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Chemistry-

General

Easy

Question

In the following reaction , $A g C l plus K I rightwards arrow K C l plus A g I$ . As KI is added , the equilibrium is shifted towards right giving more AgI precipitate , because

Both AgCl and AgI are sparingly soluble.
The K_sp of AgI is lower than K_sp of AgCl
The K_sp of AgI is higher than K_sp of AgCl
Both AgCl and AgI have same solubility product.

The correct answer is: The K_sp of AgI is lower than K_sp of AgCl

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Which of the following is most soluble ?

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Which of the following has the lowest value of K_sp at 25°C ?

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The correct representation for solubility product of SnS₂ is

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Which pair will show common ion effect ?

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Out of Ca²⁺ , Al³⁺ , Bi³⁺ , Mg²⁺ and Zn²⁺ the reagents NH₄Cl and aqueous NH₃ will precipitate

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The addition of NaCl to AgCl decreases the solubility of AgCl , because

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How do we differentiate between Fe³⁺ and Cr³⁺ in group III ?

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In qualitative analysis , to identify the IIA group sulphides , HCl is added to salt solution before the addition of H₂S . Because

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If $c o s invisible function application alpha plus 2 c o s invisible function application beta plus 3 c o s invisible function application gamma equals s i n invisible function application alpha plus 2 s i n invisible function application beta plus 3 s i n invisible function application gamma equals 0$ then $s i n invisible function application 3 alpha plus 8 s i n invisible function application 3 beta plus 27 s i n invisible function application 3 gamma equals$

If $c o s invisible function application alpha plus 2 c o s invisible function application beta plus 3 c o s invisible function application gamma equals s i n invisible function application alpha plus 2 s i n invisible function application beta plus 3 s i n invisible function application gamma equals 0$ then $s i n invisible function application 3 alpha plus 8 s i n invisible function application 3 beta plus 27 s i n invisible function application 3 gamma equals$

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$1 comma alpha subscript 1 end subscript comma alpha subscript 2 end subscript comma alpha subscript 3 horizontal ellipsis horizontal ellipsis end subscript alpha subscript n minus 1 end subscript$ are the n^th roots of unity and n is an even natural number, then $open parentheses 1 plus alpha subscript 1 end subscript close parentheses open parentheses 1 plus alpha subscript 2 end subscript close parentheses open parentheses 1 plus alpha subscript 3 end subscript close parentheses horizontal ellipsis horizontal ellipsis open parentheses 1 plus alpha subscript n minus 1 end subscript close parentheses equals$

$1 comma alpha subscript 1 end subscript comma alpha subscript 2 end subscript comma alpha subscript 3 horizontal ellipsis horizontal ellipsis end subscript alpha subscript n minus 1 end subscript$ are the n^th roots of unity and n is an even natural number, then $open parentheses 1 plus alpha subscript 1 end subscript close parentheses open parentheses 1 plus alpha subscript 2 end subscript close parentheses open parentheses 1 plus alpha subscript 3 end subscript close parentheses horizontal ellipsis horizontal ellipsis open parentheses 1 plus alpha subscript n minus 1 end subscript close parentheses equals$

Statement-I : If $x subscript n end subscript equals c o s invisible function application open parentheses pi divided by 2 to the power of n end exponent close parentheses plus i s i n invisible function application open parentheses pi divided by 2 to the power of n end exponent close parentheses$ then $x subscript 1 end subscript times x subscript 2 end subscript times x subscript 3 end subscript horizontal ellipsis horizontal ellipsis equals negative 1$
Statement-II : $x subscript n end subscript equals c o s invisible function application open parentheses pi divided by 4 to the power of n end exponent close parentheses plus i s i n invisible function application open parentheses pi divided by 4 to the power of n end exponent close parentheses$ then $x subscript 1 end subscript times x subscript 2 end subscript times x subscript 3 end subscript horizontal ellipsis horizontal ellipsis infinity equals negative 1$

Statement-I : If $x subscript n end subscript equals c o s invisible function application open parentheses pi divided by 2 to the power of n end exponent close parentheses plus i s i n invisible function application open parentheses pi divided by 2 to the power of n end exponent close parentheses$ then $x subscript 1 end subscript times x subscript 2 end subscript times x subscript 3 end subscript horizontal ellipsis horizontal ellipsis equals negative 1$
Statement-II : $x subscript n end subscript equals c o s invisible function application open parentheses pi divided by 4 to the power of n end exponent close parentheses plus i s i n invisible function application open parentheses pi divided by 4 to the power of n end exponent close parentheses$ then $x subscript 1 end subscript times x subscript 2 end subscript times x subscript 3 end subscript horizontal ellipsis horizontal ellipsis infinity equals negative 1$

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Reason:- Glycolysis can takes place in presence and absence of O₂

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Reason:- Glycolysis can takes place in presence and absence of O₂

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Reason:- Glycolysis splits the pyruvate in to H₂O & CO₂

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Reason:- Glycolysis splits the pyruvate in to H₂O & CO₂

Light wave is travelling along $y$ -direction. If the corresponding E vector at any time is along the $x$ -axis, the direction of B vector at that time is along

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A flag staff stands upon the top of a building. At a distance d mt the angles of elevation of the tops of the flag staff and building are α and β respectively. The height of the flag staff is

A flag staff stands upon the top of a building. At a distance d mt the angles of elevation of the tops of the flag staff and building are α and β respectively. The height of the flag staff is

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