Illustrate how copper metal can give different products on reaction
Question: Illustrate how copper metal can give different products on reaction with HNO3. Solution: Concentrated nitric acidis a strong oxidizing agent. It is used for oxidizing most metals. The products of oxidation depend on the concentration of the acid, temperature, and also on the material undergoing oxidation. $3 \mathrm{Cu}+8 \mathrm{HNO}_{3 \text { (dilute) }} \longrightarrow 3 \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}+2 \mathrm{NO}+4 \mathrm{H}_{2} \mathrm{O}$ $\mathrm{Cu}+4 \mathrm{HN...
Read More →The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively.
Question: The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively. Find the ratio of their areas. Solution: Let the two triangles be ABC and DEF with altitudes AP and DQ, respectively. It is given that $\triangle A B C \sim \triangle D E F$. We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes. $\therefore \frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle D E F)}=\frac{(A P)^{2}...
Read More →How is ammonia manufactured industrially?
Question: How is ammonia manufactured industrially? Solution: Ammonia is prepared on a large-scale by the Habers process. $\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \leftrightharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) \quad \Delta_{\mathrm{f}} \mathrm{H}^{\circ}=-46.1 \mathrm{~kJ} / \mathrm{mol}$ The optimum conditions for manufacturing ammonia are: (i)Pressure (around 200 105Pa) (ii)Temperature (700 K) (iii)Catalyst such as iron oxide with small amounts of Al2O3andK2O...
Read More →A horse is placed for grazing inside a rectangular
Question: A horse is placed for grazing inside a rectangular field 40 m by 36 m and is tethered to one corner by a rope 14 m long. Over how much area can it graze? (Take = 22/7) Solution: It is given that a horse is tethered to one corner of a rectangular field (40 m 36 m) by a 14 m long rope. Let r m be the radius of a circle. Then area A of circle is $A=\pi \mathrm{r}^{2} \mathrm{~m}^{2}$ $=\frac{22}{7} \times 14 \times 14 \mathrm{~m}^{2}$ $=616 \mathrm{~m}^{2}$ Since the horse can graze insid...
Read More →How is nitrogen prepared in the laboratory?
Question: How is nitrogen prepared in the laboratory? Write the chemical equations of the reactions involved. Solution: An aqueous solution of ammonium chloride is treated with sodium nitrite. NO and HNO3are produced in small amounts. These are impurities that can be removed on passing nitrogen gas through aqueous sulphuric acid, containing potassium dichromate....
Read More →A steel wire when bent in the form of a square
Question: A steel wire when bent in the form of a square encloses an area of $121 \mathrm{~cm}^{2}$. If the same wire is bent in the form of a circle, find the area of the circle. Solution: Letacm be the side of square. Then area of square is $a^{2}=121 \mathrm{~cm}^{2}$ $a=\sqrt{121 \mathrm{~cm}^{2}}$ $a=11 \mathrm{~cm}$ We have, length of wire $=$ perimeter of square $=4 a \mathrm{~cm}$ $=4 \times 11 \mathrm{~cm}$ $=44 \mathrm{~cm}$ Let the radius of circle bercm. Then, circumference of circle...
Read More →Solve the following
Question: Why does NH3form hydrogen bond but PH3does not? Solution: Nitrogen is highly electronegative as compared to phosphorus. This causes a greater attraction of electrons towards nitrogen in NH3than towards phosphorus in PH3. Hence, the extent of hydrogen bonding in PH3is very less as compared to NH3....
Read More →Discuss the trends in chemical reactivity of group 15 elements.
Question: Discuss the trends in chemical reactivity of group 15 elements. Solution: General trends in chemical properties of group 15 (i)Reactivity towards hydrogen: The elements of group 15 react with hydrogen to form hydrides of type EH3, where E = N, P, As, Sb, or Bi. The stability of hydrides decreases on moving down from NH3to BiH3. (ii)Reactivity towards oxygen: Theelements of group 15 form two types of oxides: E2O3and E2O5, where E = N, P, As, Sb, or Bi. The oxide with the element in the ...
Read More →∆ABC ∼ ∆DEF and their areas are respectively 100 cm2 and 49 cm2.
