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A balanced chemical equation is in accordance with
A balanced chemical equation satisfies the law of conservation of mass, which states that mass can neither be created nor destroyed in a chemical reaction. The total mass of the reactants must equal the total mass of the products in a balanced chemical equation.
The atomic weights of two elements X and Y are 20 and 40 respectively. If ‘a’ gm of X contains ‘b’ atoms, how many atoms are present in ‘2a’ gm of Y ?
$ No of moles of X = { a \over 20} $ $ \therefore No.of atoms of X = {a \over 20 } \times N =b (given ) $ $ \therefore a = { 20b \over N} $ $ No.of moles of Y = { 2a \over 40 } $ $ \therefore No.of atoms of Y = { 2a \over 40 } \times N $ $ = { 2 \over 40 } \times {2o b \over N } \times N = b $
If the components of air are $ N_2 ,78 \%, O_2, 21 \% ; Ar, 0.9 \% and CO_2, 0.1 \% $ by volume, what will be the molecular weight of air ?
$ Mol. wt. of air = { 78 \times 28 + 21 \times 32 + 0.9 \times 40 + 0.1 \times 44 \over 78 + 21 + 0.9 + 0.1 } $
Calculate the molarity of a solution obtained by mixing $ 50mL of 0.5M H_2SO_4 and 75 mL of 0.25M H_2SO_4 $ .
$ No.of moles of 0.05L H_2SO_4 = 0.5 \times 0.05 = 0.025 $ $ No.of moles of 0.075L H_2SO_4 = 0.25 \times 0.075 = 0.01875 $ $ \therefore Total no. of moles = 0.025 + 0.01875 = 0.04375 $ Total vol = 0.05L + 0.075L = 0.125L $ \therefore Molarity = { 0.04375 \over 0.125 } = 0.35 M $
Which of the following has the highest normality ?
Normality is a measure of the concentration of a solution and is defined as the number of gram-equivalent weights of the solute present per liter of the solution. Among the given options, 1 M H3PO4 has the highest normality of 3 as it can provide 3 H+ ions per molecule upon dissociation.
In an experiment, 4 gm of $ M_2O_x $ oxide was reduced to 2.8 gm of the metal. If the atomic mass of the metal is 56 gm/mol, the number of oxygen atoms in the oxide is (AFMC 2010)
$ 1 Mol M_2O_x = (2 \times 56 + 16x) gm$ $ Now, (2 \times 56 + 16x) gm of oxide = 112 gm metal $ $ \therefore 4 gm of oxide = { 112 \times 4 \over 112 + 16x } gm metal $ $ But { 112 \times 4 \over 112 + 16 x } = 2.8 (given ) \therefore x = 3 $
Match the following Column - I (i) femto (ii) yotta (iii) giga (iv) atto Column - II (P) $ 10^9 $ (q) $ 10 ^{-15} $ (r) $ 10 ^ {-18} $ (s) $ 10 ^ {24} $
The total number of atoms of all elements present in mole of ammonium dichromate is
Molecular formula of ammonium dichromate is $ (NH_4 )_2 Cr_2O_7 $
0.32 gm of a metal on treatment with an acid gave 112 mL of hydrogen at STP. Calculate the equivalent weight of the metal
$$ Eq. wt of metal = { wt of metal \times 11200 \over vol.of H_2 in ml displaced at STP } $$
For a reaction A + 2B ï‚® C, the amount of C formed by starting the reaction with 5 moles of A and 8 moles of B is
In the balanced chemical equation A + 2B → C, the stoichiometric coefficients indicate that 1 mole of A reacts with 2 moles of B to produce 1 mole of C. Given that the reaction starts with 5 moles of A and 8 moles of B, the limiting reactant is A (5 moles). Therefore, the maximum amount of C that can be formed is 4 moles.
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