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Chemistry-

General

Easy

Question

An aqueous solution of a mixture of two inorganic salts, when treated with diluteHCl, gave aprecipitate (P) and a filtrate (Q). The precipitate P was found todissolve in hot water. The filtrate (Q) remained unchanged,whentreatedwithH₂Sinadilute mineral acidmedium. However, it gave a precipitate(R)with H₂S in an ammoniacal medium. The precipitate R gave acolored solution(S),whentreatedwithH₂O₂inan aqueous NaOH medium. The colored solution Scontains:

Fe₂(SO₄)₃
CuSO₄
ZnSO₄
Na₂CrO₄

The correct answer is: Na₂CrO₄

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Statement-1 : A transistor amplifier in common emitter configuration has a low input impedence.
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The current flowing through the zener diode in fig. is-

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Consider the following statements 1 and 2 and identify the correct answer- 1] A zener diode is always connected in reverse bias. 2] The potential barrier of a P-N junction lies between 0.1 to 0.3 V approximately.

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Let ABC be a triangle with equations of the sides AB, BC and CA respectively x – 2 = 0, y – 5 = 0 and 5x + 2y –10 = 0. Then the orthocentre of the triangle lies on the line

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Statement I : $blank to the power of 21 end exponent C subscript 0 end subscript plus blank to the power of 21 end exponent C subscript 1 end subscript plus horizontal ellipsis plus blank to the power of 21 end exponent C subscript 10 end subscript equals 2 to the power of 20 end exponent$ because
Statement II : $blank to the power of 2 n plus 1 end exponent C subscript 0 end subscript plus blank to the power of 2 n plus 1 end exponent C subscript 1 end subscript plus horizontal ellipsis to the power of 2 n plus 1 end exponent C subscript 2 n plus 1 end subscript equals 2 to the power of 2 n plus 1 end exponent$ and ⁿC_r = ⁿC_n_r

Statement I : $blank to the power of 21 end exponent C subscript 0 end subscript plus blank to the power of 21 end exponent C subscript 1 end subscript plus horizontal ellipsis plus blank to the power of 21 end exponent C subscript 10 end subscript equals 2 to the power of 20 end exponent$ because
Statement II : $blank to the power of 2 n plus 1 end exponent C subscript 0 end subscript plus blank to the power of 2 n plus 1 end exponent C subscript 1 end subscript plus horizontal ellipsis to the power of 2 n plus 1 end exponent C subscript 2 n plus 1 end subscript equals 2 to the power of 2 n plus 1 end exponent$ and ⁿC_r = ⁿC_n_r

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Statement 2: Root mean square of n distinct real numbers is always greater than their arithmetic mean, n ≥ 2

If C_r denotes the combinatorial coefficient in the expansion of (1 + x)ⁿ then (nN
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Statement 2: Root mean square of n distinct real numbers is always greater than their arithmetic mean, n ≥ 2

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$stretchy sum from i equals 0 to n of stretchy sum from j equals 0 to n of stretchy sum from k equals 0 to n of open parentheses fraction numerator n over denominator i end fraction close parentheses open parentheses fraction numerator n over denominator k end fraction close parentheses comma open parentheses fraction numerator n over denominator r end fraction close parentheses equals blank to the power of n end exponent C subscript r end subscript$

$stretchy sum from i equals 0 to n of stretchy sum from j equals 0 to n of stretchy sum from k equals 0 to n of open parentheses fraction numerator n over denominator i end fraction close parentheses open parentheses fraction numerator n over denominator k end fraction close parentheses comma open parentheses fraction numerator n over denominator r end fraction close parentheses equals blank to the power of n end exponent C subscript r end subscript$

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Which of the following pairs of ions cannot beseparated by H₂Sindilute HCl?

Which of the following pairs of ions cannot beseparated by H₂Sindilute HCl?

Chemistry-General

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