Water or hydride of oxygen Questions in English

(N/A) Water is amphoteric in nature,meaning it can act as both an acid and a base.
$1$. As an acid (reacting with a base like $NH_3$):
$H_2O(l) + NH_3(aq) \rightleftharpoons OH^-(aq) + NH_4^+(aq)$
$2$. As a base (reacting with an acid like $H_2S$):
$H_2O(l) + H_2S(aq) \rightleftharpoons H_3O^+(aq) + HS^-(aq)$
These reactions demonstrate that water can donate a proton $(H^+)$ to act as an acid or accept a proton to act as a base.

(C) Temporary hardness of water is caused by the presence of magnesium and calcium bicarbonates, which can be removed by Clark's method (adding lime) or boiling.
Permanent hardness of water is caused by the presence of chlorides and sulfates of magnesium and calcium.
Methods used for the removal of permanent hardness include treatment with sodium carbonate $(Na_{2}CO_{3})$, Calgon's method, and the ion-exchange method.
Therefore, Clark's method is not suitable for the removal of permanent hardness.

(D) Hydrolysis is a reaction where water molecules break down a compound without changing the oxidation states of the elements involved.
In the reaction $2F_{2(g)} + 2H_{2}O_{(l)} \rightarrow 4HF_{(aq)} + O_{2(g)}$,fluorine $(F_2)$ acts as a strong oxidizing agent.
It oxidizes oxygen in water from $-2$ to $0$ (in $O_2$) and itself gets reduced from $0$ to $-1$ (in $HF$).
Since this is a redox reaction and not a simple hydrolysis,option $D$ is the correct answer.

(C) Heavy water $(D_2O)$ is used as a moderator in nuclear reactors to slow down fast neutrons.
It is obtained as a by-product in the fertilizer industry during the production of hydrogen.
It is used in the study of reaction mechanisms by using $D$ as a tracer.
The dielectric constant of ordinary water $(H_2O)$ is $78.39$,while that of heavy water $(D_2O)$ is $78.06$ at $298 \ K$.
Therefore,the statement that heavy water has a higher dielectric constant than water is $INCORRECT$.

(D) For temporary hardness,the reaction on heating is:
$Mg(HCO_3)_2 \xrightarrow{\Delta} Mg(OH)_2 \downarrow + 2CO_2 \uparrow$
Thus,Assertion $A$ is false because $Mg(HCO_3)_2$ converts to $Mg(OH)_2$,not $MgCO_3$.
Regarding the solubility products $(K_{sp})$:
$K_{sp}(MgCO_3) = 3.5 \times 10^{-8}$
$K_{sp}(Mg(OH)_2) = 1.8 \times 10^{-11}$
Since $3.5 \times 10^{-8} > 1.8 \times 10^{-11}$,the solubility product of $MgCO_3$ is greater than that of $Mg(OH)_2$.
Therefore,Reason $R$ is also false.
Both statements are false.

(A) Statement $A$ is correct: Heavy water $(D_2O)$ is used in exchange reactions to study reaction mechanisms.
Statement $B$ is correct: Heavy water is prepared by the exhaustive electrolysis of ordinary water.
Statement $C$ is correct: The boiling point of $D_2O$ $(374.4 \ K)$ is higher than that of $H_2O$ $(373 \ K)$ due to stronger hydrogen bonding.
Statement $D$ is incorrect: The viscosity of $D_2O$ $(1.107 \ \text{centipoise})$ is greater than that of $H_2O$ $(0.89 \ \text{centipoise})$.
Therefore,statements $A, B,$ and $C$ are correct.

(A) The correct answer is $(A)$.
Water reacts with carbon and hydrocarbons at high temperatures to produce syngas $(CO + H_2)$.
$C + H_2O \xrightarrow{1270 \ K} CO + H_2$
$CH_4 + H_2O \xrightarrow{1270 \ K, \ Ni} CO + 3H_2$
$C_3H_8 + 3H_2O \xrightarrow{\Delta, \ Ni} 3CO + 7H_2$
Reaction with $CO_2$:
$CO_2 + H_2O \rightleftharpoons H_2CO_3$ (Carbonic acid).
No $CO$ is formed in this reaction.

