Hydrogen practice · Easy

Option A ($\mathrm{N}{\mathrm{a}_2}[ {\mathrm{N}{\mathrm{a}_4}{{\left( {\mathrm{P}{\mathrm{O}_3}} \right)}_6}} ]$) is correct because it represents the chemical formula of Calgon, which is sodium hexametaphosphate.

Option C (Beta (${{\beta }}^ -$)) is correct because tritium undergoes beta decay, emitting a beta particle (electron).

Option D ($\mathrm{PbS}+4 \mathrm{H}_2 \mathrm{O}_2 \longrightarrow \mathrm{PbSO}_4+4 \mathrm{H}_2 \mathrm{O}$) is correct because hydrogen peroxide oxidizes PbS to PbSO4, indicating its oxidizing nature.

Option B (Ni) is correct because nickel (Ni) is a commonly used catalyst in the hydrogenation of oils, facilitating the addition of hydrogen to unsaturated fatty acids.

Option A (3) is correct because tritium, an isotope of hydrogen, has one proton and two neutrons, summing to 3.

Option D ($\mathrm{N}{\mathrm{a}_2}\mathrm{S}{\mathrm{O}_4}$) is correct because it is not used as a water softener, unlike Calgon, Permutit, and $\mathrm{N}{\mathrm{a}_2}\mathrm{C}{\mathrm{O}_3}$.

Option B (Clark's method) is correct because Clark's method, which involves adding lime, specifically removes temporary hardness from water.

Option A (1) is correct because the molarity of a commercial 11.2 volume ${H_2}{O_2}$ solution is calculated using the formula: Volume strength = 11.2 × molarity.

Option C (${111.5^0}$) is correct because the dihedral angle in gaseous ${H_2}{O_2}$ is experimentally determined to be approximately 111.5 degrees.

Option A (3) is correct because tritium, an isotope of hydrogen, has one proton and two neutrons, summing to 3.

Option C (spin of protons) is correct because para and ortho hydrogen differ in the relative spins of their protons.

Option D (All of these) is correct because H2(l), O2(l), and N2H4(l) are all used as rocket propellants, either as fuel or oxidizer.

Option A (1) is correct because an 11.2 volume solution of H₂O₂ has a molarity of approximately 1M.

Option D (Silicol process) is correct because the Silicol process is the method by which hydrogen is obtained by the action of an alloy of silicon and iron with $\mathrm{NaOH}\mathrm{.}$

Option A (Less stable than ortho hydrogen) is correct because para hydrogen is less stable at room temperature and tends to convert to the more stable ortho form.

Option C (Blue flame) is correct because hydrogen gas burns with a pale blue, almost invisible flame in the presence of sufficient oxygen.

Option D is correct because the electronegativity difference between hydrogen and chlorine is less than 2.1, leading to a covalent bond in hydrogen chloride, whereas sodium chloride has a larger electronegativity difference, resulting in an ionic bond.

Option B ($^2{H_2}{O^{16}}$) is correct because heavy water is deuterium oxide, where deuterium ($^2H$) is an isotope of hydrogen.

Option C (100) is correct because perhydrol is a 30% solution of $H_2O_2$, which corresponds to a volume strength of 100.

Option C (Occluded H) is correct because hydrogen adsorbed on palladium is specifically referred to as occluded hydrogen.

Option B (5.6 V) is correct because the volume strength of 1N ${H_2}{O_2}$ is calculated by multiplying the normality by 5.6.

Option C (Metallic hydride) is correct because transition metals like titanium form metallic hydrides, which are non-stoichiometric compounds.

Option C ($\mathrm{Zn}{\mathrm{H}_2}$) is correct because interstitial hydrides are formed by transition metals, and $\mathrm{Zn}{\mathrm{H}_2}$ is an example of such a hydride.

Option D ($\mathrm{CH}_4$) is correct because it is an electron-precise hydride, meaning it has the exact number of electrons required for normal covalent bonds.

Option C ($\mathrm{N}{\mathrm{a}_2}\mathrm{A}{\mathrm{l}_2}\mathrm{S}{\mathrm{i}_2}{\mathrm{O}_8}$) is correct because this is the general formula for sodium zeolite, a hydrated aluminosilicate.

Option C (nascent $\mathrm{H}$) is correct because nascent hydrogen is a strong reducing agent and can reduce $\mathrm{FeC}{\mathrm{l}_3}$ to $\mathrm{FeC}{\mathrm{l}_2}$.

The correct answer is 4 because the heaviest isotope of hydrogen is tritium ($_1{\mathrm{H}^3}$), which has 1 proton, 2 neutrons, and 1 electron, summing to 4.

Option B (non - planar) is correct because the structure of H2O2 is a non-planar open book structure due to repulsion between lone pairs.

Option A ($\mathrm{N}{\mathrm{a}_2}{\mathrm{O}_2}$) is correct because hydrogen peroxide (${\mathrm{H}_2}{\mathrm{O}_2}$) is formed by the reaction of sodium peroxide ($\mathrm{N}{\mathrm{a}_2}{\mathrm{O}_2}$) with dilute sulfuric acid.

Option B (Almost pure hydrogen peroxide) is correct because atomic hydrogen reacts with oxygen to produce almost pure hydrogen peroxide ($\mathrm{H}_2\mathrm{O}_2$).