Question: ∆ABC ∆DEFand their areas are respectively 100 cm2and 49 cm2. If the altitude of ∆ABCis 5 cm, find the corresponding altitude of ∆DEF. Solution: It is given that∆ABC ∆DEF.Therefore, theratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.Also, the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.Let the altitude of∆ABCbeAP,drawn from A to BC to meet BC at P and the altitude of ∆D...
Read More →Why does the reactivity of nitrogen differ from phosphorus?
Question: Why does the reactivity of nitrogen differ from phosphorus? Solution: Nitrogen is chemically less reactive. This is because of the high stability of its molecule, N2. In N2, the two nitrogen atoms form a triple bond. This triple bond has very high bond strength, which is very difficult to break. It is because of nitrogens small size that it is able to form ppbonds with itself. This property is not exhibited by atoms such as phosphorus. Thus, phosphorus is more reactive than nitrogen....
Read More →A horse is tied to a pole with 28 m long string.
Question: A horse is tied to a pole with 28 m long string. Find the area where the horse can graze. (Take = 22/7). Solution: We know that the horse is tied to a pole with 28 m long string. So the horse can graze the area of a circle of radius 28 m. Area of circle is $A=\pi r^{2}$ $=\frac{22}{7} \times 28 \times 28 \mathrm{~m}^{2}$ $=2464 \mathrm{~m}^{2}$ Hence the horse can graze $2464 \mathrm{~m}^{2}$ area....
Read More →Discuss the general characteristics of Group 15 elements with reference to their electronic configuration,
Question: Discuss the general characteristics of Group 15 elements with reference to their electronic configuration, oxidation state, atomic size, ionisation enthalpy and electronegativity. Solution: General trends in group15 elements (i)Electronic configuration:All the elements in group 15 have 5 valence electrons. Their general electronic configuration isns2np3. (ii)Oxidation states:All these elements have 5 valence electrons and require three more electrons to complete their octets. However, ...
Read More →The circumference of a circle exceeds
Question: The circumference of a circle exceeds the diameter by 16.8 cm. Find the circumference of the circle. Solution: Let the radius of a circle be $r \mathrm{~cm}$, then diameter of circle is $2 r \mathrm{~cm}$ and Circumference is $C=2 \pi r \mathrm{~cm}$. It is given that the circumference exceeds the diameter of circle by $16.8 \mathrm{~cm}$. So, circumference $=16.8+$ diameter $2 \pi r=16.8+2 r \mathrm{~cm}$ $2 \times \frac{22}{7} \times r=16.8+2 r \mathrm{~cm}$ $44 r=117.6+14 r \mathrm{...
Read More →Why has it been difficult to study the chemistry of radon?
Question: Why has it been difficult to study the chemistry of radon? Solution: It is difficult to study the chemistry of radon because it is a radioactive substance having a half-life of only 3.82 days. Also, compounds of radon such as RnF2have not been isolated. They have only been identified....
Read More →The areas of two similar triangles are 169 cm2 and 121 cm2 respectively.
Question: The areas of two similar triangles are 169 cm2and 121 cm2respectively. If the longest side of the larger triangle is 26 cm. find the longest side of the smaller triangle. Solution: It is given that the triangles are similar.Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.Let the longest side of smaller triangle bexcm. $\frac{\text { ar(Larger triangle) }}{\operatorname{ar}(\text { Smaller triangle })}=\frac{\text {...
Read More →Balance the following equation:
Question: Balance the following equation: XeF6+ H2OXeO2F2+ HF Solution: Balanced equation XeF6+ 2 H2OXeO2F2+ 4 HF...
Read More →Why is helium used in diving apparatus?
Question: Why is helium used in diving apparatus? Solution: Air containsa large amount of nitrogen and the solubility of gases in liquids increases with increase in pressure. When sea divers dive deep into the sea, large amount of nitrogen dissolves in their blood. When they come back to the surface, solubility of nitrogen decreases and it separates from the blood and forms small air bubbles. This leads to a dangerous medical condition called bends. Therefore, air in oxygen cylinders used for di...
