Find the ratio by mass of the combining elements in the following compounds. -Maths 9th

Find the ratio by mass of the combining elements in the following compounds. -Maths 9th
by

1 Answer

Answer :

(a) CaCOg → Ca: C :0 = 40:12 : 48= 10 : 3:12 (b) MgCI2 →  Mg : Cl = 24 : 2 x 35.5 = 24 : 71 (c) H2SO4 → H : S : O = 2 x 1 : 32 : 4 x 16 = 2 : 32 : 64= 1:16: 32 (d) C2H5OH → C : H : 0 = 2 x 12 : 6 x 1 : 16 = 24: 6:16= 12 : 3: 8 (e) NH3 → N : H = 14 : 3 x 1=14: 3 (f) Ca (OH)2 → Ca : 0 : H = 40 : 2 x 16 : 2 x 1= 40: 32 :2 = 20 :16:1
by

Related questions

Find the ratio by mass of the combining elements in the following compounds. -Maths 9th

Description : Find the ratio by mass of the combining elements in the following compounds. -Maths 9th

Last Answer : Ratio by mass of the following compounds given below : 1) CaCO : Molar mass of calcium = 40 g Molar mass of oxygen = 16 g Molar mass of carbon = 12 g Hence , ratio by mass will be = 40 : 12 : 48 = 10 : 3 : 12 2) ... g Ratio by mass = 14 : 3 6) Ca(OH) : Ratio by mass = 40 : 32 : 2 = 20 : 16 : 1

Simplify by combining similar terms: -Maths 9th

Description : Simplify by combining similar terms: -Maths 9th

Last Answer : 5√2 + 20√2 = (5+20)√2 = 25√2

Simplify by combining similar terms : -Maths 9th

Description : Simplify by combining similar terms : -Maths 9th

Last Answer : Reducing into simplest form √12 = √22 × 3 = 2√3 √75 = √3 × 52 = 5√3 Therefore 4√3 - 3√12 + 2√75 = 4√3 - 3 × 2√3 + 2 × 5√3 = 4√3 - 6√3 + 10√3 = (4 - 6 + 10) √3 = 8√3

Simplify by combining similar terms : -Maths 9th

Description : Simplify by combining similar terms : -Maths 9th

Last Answer : Reducing into simplest form √8 = √(22× 2) = 2 × √2 √32 = √(22 × 22 × 2) = 2 × 2√2 = 4√2 ∴ √8 + √32 - √2 = 2√2 + 4√2 - √2 = (2 + 4 -1) √2 = 5√2 .

Simplifying by combining similar terms : -Maths 9th

Description : Simplifying by combining similar terms : -Maths 9th

Last Answer : Reducing into simplest form : √45 = √32 × 5 = 3√5 √20 = √22 × 5 = 2√5 ∴ √45 -3√20 + 4√5 = 3√5 - 3 × 2√5 + 4√5 = 3√5 - 6√5 + 4√5 = √5

Simplify by combining similar terms : -Maths 9th

Description : Simplify by combining similar terms : -Maths 9th

Last Answer : √12 = √(22 × 3) = 2√3 √50 = √(2 × 52) = (5 √2) √48 = √(22 × 22 × 3) = (2 × 2√3)= 4√3 ∴ 4√12 - √50 - 7 √48 = 4 × 2√3 - 5√2 - 7 × 4√3 = 8√3 - 5√2 - 28√3 = (8 -28)√3 - 5√2 = -20√3 - 5√2

Simplifying by combining similar terms: -Maths 9th

Description : Simplifying by combining similar terms: -Maths 9th

Last Answer : Reducing into simplest form

Simplify by combining similar terms : -Maths 9th

Description : Simplify by combining similar terms : -Maths 9th

Last Answer : Reducing into simplest form

Simplify by combining similar terms: -Maths 9th

Description : Simplify by combining similar terms: -Maths 9th

Last Answer : 5√2 + 20√2 = (5+20)√2 = 25√2

Simplify by combining similar terms : -Maths 9th

Description : Simplify by combining similar terms : -Maths 9th

Last Answer : Reducing into simplest form √12 = √22 × 3 = 2√3 √75 = √3 × 52 = 5√3 Therefore 4√3 - 3√12 + 2√75 = 4√3 - 3 × 2√3 + 2 × 5√3 = 4√3 - 6√3 + 10√3 = (4 - 6 + 10) √3 = 8√3

Simplify by combining similar terms : -Maths 9th

Description : Simplify by combining similar terms : -Maths 9th

Last Answer : Reducing into simplest form √8 = √(22× 2) = 2 × √2 √32 = √(22 × 22 × 2) = 2 × 2√2 = 4√2 ∴ √8 + √32 - √2 = 2√2 + 4√2 - √2 = (2 + 4 -1) √2 = 5√2 .

