16 g of oxygen has the same number of molecules as in:
(a) 16 g of CO
(b) 28 g of N2
(c) 14 g of N2
(d) 1.0 g of H2
1. (a), (b)
2. (b), (c)
3. (c), (d)
4. (b), (d)
Unitless terms among the following are:
(a) | Molality | (b) | Molarity |
(c) | Mole fraction | (d) | Mass percent |
1. | (a), (b) | 2. | (b), (c) |
3. | (c), (d) | 4. | (b), (d) |
List I | List II | ||
a. | Micro | i. | m |
b. | Mega | ii. | m |
c. | Giga | iii. | m |
d. | Femto | iv. | m |
a | b | c | d | |
1. | i | iv | iii | ii |
2. | iii | iv | ii | i |
3. | ii | iii | iv | i |
4. | i | iii | iv | ii |
1. | 16 g of gas |
2. | 22.9 L of gas |
3. | dioxygen molecules |
4. | 11.2 L of gas |
The incorrect statements regarding 1 mole of dioxygen gas at STP from the above four statements are:
1. | (a, b, c) | 2. | (a, b, d) |
3. | (b, c, d) | 4. | (a, c, d) |
The numbers 234,000 and 6.0012 can be represented in scientific notation as:
1.
2. 0.234 and
3.
4. 2.34 and 6.0012
0.50 mol Na2CO3 and 0.50 M Na2CO3 are different because:
1. | Both have different amounts of Na2CO3. |
2. | 0.50 mol is the number of moles and 0.50 M is the molarity. |
3. | 0.50 mol Na2CO3 will generate more ions. |
4. | None of the above. |
Round up the following number into three significant figures:
i. 10.4107 ii. 0.04597 respectively are
1. | 10.4, 0.0460 | 2. | 10.41, 0.046 |
3. | 10.0, 0.04 | 4. | 10.4, 0.0467 |
The following data was obtained when dinitrogen and dioxygen react together to form different compounds:
Mass of dinitrogen | Mass of dioxygen | |
i. | 14 g | 16 g |
ii. | 14 g | 32 g |
iii. | 28 g | 32 g |
iv. | 28 g | 80 g |
The law of chemical combination applicable to the above experimental data is:
1. Law of reciprocal proportions
2. Law of multiple proportions
3. Law of constant composition
4. None of the above.
An organic compound contains 80 % (by wt.) carbon and the remaining percentage of hydrogen. The empirical formula of this compound is:
[Atomic wt. of C is 12, H is 1]
1. CH3
2. CH4
3. CH
4. CH2