It is known that an atom contains protons, neutrons, and electrons. If the mass of the neutron is assumed to be half of its original value, and that of the proton is assumed to be twice its original value, then the atomic mass of will:
1. | Remain the same | 2. | be 14.28% less |
3. | be 14.28% more | 4. | be 28.56% less |
Nitric acid can be produced in three steps process
(I)
(II)
(III)
percent yield Ist, IInd and IIIrd are respectively 50%, 60% and 80% respectively then what volume of at 1 atm and 0C required to produces 1575 g of
1. 156.25
2. 350 L
3. 3500 L
4. None of these
The density of dry air containing only and is 1.15 g/L at 740 mm and 300 K. What is the percentage
composition of by weight in the air?
1. | 78.00% | 2. | 75.50% |
3. | 70.02% | 4. | 72.75% |
In preparation of iron from haematite by the reaction with carbon:
How much 80% pure iron could be produced from 120 kg of 90% pure ?
1. 94.5 kg
2. 60.48 kg
3. 116.66 kg
4. 120 kg
A gaseous mixture of and gas contains 66 mass % of . The vapour density of the mixture is:
1. 6.1
2. 5.4
3. 2.7
4. 10.8
How much volume of 75% alcohol by weight (d=0.80 g/cm³) is required to prepare 150 cm³ of 30% alcohol by weight (density = 0.90 g/cm³)?
1. | 67.5 mL | 2. | 56.25 mL |
3. | 44.44 mL | 4. | None of the above |
The average atomic mass of magnesium is 24.31 a.m.u. Magnesium consists of three isotopes: \(^{24}Mg\), \(^{25}Mg \), and \(^{26}Mg\). The mole percentage of \(^{24}Mg\) is 79%, and the remaining 21% is a mixture of \(^{25}Mg \) and \(^{26}Mg\).
Calculate the mole percentage of \(^{26}Mg\).
1. 10 %
2. 11 %
3. 15 %
4. 16 %
A mixture of contain 30.40% mass per cent of nitrogen. What is the mass ratio of the two components in the mixture?
1. 2 : 1
2. 1 : 2
3. 3 : 4
4. 4 : 1
A mixture of and gas "Y" (mol. wt. 80) in the mole ratio a : b has a mean molecular weight of 40. What would be the mean molecular weight, if the gases are mixed in the ratio b : a under identical conditions? (assume that these gases do not react)
1. | 40 | 2. | 48 |
3. | 62 | 4. | 72 |
If 2.74 g of the metal oxide contains 1.53 g of metal, then the empirical formula of vanadium oxide is:
(Atomic Mass of V = 52)
1.
2.
3.
4.