Chemistry-
General
Easy

Question

The eq. wt. of Na subscript 2 straight S subscript 2 straight O subscript 3 in the reaction, Na subscript 2 straight S subscript 2 straight O subscript 3 plus 5 straight H subscript 2 straight O plus 4 Cl subscript 2 not stretchy rightwards arrow 2 NaHSO subscript 4 plus 8 HClis/are:

  1. straight M divided by 1
  2. straight M divided by 2
  3. straight M divided by 4
  4. 3 straight M divided by 8

The correct answer is: 3 straight M divided by 8

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To solve a trigonometric inequation of the type sin x ≥ a where |a| ≤ 1, we take a hill of length 2straight pi  in the sine curve and write the solution within that hill. For the general solution, we add 2nstraight pi. For instance, to solve sin space x greater or equal than negative 1 half, we take the hill open square brackets negative pi over 2 comma fraction numerator 3 pi over denominator 2 end fraction close square brackets over which solution is negative pi over 6 less than X less than fraction numerator 7 pi over denominator 6 end fraction The general solution is 2 straight n pi minus pi over 6 less than straight x less than 2 straight n pi plus fraction numerator 7 pi over denominator 6 end fraction, n is any integer. Again to solve an inequation of the type sin x ≤ a, where |a| ≤ 1, we take a hollow of length 2straight pi in the sine curve. (since on a hill, sinx ≤ a is satisfied over two intervals). Similarly cos x ≥ a or cosx ≤a, |a| ≤ 1 are solved.

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Cr subscript 2 straight O subscript 7 superscript 2 minus end superscript plus straight X not stretchy ⟶ with straight H to the power of ⊖ on top Cr to the power of 3 plus end exponent plus straight H subscript 2 straight O plus oxidised product of , in the above reaction cannot be

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Let a, b, c, d element of R. Then the cubic equation of the type a x cubed plus b x squared plus c x plus d equals 0 has either one root real or all three roots are real. But in case of trigonometric equations of the type sin cubed space x plus b sin squared space x plus c sin space x plus d equals 0 can possess several solutions depending upon the domain of x. To solve an equation of the type a cos space theta plus b sin space theta equals c. The equation can be written as cos space left parenthesis theta minus alpha right parenthesis equals c divided by square root of open parentheses a squared plus b squared close parentheses end root The solution is theta equals 2 straight n pi plus alpha plus-or-minus beta where tan space alpha = b divided by a comma cos space beta equals c divided by square root of open parentheses a squared plus b squared close parentheses end root

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