized further</u></em><u>.</u>