(A) Calgon is sodium hexametaphosphate,with the formula $(NaPO_{3})_{6}$ or $Na_{6}P_{6}O_{18}$.
$1$. The $2^{nd}$ most abundant element in the Earth's crust by weight is Silicon $(Si)$,which is not present in Calgon. Thus,statement $A$ is false.
$2$. Calgon is a polymeric compound and is water-soluble. Statement $B$ is true.
$3$. It is commonly known as Graham's salt. Statement $C$ is true.
$4$. It removes $Ca^{2+}$ and $Mg^{2+}$ ions by forming soluble complex ions,not by precipitation. Statement $D$ is true.
Therefore,the statement that is $NOT$ true is $A$.

(A) Pure demineralised (de-ionized) water,free from all soluble mineral salts,is obtained by passing water successively through a cation exchange resin (in the $H^{+}$ form) and an anion exchange resin (in the $OH^{-}$ form).

(A) Temporary hardness in water is due to the presence of magnesium and calcium hydrogen carbonates.
When water is boiled,magnesium hydrogen carbonate is converted into insoluble magnesium hydroxide:
$Mg(HCO_3)_2 \xrightarrow{\text{Boil}} Mg(OH)_2 \downarrow + 2CO_2 \uparrow$
Calcium hydrogen carbonate is converted into insoluble calcium carbonate:
$Ca(HCO_3)_2 \xrightarrow{\text{Boil}} CaCO_3 \downarrow + H_2O + CO_2 \uparrow$
Thus,the temporary hardness is removed by converting them into $CaCO_3$ and $Mg(OH)_2$.

(B) $Clark's$ method is used to remove temporary hardness of water by adding calculated amounts of lime,$Ca(OH)_2$.
The chemical reactions involved are:
$Ca(HCO_3)_2 + Ca(OH)_2 \rightarrow 2CaCO_3(s) + 2H_2O(l)$
$Mg(HCO_3)_2 + 2Ca(OH)_2 \rightarrow 2CaCO_3(s) + Mg(OH)_2(s) + 2H_2O(l)$
As seen from the reactions,the insoluble metal salts formed are $CaCO_3$ and $Mg(OH)_2$.

(C) Clark's method is used to remove temporary hardness of water by adding calculated amounts of lime,$Ca(OH)_{2}$.
The chemical reactions are:
$Ca(HCO_{3})_{2} + Ca(OH)_{2} \rightarrow 2CaCO_{3} \downarrow + 2H_{2}O$
$Mg(HCO_{3})_{2} + 2Ca(OH)_{2} \rightarrow 2CaCO_{3} \downarrow + Mg(OH)_{2} \downarrow + 2H_{2}O$
Thus,the products formed are $CaCO_{3}$ and $Mg(OH)_{2}$.

(D) Sodium hexametaphosphate $(Na_{6}P_{6}O_{18})$,commercially known as $Calgon$,is used to remove permanent hardness of water.
It works by sequestering $Ca^{2+}$ and $Mg^{2+}$ ions into a soluble complex,preventing them from forming precipitates.
The reaction is:
$Na_{6}P_{6}O_{18} \longrightarrow 2Na^{+} + Na_{4}P_{6}O_{18}^{2-}$
$Ca^{2+} + Na_{4}P_{6}O_{18}^{2-} \longrightarrow 2Na^{+} + CaNa_{2}P_{6}O_{18}^{2-}$

(B) .
Zeolite is used in the Permutit process for water softening.
It is represented as $Na_2Al_2Si_2O_8 \cdot xH_2O$.
When hard water containing $Ca^{2+}$ or $Mg^{2+}$ ions is passed through it,the $Na^+$ ions in the zeolite are exchanged with these hardness-causing ions ($Ca^{2+}$ or $Mg^{2+}$),thereby softening the water.

(B) The correct order of hydrogen bonding is $Ice > Liquid \ water > Impure \ water$.
In $Ice$,each water molecule is hydrogen-bonded to four other water molecules in a rigid tetrahedral structure,leading to maximum hydrogen bonding.
In $Liquid \ water$,the structure is less ordered and molecules are in constant motion,resulting in fewer hydrogen bonds compared to ice.
In $Impure \ water$,the presence of solute particles disrupts the hydrogen-bonded network of water molecules,further decreasing the extent of hydrogen bonding.