Simplifying by combining similar terms : -Maths 9th

Description : Simplifying by combining similar terms : -Maths 9th

Last Answer : Reducing into simplest form : √45 = √32 × 5 = 3√5 √20 = √22 × 5 = 2√5 ∴ √45 -3√20 + 4√5 = 3√5 - 3 × 2√5 + 4√5 = 3√5 - 6√5 + 4√5 = √5

Simplify by combining similar terms : -Maths 9th

Description : Simplify by combining similar terms : -Maths 9th

Last Answer : √12 = √(22 × 3) = 2√3 √50 = √(2 × 52) = (5 √2) √48 = √(22 × 22 × 3) = (2 × 2√3)= 4√3 ∴ 4√12 - √50 - 7 √48 = 4 × 2√3 - 5√2 - 7 × 4√3 = 8√3 - 5√2 - 28√3 = (8 -28)√3 - 5√2 = -20√3 - 5√2

Simplifying by combining similar terms: -Maths 9th

Description : Simplifying by combining similar terms: -Maths 9th

Last Answer : Reducing into simplest form

Simplify by combining similar terms : -Maths 9th

Description : Simplify by combining similar terms : -Maths 9th

Last Answer : Reducing into simplest form

(a) Show the formation of Na2O by the transfer of electrons between the combining atoms. (b) Why are ionic compounds usually hard? (c) How is it that ionic compounds in the solid state do not conduct electricity but they do so when in molten state? -Chemistry

Description : (a) Show the formation of Na2O by the transfer of electrons between the combining atoms. (b) Why are ionic compounds usually hard? (c) How is it that ionic compounds in the solid state do not conduct electricity but they do so when in molten state? -Chemistry

Last Answer : (b) It is due to strong force of attraction between oppositely charged ions. (c) In solid state, ions are not free to move whereas in molten state ions are free to move, therefore, they conduct electricity in molten state.

Write the molecular formulae for the following compounds -Maths 9th

Description : Write the molecular formulae for the following compounds -Maths 9th

Last Answer : NEED ANSWER

Write the molecular formulae for the following compounds -Maths 9th

Description : Write the molecular formulae for the following compounds -Maths 9th

Last Answer : molecular formulae of the following compounds.

Where can I read solid theories that claim life came from a meteorite combining with the elements on Earth?

Description : Where can I read solid theories that claim life came from a meteorite combining with the elements on Earth?

Last Answer : Panspermia describes such theories. A related term is exogenesis.

What is the percentage by mass concentration of a solution made by combining 12.0 g of NaNO3 with 100.0 g of water?

Description : What is the percentage by mass concentration of a solution made by combining 12.0 g of NaNO3 with 100.0 g of water?

Last Answer : The percentage concentration is 12 %.

Which of the following equations is obtained on combining 1st and 2nd law of thermodynamics, for a system of constant mass?(A) dE = Tds - PdV (B) dQ = CvdT + PdV (C) dQ = CpdT + Vdp (D) Tds = dE - PdV

Description : Which of the following equations is obtained on combining 1st and 2nd law of thermodynamics, for a system of constant mass? (A) dE = Tds - PdV (B) dQ = CvdT + PdV (C) dQ = CpdT + Vdp (D) Tds = dE - PdV

Last Answer : (A) dE = Tds - PdV

In how many chapters did Euclid divide his famous treatise ' The elements' ? -Maths 9th

Description : In how many chapters did Euclid divide his famous treatise ' The elements' ? -Maths 9th

Last Answer : Solution :- 13 chapters.

Let X be the set of all graduates in India. Elements x and y in X are said to be related, if they are graduates of the same university. -Maths 9th

Description : Let X be the set of all graduates in India. Elements x and y in X are said to be related, if they are graduates of the same university. -Maths 9th

Last Answer : R = {(x, y)}: x and y are graduates of same university, x, y ∈ {All graduates of India}. R is reflexive as (x, x) ∈ R, since x and x are graduates from the same university. R ... graduates from the same university ⇒ (x, z) ∈ R. R being reflexive, symmetric and transitive is an equivalence relation.