Read More →Why is ICl more reactive than
Question: Why is ICl more reactive than I2? Solution: ICl is more reactive than I2because ICl bond in ICl is weaker than II bond in I2....
Read More →Name two poisonous gases which can be prepared from chlorine gas.
Question: Name two poisonous gases which can be prepared from chlorine gas. Solution: Two poisonous gases that can be prepared from chlorine gas are (i)Phosgene (COCl2) (ii)Mustard gas (ClCH2CH2SCH2CH2Cl)...
Read More →∆ABC ∼ ∆PQR and ar(∆ABC) = 4 ar(∆PQR).
Question: ∆ABC ∆PQRand ar(∆ABC) = 4 ar(∆PQR). IfBC= 12 cm, findQR. Solution: Given : $\operatorname{ar}(\triangle A B C)=4 \operatorname{ar}(\Delta P Q R)$ $\frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\Delta P Q R)}=\frac{4}{1}$ $\because \Delta A B C-\Delta P Q R$ $\therefore \frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\Delta P Q R)}=\frac{B C^{2}}{Q R^{2}}$ $\therefore \frac{B C^{2}}{Q R^{2}}=\frac{4}{1}$ $\Rightarrow Q R^{2}=\frac{12^{2}}{4}$ $\Rightarrow Q R^{...
Read More →Give the reason for bleaching action of
Question: Give the reason for bleaching action of Cl2. Solution: When chlorine reacts with water, it produces nascent oxygen. This nascent oxygen then combines with the coloured substances present in the organic matter to oxide them into colourless substances. Coloured substances + [O]Oxidized colourless substance...
Read More →Find the area of a circle whose circumference is 44 cm.
Question: Find the area of a circle whose circumference is 44 cm. Solution: Letrbe the radius of the circle. Then Circumference of the circle $C=2 \times \pi \times r$ $44 \mathrm{~cm}=2 \times \frac{22}{7} \times r$ $44 \mathrm{~cm}=\frac{44}{7} \times r$ $r=7 \mathrm{~cm}$ We know that the area of a circle of radiusris $A=\pi r^{2}$ Substituting the value ofrin above formula $A=\pi \times 7 \times 7 \mathrm{~cm}^{2}$ $=\frac{22}{7} \times 49 \mathrm{~cm}^{2}$ $=154 \mathrm{~cm}^{2}$...
Read More →Sea is the greatest source of some halogens.
Question: Sea is the greatest source of some halogens. Comment. Solution: Sea water contains chlorides, bromides, and iodides of Na, K, Mg, and Ca. However, it primarily contains NaCl. The deposits of dried up sea beds contain sodium chloride and carnallite, KCl.MgCl2.6H2O. Marine life also contains iodine in their systems. For example, sea weeds contain upto 0.5% iodine as sodium iodide. Thus, sea is the greatest source of halogens....
Read More →The areas of two similar triangles ABC and PQR are in the ratio
Question: The areas of two similar trianglesABCandPQRare in the ratio 9 : 16. If BC = 4.5 cm, find the length ofQR. Solution: It is given that $\triangle A B C \sim \triangle P Q R$. Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides. $\frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle P Q R)}=\frac{B C^{2}}{Q R^{2}}$ $\Rightarrow \frac{9}{16}=\frac{4.5^{2}}{Q R^{2}}$ $\Rightarrow Q R^{2}=\frac{4.5 \times 4.5...
Read More →Find the circumference of a circle
Question: Find the circumference of a circle whose area is 301.84 cm2. Solution: Letrcm be the radius of the circle. Then Area of a circle is $A=\pi r^{2} \mathrm{~cm}^{2}$ $301.84 \mathrm{~cm}^{2}=\pi \times r^{2}$ $301.84 \mathrm{~cm}^{2}=\frac{22}{7} \times r^{2}$ $r^{2}=96.04 \mathrm{~cm}^{2}$ $r=9.8 \mathrm{~cm}$ We know that the Circumference of circle of radiusris $C=2 \pi r \mathrm{~cm}$ So substituting the value of r in above formula $C=2 \times \frac{22}{7} \times 9.8 \mathrm{~cm}$ $=6...
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