(B) The balanced chemical equations are:
$Na_2O + H_2O \rightarrow 2 NaOH$
Here,$X = NaOH$. The number of oxygen atoms in one molecule of $X$ is $1$.
$Cl_2O_7 + H_2O \rightarrow 2 HClO_4$
Here,$Y = HClO_4$. The number of oxygen atoms in one molecule of $Y$ is $4$.
The total number of oxygen atoms in $X$ and $Y$ is $1 + 4 = 5$.

(D) Statement-$I$ is incorrect because the synthetic ion-exchange resin method is more efficient than the Permutit (zeolite) process for water softening.
Statement-$II$ is incorrect because the synthetic resin method removes all cations and anions,resulting in demineralized water,whereas the Permutit process involves the exchange of $Ca^{2+}$ and $Mg^{2+}$ ions with $Na^+$ ions,which leads to the formation of soluble sodium salts in the treated water.
Therefore,both statements are incorrect.

(B) Temporary hardness of water is due to the presence of magnesium and calcium bicarbonates.
$1$. $Ca(OH)_2$ (Clarke's method) is used to remove temporary hardness by converting soluble bicarbonates into insoluble carbonates:
$Ca(HCO_3)_2 + Ca(OH)_2 \longrightarrow 2 CaCO_3 \downarrow + 2 H_2O$
$2$. $Na_2CO_3$ (Washing soda) is used to remove both temporary and permanent hardness by precipitating calcium and magnesium ions as carbonates:
$Ca(HCO_3)_2 + Na_2CO_3 \longrightarrow CaCO_3 \downarrow + 2 NaHCO_3$
Therefore,both $(B)$ and $(C)$ are correct reagents for softening temporary hardness.

(D) $BaCl_{2} \cdot 2H_{2}O$ contains interstitial water molecules.
In this salt,the water molecules are not coordinated to the metal ion but are trapped within the crystal lattice structure,specifically in the interstitial sites.
In contrast,in $CuSO_{4} \cdot 5H_{2}O$,four water molecules are coordinated to the $Cu^{2+}$ ion,and one is hydrogen-bonded.
Therefore,$BaCl_{2} \cdot 2H_{2}O$ is the correct example of a salt with interstitial water.

(B) In the crystalline structure of ice,each oxygen atom is surrounded by four hydrogen atoms.
Two of these hydrogen atoms are covalently bonded to the oxygen atom,while the other two are linked through hydrogen bonding.
This arrangement creates a tetrahedral geometry around each oxygen atom.

(A) The boiling point of heavy water $(D_2O)$ is $101.4^{\circ} C$ $(374.4 \ K)$.
This is higher than the boiling point of ordinary water $(H_2O)$,which is $100^{\circ} C$ $(373 \ K)$,due to stronger hydrogen bonding in $D_2O$.

(D) The molecular weight of water isotopes depends on the mass of the hydrogen isotopes $(H=1, D=2, T=3)$ and oxygen isotopes $(^{16}O=16, ^{17}O=17, ^{18}O=18)$.
Calculating the molar mass for each option:
$A$. $T_{2}{ }^{17}O = (3 \times 2) + 17 = 23 \ g/mol$
$B$. $T_{2}{ }^{16}O = (3 \times 2) + 16 = 22 \ g/mol$
$C$. $HD^{18}O = (1 + 2) + 18 = 21 \ g/mol$
$D$. $T_{2}{ }^{18}O = (3 \times 2) + 18 = 24 \ g/mol$
Comparing the values,$T_{2}{ }^{18}O$ has the highest molecular weight,making it the heaviest isotope of water.

(C) Permanent hardness is caused by the presence of soluble salts of magnesium and calcium in the form of chlorides and sulphates in water.
Clark's method involves the addition of calculated amounts of lime $(Ca(OH)_2)$ to water to remove temporary hardness caused by calcium bicarbonate $(Ca(HCO_3)_2)$ and magnesium bicarbonate $(Mg(HCO_3)_2)$.
It is ineffective against permanent hardness,which requires methods like washing soda treatment,Calgon's method,or ion exchange method.

(A) The hardness of water is primarily caused by the presence of dissolved calcium $(Ca^{2+})$ and magnesium $(Mg^{2+})$ ions in the form of their hydrogen carbonates,chlorides,and sulphates. Among the given options,$Ca^{2+}$ is responsible for causing hardness in water.