If R is a relation on a finite set A having n elements, then the number of relations on A is -Maths 9th

Description : If R is a relation on a finite set A having n elements, then the number of relations on A is -Maths 9th

Last Answer : (d) \(2^{n^2}\)Set A has n elements ⇒ n(A) = n ⇒ A × A has n × n = n2 elements ∴ Number of relations on A = Number of subsets of A × A = \(2^{n^2}\)

A determinant of second order is made with the elements 0, 1. What is the probability that the determinant is positive? -Maths 9th

Description : A determinant of second order is made with the elements 0, 1. What is the probability that the determinant is positive? -Maths 9th

Last Answer : (c) \(rac{3}{16}\)Total number of determinants that can be formed using 0 and 1 = 16 (4 4)The positive determinants are \(\begin{bmatrix}1&0\[0.3em]1&1nd{bmatrix}\),\(\begin{bmatrix}1&0\[0.3em]0&1nd{ ... bmatrix}1&1\[0.3em]0&1nd{bmatrix}\), i.e, 3 in number.∴ Required probability = \(rac{3}{16}.\)

What elements will always form ionic compounds in a 12 ratio with zinc.?

Description : What elements will always form ionic compounds in a 12 ratio with zinc.?

Last Answer : Need answer

Force applied on a body of mass 5 kg is directly proportional to the acceleration produced in the body. Represent this solution as a linear equation in two variables. -Maths 9th

Description : Force applied on a body of mass 5 kg is directly proportional to the acceleration produced in the body. Represent this solution as a linear equation in two variables. -Maths 9th

Last Answer : Solution :- Let the force be x and acceleration due to force be y.

Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes. -Maths 9th

Description : Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes. -Maths 9th

Last Answer : Let each side of a cube = a cm Then surface area = 6a² cm² and surface area of 3 such cubes = 3 x 6a² = 18a² cm² By placing three cubes side by side we get a cuboid whose ... + 3a²] = 14 a² ∴ Ratio between their surface areas = 14a² : 18a² = 7 : 9

Find the ratio of the total surface area and lateral surface area of a cube. -Maths 9th

Description : Find the ratio of the total surface area and lateral surface area of a cube. -Maths 9th

Last Answer : Let a be the edge of the cube, then Total surface area = 6a2² and lateral surface area = 4a² Now ratio between total surface area and lateral surface area = 6a² : 4a² = 3 : 2

Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area. -Maths 9th

Description : Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area. -Maths 9th

Last Answer : Let the sides of the triangle be a = 12x cm, b = 17x cm, c = 25x cm Perimeter of the triangle = 540 cm Now, 12x + 17x + 25x = 540 ⇒ 54x = 54 ⇒ x = 10 ∴ a = (12 x10)cm = 120cm, b = (17 x 10) cm = 170 cm and c = (25 x 10)cm = 250 cm Now, semi-perimeter, s = 5402cm = 270 cm

The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas -Maths 9th

Description : The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas -Maths 9th

Last Answer : If diameter of earth is said d, then the diameter of moon will be d/4 (as per given statement) Radius of earth = d/2 Radius of moon = ½×d/4 = d/8 Surface area of moon = 4π(d/8)2 Surface area of earth = 4π(d/2)2 Ncert solutions class 9 chapter 13-6

The radius of a spherical balloon increases from 7cm to 14cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases. -Maths 9th

Description : The radius of a spherical balloon increases from 7cm to 14cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases. -Maths 9th

Last Answer : Let r1 and r2 be the radii of spherical balloon and spherical balloon when air is pumped into it respectively. So r1 = 7cm r2 = 14 cm Now, Required ratio = (initial surface area)/(Surface area after pumping air into ... = (7/14)2 = (1/2)2 = ¼ Therefore, the ratio between the surface areas is 1:4.

The sides of a triangle are in the ratio of 3 : 4 : 5 and its perimeter is 510 m. What is the measure of its greatest side? -Maths 9th

Description : The sides of a triangle are in the ratio of 3 : 4 : 5 and its perimeter is 510 m. What is the measure of its greatest side? -Maths 9th

Last Answer : Let the sides of triangle be 3x,4x,5x Perimeter =3x + 4x + 5x=144 cm 12x=144 ∴x=12 Then sides of triangle are 3x=3 12=36 cm, 4x=4 12=48 cm, 5x=5 12=60 cm. Now, Semi perimeter, s=2 Sum of sides of ... , Area of triangle =s (s−a)(s−b)(s−c) = 72(72−36)(72−48)(72−60) = 72 36 24 12 = 864 cm2

Dhondoop's field has adjesent angles in the ratio 4:5. Find all the angles of his field. -Maths 9th

Description : Dhondoop's field has adjesent angles in the ratio 4:5. Find all the angles of his field. -Maths 9th

Last Answer : This answer was deleted by our moderators...

