Number of moles of Oxygen needed to produce 4 moles of water on reacting with 4 moles of Hydrogen gas is …………….
A) 1 mole
B) 2 moles
C) 3 moles
D) 4 moles
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At constant temperature and pressure, two litres of hydrogen gas react with one litre of oxygen gas to produce two litres of water vapour.
(D) law of gaseous volumes \(\underset{2L}{H_2(g)}+\underset{2L}{O_2(g)}\longrightarrow \underset{2L}{H_2O(g)}\) Gay-Lussac's states that "At same temperature and pressure, when gases combine or produced in a chemical reaction, they do so in a simple ratio by volume. Here, \(H_2\), \(O_2\) and \(H_2O(g)\) are...
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At constant temperature and pressure, two litres of hydrogen gas react with one litre of oxygen gas to produce two litres of water vapour.
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11.2 cm^3 of hydrogen gas at STP, contains …………….. moles.
(A) 0.0005 As we know, \(1\,cm^3=1\,ml\) \(\therefore 11.2\,cm^3=11.2\,ml\) The volume of 1 mole of any gas at STP contain = 22400 ml/mol \(\therefore\) Number moles of hydrogen gas = \(\frac{11.2\,ml}{22400\,ml/mol}\) = \(5\times10^{-4}\) moles or 0.0005 moles.
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11.2 cm^3 of hydrogen gas at STP, contains …………….. moles.
2 Answers
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The ratio of molecular weight of hydrogen and oxygen in water is 1 : 8. (The relation between hydrogen and oxygen in linear
Correct option is (A) y = 8x \(\because\) Quantities of hydrogen and oxygen are x and y and their quantity are in ratio 1 : 8. \(\therefore\) x : y = 1 : 8 \(\Rightarrow\) \(\frac xy=\frac18\) \(\Rightarrow\) y =...
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The ratio of molecular weight of hydrogen and oxygen in water is 1 : 8. (The relation between hydrogen and oxygen in linear
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In a senior secondary school,80 students play football or hockey.The number that play football is 5 more than twice the number that play hockey.If 15 students play both games and every student in the school plays at least one game,find (i)the number of students that play football (ii)the number of students that play football but not hockey (iii)the number of students that play hockey but not football.
I thought F or H = 80 is giving by n(F u H) = n(F) + n(H) - n(F n H)
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The oxidation number of oxygen is …………. in oxygen difluoride.
Correct option is: (C) +2 The oxidation number of oxygen is +2 in oxygen difluoride because fluorine is more electronegative than oxygen.
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The oxidation number of oxygen is …………. in oxygen difluoride.
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