(D) Clark's process is a commercial method for the removal of temporary hardness from water.
In this process,a calculated amount of lime $(Ca(OH)_2)$ is added to the hard water.
The lime reacts with the soluble bicarbonates of calcium and magnesium to form insoluble carbonates,which are then removed by filtration.
The chemical reactions are as follows:
$CaO + H_2O \rightarrow Ca(OH)_2$
$Ca(HCO_3)_2 + Ca(OH)_2 \rightarrow 2 CaCO_3 \downarrow + 2 H_2O$
$Mg(HCO_3)_2 + 2 Ca(OH)_2 \rightarrow Mg(OH)_2 \downarrow + 2 CaCO_3 \downarrow + 2 H_2O$

(A) In the gas phase, the water molecule $(H_2O)$ has a bent geometry.
Experimental data shows that the $O-H$ bond length is $95.7 \ pm$ and the bond angle is $104.5^{\circ}$.

(A) The physical properties of $D_2O$ (heavy water) are generally higher than those of $H_2O$ due to stronger hydrogen bonding and higher molecular mass.
Specifically,the dielectric constant of $D_2O$ is $78.06$ at $298 \ K$,whereas for $H_2O$ it is $78.39$ at $298 \ K$.
Thus,the dielectric constant of $D_2O$ is less than that of $H_2O$.

(D) Lime $(Ca(OH)_2)$ is typically used to remove temporary hardness of water,a process known as Clark's method.
$Ca(HCO_3)_2 + Ca(OH)_2 \rightarrow 2CaCO_3 \downarrow + 2H_2O$
Permanent hardness is caused by the presence of chlorides and sulfates of magnesium and calcium,which cannot be removed by adding lime.

(D) Incorrect: In the structure of ice,each oxygen atom is surrounded by $4$ hydrogen atoms,but only $2$ are attached via hydrogen bonding,while $2$ are attached via covalent bonds.
b) Correct: Temporary hardness is indeed due to bicarbonates of $Ca^{2+}$ and $Mg^{2+}$,not $NaHCO_3$.
c) Correct: In the reaction $2MnO_4^- + 6H^+ + 5H_2O_2 \rightarrow 2Mn^{2+} + 8H_2O + 5O_2$,$H_2O_2$ acts as a reducing agent.
d) Correct: $100$ volume $H_2O_2$ corresponds to approximately $303.7 \ g \ L^{-1}$,so $3 \ g \ L^{-1}$ is not equal to $100$ volume $H_2O_2$.
The only incorrect statement is $(a)$.

(C) Calgon is a commercial name for sodium hexametaphosphate,which has the chemical formula $Na_6P_6O_{18}$.
It is used in water softening to sequester calcium and magnesium ions.

(D) Temporary hardness of water is caused by the presence of magnesium and calcium bicarbonates,$Mg(HCO_3)_2$ and $Ca(HCO_3)_2$.
Clark's method involves the addition of calculated amounts of lime,$Ca(OH)_2$,to the water.
This process precipitates the bicarbonates as insoluble carbonates: $Ca(HCO_3)_2 + Ca(OH)_2 \rightarrow 2CaCO_3(s) + 2H_2O(l)$ and $Mg(HCO_3)_2 + 2Ca(OH)_2 \rightarrow 2CaCO_3(s) + Mg(OH)_2(s) + 2H_2O(l)$.
Therefore,Clark's method is specifically used for the removal of temporary hardness.

(D) Hydrolysis is a reaction where water acts as a reactant to break bonds in a compound.
In the reaction $2F_2 + 2H_2O \rightarrow 4HF + O_2$,fluorine is being oxidized from $0$ to $-1$ and oxygen is being oxidized from $-2$ to $0$.
This is a redox reaction,not a simple hydrolysis reaction.
In the other options,water acts as a nucleophile to break bonds in the substrate without changing the oxidation states of the central atoms.

(A) In photosynthesis $(I)$: $6CO_2 + 6H_2O \xrightarrow{h\nu} C_6H_{12}O_6 + 6O_2$. Here,the oxidation state of oxygen changes from $-2$ in $H_2O$ to $0$ in $O_2$,so $H_2O$ is oxidised.
In the reaction of $H_2O$ with fluorine $(II)$: $2H_2O + 2F_2 \rightarrow 4HF + O_2$. Here,the oxidation state of oxygen changes from $-2$ in $H_2O$ to $0$ in $O_2$,so $H_2O$ is oxidised.
In the reaction of $H_2O$ with hydrogen $(III)$: No reaction occurs.
In the reaction of $H_2O$ with $P_4O_{10}$ $(IV)$: $P_4O_{10} + 6H_2O \rightarrow 4H_3PO_4$. This is a hydrolysis reaction,not a redox reaction.
Therefore,$H_2O$ is oxidised in reactions $I$ and $II$.