The outer curved surface areas of the hemisphere and sphere are in ratio 2:9. find their ratio of their raddii -Maths 9th

Description : The outer curved surface areas of the hemisphere and sphere are in ratio 2:9. find their ratio of their raddii -Maths 9th

Last Answer : This answer was deleted by our moderators...

If the lengths of the sides of a triangle are in the ratio 6:11:15 and it's perimeter is 96cm , then the height corresponding to the longest side is -Maths 9th

Description : If the lengths of the sides of a triangle are in the ratio 6:11:15 and it's perimeter is 96cm , then the height corresponding to the longest side is -Maths 9th

Last Answer : LET EACH SIDE BE X 6X+11X+15X=96 32X=96 X=3 SIDES=6 3=18 11 3=33 15 3=45 AREA OF TRIANGLE BY HERONS FORMULA=S=96/2=48 WHOLE UNDERROOT 48 48-18 48-33 48-45 UNDERROOT=12 4 30 15 3 4 3 15ROOT2 180 ... bh/2 180root2=18 h/2 360root2=18h h=20 root2 But root 2=1.4(approx) h=20 1.4(approx) h=28cm(approx).

The whole surface area of a rectangular block is 1300 cm2. Find its volume, if their dimensions are in the ratio of 4 : 3 : 2. -Maths 9th

Description : The whole surface area of a rectangular block is 1300 cm2. Find its volume, if their dimensions are in the ratio of 4 : 3 : 2. -Maths 9th

Last Answer : Let the length, breadth and height of the rectangular box be 4x, 3x and 2x, respectively. ∵ Total surface area = 1300 cm2 2(4x × 3x + 3x × 2x + 4x × 2x) = 1300 52x2 =1300x2 = 25x = 5 ∴ Volume of rectangular box = 4x × 3x × 2x = 24(5)2 = 3000 cm3

If the ratio of curved surface area and total surface area of a cylinder is 1 : 3, then find the volume of cylinder when the height is 2 cm. -Maths 9th

Description : If the ratio of curved surface area and total surface area of a cylinder is 1 : 3, then find the volume of cylinder when the height is 2 cm. -Maths 9th

Last Answer : Let the radius and height of the cylinder be r and h, respectively . Given that, Curved surface area / Total surface area = 1/3 ⇒ 2πrh / 2πr(h + r) = 1/3 ⇒ 3h = h + r ⇒ r = 2h = 4cm ∴ volume of cylinder πr2h = π × (4)2 × 2 = 32π cm3

If a sphere is inscribed in a cube, then find the ratio of the volume of the cube to the volume of the sphere. -Maths 9th

Description : If a sphere is inscribed in a cube, then find the ratio of the volume of the cube to the volume of the sphere. -Maths 9th

Last Answer : The ratio of the volume of the cube to the volume of the sphere are as

If a triangle and a parallelogram are on same base and between same parallels, then find the ratio of the area of the triangle to the area of parallelogram. -Maths 9th

Description : If a triangle and a parallelogram are on same base and between same parallels, then find the ratio of the area of the triangle to the area of parallelogram. -Maths 9th

Last Answer : If a triangle and a parallelogram are on the same base and between the same parallels, the area of the triangle is equal to half of the parallelogram.

The radii of two cylinders of the same height are in the ratio 4 :5, then find the ratio of their volumes. -Maths 9th

Description : The radii of two cylinders of the same height are in the ratio 4 :5, then find the ratio of their volumes. -Maths 9th

Last Answer : Let r1 and r2 be radii of two cyclinder and V1, V2 be their volume . Let h be height of the two cyclinders, then V1 = πr2h and V2 = πr22h ∴ V1 / V2 = πr12h / πr22h = r12 / r22 = 16 / 25 .

If the angles of a triangle are in the ratio 5:3:7, then the triangle is -Maths 9th

Description : If the angles of a triangle are in the ratio 5:3:7, then the triangle is -Maths 9th

Last Answer : (a) Given, the ratio of angles of a triangle is 5 : 3 : 7. Let angles of a triangle be ∠A,∠B and ∠C. Then, ∠A = 5x, ∠B = 3x and ∠C = 7x In ΔABC, ∠A + ∠B + ∠C = 180° [since, sum of ... 36° and ∠C =7x = 7 x 12° = 84° Since, all angles are less than 90°, hence the triangle is an acute angled triangle.

