Chemistry-
General
Easy

Question

Statement 1:Iodimetric titration are redox titrations.
Statement 2:The iodine solution acts as an oxidant to reduce the reductant straight I subscript 2 plus 2 straight e not stretchy rightwards arrow 2 straight I to the power of minus

  1. Statement 1 is True, Statement 2 is True; Statement 2 is correct explanation for Statement 1
  2. Statement 1 is True, Statement 2 is True; Statement 2 is not correct explanation for Statement 1
  3. Statement 1 is True, Statement 2 is False
  4. Statement 1 is False, Statement 2 is True

The correct answer is: Statement 1 is True, Statement 2 is False

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Statement 1:Diisopropyl ketone on reaction with isopropyl magnesium bromide followed by hydrolysis gives alcohol
Statement 2:Grignard reagent acts as a reducing agent

Statement 1:Diisopropyl ketone on reaction with isopropyl magnesium bromide followed by hydrolysis gives alcohol
Statement 2:Grignard reagent acts as a reducing agent

chemistry-General
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The first noble gas compound obtained was:

The first noble gas compound obtained was:

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Maths-

Given that log subscript p space x equals alpha and log subscript g space x equals beta then value of log subscript p divided by q end subscript space x equals –

These four basic properties all follow directly from the fact that logs are exponents. 
logb(xy) = logbx + logby.
logb(x/y) = logbx - logby.
logb(xn) = n logbx.
logbx = logax / logab.

Given that log subscript p space x equals alpha and log subscript g space x equals beta then value of log subscript p divided by q end subscript space x equals –

Maths-General

These four basic properties all follow directly from the fact that logs are exponents. 
logb(xy) = logbx + logby.
logb(x/y) = logbx - logby.
logb(xn) = n logbx.
logbx = logax / logab.

parallel
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chemistry-

Statement 1:Change in colour of acidic solution of potassium dichromate by breath is used to test drunk drivers.

Statement 2:Change in colour is due to the complexation of alcohol with potassium dichromate.

Statement 1:Change in colour of acidic solution of potassium dichromate by breath is used to test drunk drivers.

Statement 2:Change in colour is due to the complexation of alcohol with potassium dichromate.

chemistry-General
General
Maths-

Total number of solutions of sin{x} = cos{x}, where {.} denotes the fractional part, in [0, 2straight pi] is equal to

In this question, we have to find the number of solutions. Here we have fractional part of x. It is {x} and {x}= x-[x]. The {x} is always belongs to 0 to 1.

Total number of solutions of sin{x} = cos{x}, where {.} denotes the fractional part, in [0, 2straight pi] is equal to

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In this question, we have to find the number of solutions. Here we have fractional part of x. It is {x} and {x}= x-[x]. The {x} is always belongs to 0 to 1.

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Statement 1:If a strong acid is added to a solution of potassium chromate it changes itscolour from yellow to orange.

Statement 2:The colour change is due to the oxidation of potassium chromate.

Statement 1:If a strong acid is added to a solution of potassium chromate it changes itscolour from yellow to orange.

Statement 2:The colour change is due to the oxidation of potassium chromate.

chemistry-General
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General
Maths-

Total number of solutions of the equation 3x + 2 tan x =fraction numerator 5 straight pi over denominator 2 end fraction in x element of [0, 2straight pi] is equal to

In this question, we use the graph of tanx . The intersection is the total number of solutions of this equation. The graph region is [ 0, 2π ].

Total number of solutions of the equation 3x + 2 tan x =fraction numerator 5 straight pi over denominator 2 end fraction in x element of [0, 2straight pi] is equal to

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In this question, we use the graph of tanx . The intersection is the total number of solutions of this equation. The graph region is [ 0, 2π ].

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chemistry-

The number of sigma bonds in P subscript 4 O subscript 10 is:

The number of sigma bonds in P subscript 4 O subscript 10 is:

chemistry-General
General
chemistry-

Statement 1:Oxidation number of Ni in is zero.

Statement 2:Nickel is bonded to neutral ligand carbonyl.

Statement 1:Oxidation number of Ni in is zero.

Statement 2:Nickel is bonded to neutral ligand carbonyl.

chemistry-General
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General
Maths-

Suppose equation is f(x) – g(x) = 0 of f(x) = g(x) = y say, then draw the graphs of y = f(x) and y = g(x). If graph of y = f(x) and y = g(x) cuts at one, two, three, ...., no points, then number of solutions are one, two, three, ...., zero respectively.

The number of solutions of sin x = fraction numerator vertical line x vertical line over denominator 10 end fraction is

In this question, we have drawn the graph. The number of intersections of both function fx and gx are the number solutions. Draw the graph carefully and find the intersection points.

Suppose equation is f(x) – g(x) = 0 of f(x) = g(x) = y say, then draw the graphs of y = f(x) and y = g(x). If graph of y = f(x) and y = g(x) cuts at one, two, three, ...., no points, then number of solutions are one, two, three, ...., zero respectively.

The number of solutions of sin x = fraction numerator vertical line x vertical line over denominator 10 end fraction is

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In this question, we have drawn the graph. The number of intersections of both function fx and gx are the number solutions. Draw the graph carefully and find the intersection points.

General
chemistry-

The first noble gas compound obtained was:

The first noble gas compound obtained was:

chemistry-General
General
chemistry-

Statement 1:The redox titrations in which liberated  straight I subscript 2 is used as oxidant are called as iodometric titrations
Statement 2:Addition of KI of CuSO subscript 4 liberates straight I subscript 2 which is estimated against hypo solution.

Statement 1:The redox titrations in which liberated  straight I subscript 2 is used as oxidant are called as iodometric titrations
Statement 2:Addition of KI of CuSO subscript 4 liberates straight I subscript 2 which is estimated against hypo solution.

chemistry-General
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General
chemistry-

N subscript 2 O subscript 4 molecule is completely changed into 2 N O subscript 2 molecules at:

N subscript 2 O subscript 4 molecule is completely changed into 2 N O subscript 2 molecules at:

chemistry-General
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Maths-

Statement-I : If sin squared space A equals sin squared space B and cos squared space A equals cos squared space B then straight A equals straight n pi plus straight B comma space straight n element of straight I
Statement-II : If sinA = sinB and cosA = cosB, then A equals n pi plus B comma n element of I

In this question, we have to find the statements are the correct or not and statement 2 is correct explanation or not, is same as assertion and reason. Here, Start solving first Statement and try to prove it. Then solve the Statement-II.

Statement-I : If sin squared space A equals sin squared space B and cos squared space A equals cos squared space B then straight A equals straight n pi plus straight B comma space straight n element of straight I
Statement-II : If sinA = sinB and cosA = cosB, then A equals n pi plus B comma n element of I

Maths-General

In this question, we have to find the statements are the correct or not and statement 2 is correct explanation or not, is same as assertion and reason. Here, Start solving first Statement and try to prove it. Then solve the Statement-II.

General
chemistry-

Arsine is:

Arsine is:

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