(A) Concentrated $H_2SO_4$ is a powerful dehydrating agent.
When it reacts with sugar $(C_{12}H_{22}O_{11})$,it removes water molecules,leaving behind a black residue of carbon.
This process is known as the charring of sugar.
$C_{12}H_{22}O_{11} \xrightarrow{\text{conc. } H_2SO_4} 12C + 11H_2O$.

(D) In photosynthesis,$H_2O$ is oxidized to $O_2$ in photosystem $II$ at the oxygen-evolving complex.
Statement $A$ is correct.
$MgH_2$ is a saline (ionic) hydride,not an interstitial hydride. Interstitial hydrides are formed by $d$-block and $f$-block elements. Statement $B$ is incorrect.
Sodium hexametaphosphate $(Na_6P_6O_{18})$,also known as Calgon,is used to remove permanent hardness of water by sequestering $Ca^{2+}$ and $Mg^{2+}$ ions. Statement $C$ is correct.
Therefore,statements $A$ and $C$ are correct.

(A) The correct matches are as follows:
$A$. $Mg(HCO_3)_2 \rightarrow Mg(OH)_2 \downarrow + 2 CO_2 \uparrow$ is the reaction occurring during the removal of temporary hardness by $III$. Boiling.
$B$. $M^{2+} + Na_4P_6O_{18}^{2-} \rightarrow [Na_2MP_6O_{18}]^{2-} + 2 Na^{+}$ is the reaction involved in $IV$. Calgon's method (using sodium hexametaphosphate).
$C$. $Ca(HCO_3)_2 + Ca(OH)_2 \rightarrow 2 CaCO_3 + 2 H_2O$ is the reaction in $I$. Clark's method (adding lime).
$D$. $2 NaZ + Ca^{2+}_{(aq)} \rightarrow 2 Na^{+} + CaZ_2$ is the reaction in $II$. Ion exchange method (using zeolite).
Therefore,the correct sequence is $A-III, B-IV, C-I, D-II$.

(A) The temperature of maximum density for $H_2O$ is approximately $277 \ K$ $(4 \ ^\circ C)$.
The temperature of maximum density for $D_2O$ is approximately $284 \ K$ $(11 \ ^\circ C)$.
Therefore,$x = 284 \ K$ and $y = 277 \ K$.
The value of $(x-y) = 284 - 277 = 7 \ K$.

(D) Calgon is the commercial name for sodium hexametaphosphate,which has the chemical formula $Na_6P_6O_{18}$.
It is commonly used in water treatment to soften water by sequestering calcium and magnesium ions.

(C) Boiling removes only temporary hardness of water.
During boiling,the soluble $Mg(HCO_3)_2$ is converted into insoluble magnesium hydroxide and $Ca(HCO_3)_2$ is changed to insoluble calcium carbonate.
In the permutit process,hydrated sodium aluminium silicate is used. For simplicity,sodium aluminium silicate $(NaAlSiO_4)$ can be written as $NaZ$. When this is added to hard water,exchange reactions take place:
$2 NaZ_{(s)} + M^{2+}_{(aq)} \rightarrow MZ_{2(s)} + 2 Na^{+}_{(aq)}$ (where $M = Mg, Ca$).
In the permutit process,only cations causing hardness are removed.
In the ion-exchange process,both the cations and anions causing permanent hardness of water can be removed by using cation and anion exchange resins. Demineralized water is obtained after this process.
Hence,the most effective method for the removal of hardness of water is the ion-exchange process.

(D) Deionized water is obtained by passing hard water through a series of ion exchange resins. First,it passes through a cation exchange resin (which removes $Ca^{2+}$,$Mg^{2+}$,etc.) and then through an anion exchange resin (which removes $Cl^-$,$SO_4^{2-}$,etc.). Thus,both cationic and anionic exchangers are used one after the other.