Angles of a triangle are in the ratio 2:4:3. The smallest angle of the triangle is -Maths 9th

Description : Angles of a triangle are in the ratio 2:4:3. The smallest angle of the triangle is -Maths 9th

Last Answer : (b) Given, the ratio of angles of a triangle is 2 : 4 : 3. Let the angles of a triangle be ∠A, ∠B and ∠C. ∠A = 2x, ∠B = 4x ∠C = 3x , ∠A+∠B+ ∠C= 180° [sum of all the angles of a triangle is 180°] 2x ... ∠B = 4x = 4 x 20° = 80° ∠C = 3x = 3 x 20° = 60° Hence, the smallest angle of a triangle is 40°.

If the angles of a triangle are in the ratio 5:3:7, then the triangle is -Maths 9th

Description : If the angles of a triangle are in the ratio 5:3:7, then the triangle is -Maths 9th

Last Answer : (a) Given, the ratio of angles of a triangle is 5 : 3 : 7. Let angles of a triangle be ∠A,∠B and ∠C. Then, ∠A = 5x, ∠B = 3x and ∠C = 7x In ΔABC, ∠A + ∠B + ∠C = 180° [since, sum of ... 36° and ∠C =7x = 7 x 12° = 84° Since, all angles are less than 90°, hence the triangle is an acute angled triangle

If angles A, B,C and D of the quadrilateral ABCD, taken in order are in the ratio 3 :7:6:4, then ABCD is a -Maths 9th

Description : If angles A, B,C and D of the quadrilateral ABCD, taken in order are in the ratio 3 :7:6:4, then ABCD is a -Maths 9th

Last Answer : (c) Given, ratio of angles of quadrilateral ABCD is 3 : 7 : 6 : 4. Let angles of quadrilateral ABCD be 3x, 7x, 6x and 4x, respectively. We know that, sum of all angles of a quadrilateral is 360°. 3x + 7x + 6x + 4x = 360° => 20x = 360° => x=360°/20° = 18°

If a triangle and a parallelogram are on the same base and between same parallels, then the ratio of the area of the triangle to the area of parallelogram is -Maths 9th

Description : If a triangle and a parallelogram are on the same base and between same parallels, then the ratio of the area of the triangle to the area of parallelogram is -Maths 9th

Last Answer : (b) We know that, if a parallelogram and a triangle are on the same base and between the same parallels, then area of the triangle is half the area of the parallelogram.

If arcs AXB and CYD of a circle are congruent, find the ratio of AB and CD. -Maths 9th

Description : If arcs AXB and CYD of a circle are congruent, find the ratio of AB and CD. -Maths 9th

Last Answer : Let AXB and CYD are arcs of circle whose centre and radius are O and r units, respectively. Hence, the ratio of AB and CD is 1:1.

Dhondoop's field has adjesent angles in the ratio 4:5. Find all the angles of his field. -Maths 9th

Description : Dhondoop's field has adjesent angles in the ratio 4:5. Find all the angles of his field. -Maths 9th

Last Answer : This answer was deleted by our moderators...

The outer curved surface areas of the hemisphere and sphere are in ratio 2:9. find their ratio of their raddii -Maths 9th

Description : The outer curved surface areas of the hemisphere and sphere are in ratio 2:9. find their ratio of their raddii -Maths 9th

Last Answer : This answer was deleted by our moderators...

If the lengths of the sides of a triangle are in the ratio 6:11:15 and it's perimeter is 96cm , then the height corresponding to the longest side is -Maths 9th

Description : If the lengths of the sides of a triangle are in the ratio 6:11:15 and it's perimeter is 96cm , then the height corresponding to the longest side is -Maths 9th

Last Answer : LET EACH SIDE BE X 6X+11X+15X=96 32X=96 X=3 SIDES=6 3=18 11 3=33 15 3=45 AREA OF TRIANGLE BY HERONS FORMULA=S=96/2=48 WHOLE UNDERROOT 48 48-18 48-33 48-45 UNDERROOT=12 4 30 15 3 4 3 15ROOT2 180 ... bh/2 180root2=18 h/2 360root2=18h h=20 root2 But root 2=1.4(approx) h=20 1.4(approx) h=28cm(approx).