(C) Temporary hardness of water is removed by Clark's process,where slaked lime,$Ca(OH)_2$,is added to hard water.
The chemical reactions involved are:
$Ca(HCO_3)_{2(aq)} + Ca(OH)_{2(aq)} \rightarrow 2CaCO_{3(s)} + 2H_2O_{(l)}$
$Mg(HCO_3)_{2(aq)} + 2Ca(OH)_{2(aq)} \rightarrow 2CaCO_{3(s)} + Mg(OH)_{2(s)} + 2H_2O_{(l)}$
Note: In the case of magnesium bicarbonate,the product is magnesium hydroxide,$Mg(OH)_2$,due to its lower solubility product compared to $MgCO_3$.
Thus,the insoluble precipitates formed are $Mg(OH)_2$ and $CaCO_3$.

(B) The correct matches are as follows:
$(A)$ Treatment with washing soda uses $Na_2CO_3$ to precipitate calcium and magnesium ions as carbonates. Thus,$A-II$.
$(B)$ Calgon's method uses sodium hexametaphosphate,$Na_6P_6O_{18}$,which is known as Calgon. Thus,$B-IV$.
$(C)$ Clark's method involves the addition of calculated amounts of lime,$Ca(OH)_2$,to remove temporary hardness. Thus,$C-I$.
$(D)$ Zeolite/permutit process uses hydrated sodium aluminium silicate,$NaAlSiO_4$ (often represented as $Na_2Ze$),for ion exchange. Thus,$D-III$.
Therefore,the correct sequence is $A-II, B-IV, C-I, D-III$.

(B) Hard water is defined as water that does not produce lather with soap.
Hardness is primarily due to the presence of dissolved bicarbonates,chlorides,and sulphates of $Ca^{2+}$ and $Mg^{2+}$ ions.
The reaction with soap (sodium stearate) is:
$M^{2+} + 2C_{17}H_{35}COONa \longrightarrow (C_{17}H_{35}COO)_2M + 2Na^+$
(where $M = Ca^{2+}, Mg^{2+}$).
These ions react with soap to form insoluble precipitates of calcium and magnesium salts of fatty acids,preventing the formation of lather.

(B) At atmospheric pressure,ice crystallises in the hexagonal form.
However,at very low temperatures (below $200 \ K$),water condenses into the cubic form.

(A) Calgon is the trade name for sodium hexametaphosphate,$\left(NaPO_3\right)_6$.
It ionizes in water to provide a complex anion:
$Na_2\left(Na_4P_6O_{18}\right) \longrightarrow 2Na^{+} + \left[Na_4P_6O_{18}\right]^{2-}$
When added to hard water,the $Ca^{2+}$ ions displace the $Na^{+}$ ions from the complex anion to form a soluble complex:
$Ca^{2+} + \left[Na_4P_6O_{18}\right]^{2-} \longrightarrow \left[CaNa_2P_6O_{18}\right]^{2-} + 2Na^{+}$

(A) Clark's method is used to remove temporary hardness of water.
In this method,a calculated amount of lime $(Ca(OH)_2)$ is added to hard water,which precipitates out the dissolved bicarbonates as carbonates.
Permanent hardness is removed by methods such as Calgon's method,Ion-exchange method,and Synthetic resins method.

(D) In Clark's process,temporary hardness of water is removed by adding a calculated amount of lime $(Ca(OH)_2)$. It reacts with soluble bicarbonates of calcium and magnesium to form insoluble carbonates and hydroxides which can be filtered out.
$Ca(HCO_3)_2 + Ca(OH)_2 \rightarrow 2CaCO_3 \downarrow + 2H_2O$
$Mg(HCO_3)_2 + 2Ca(OH)_2 \rightarrow 2CaCO_3 \downarrow + Mg(OH)_2 \downarrow + 2H_2O$

(A) The solubility of an ionic compound like $NaCl$ depends on the dielectric constant of the solvent.
Higher dielectric constant leads to better solvation of ions,which increases solubility.
The dielectric constant of ordinary water $(H_2O)$ is approximately $81$,while that of heavy water $(D_2O)$ is approximately $80$.
Since the dielectric constant of ordinary water is higher,$NaCl$ is more soluble in ordinary water and less soluble in heavy water.
Therefore,both $(A)$ and $(R)$ are true and $(R)$ is the correct explanation of $(A)$.