Home
Class 12
CHEMISTRY
XeF6 + x H2O to XeO3 + y HF, the y/x =...

`XeF_6 + x H_2O to XeO_3 + y HF`, the `y/x =`________

Text Solution

Verified by Experts

The correct Answer is:
2
`XeF_6 + 3H_2O to XeO_3 + 6HF , y/x = 3`
Promotional Banner

Topper's Solved these Questions

  • NOBLE GASES

    AAKASH SERIES|Exercise PRACTICE SHEET - 1 (Single more than one option questions)|16 Videos

  • NOBLE GASES

    AAKASH SERIES|Exercise PRACTICE SHEET - 1 (Linked Comprehension type questions )|6 Videos

  • NOBLE GASES

    AAKASH SERIES|Exercise LEVEL - II (LECTURE SHEET) (EXERCISE - III MATCH THE FOLLOWING QUESTIONS)|5 Videos

  • NITROGEN CONTAINING COMPOUNDS

    AAKASH SERIES|Exercise Conversions|18 Videos

  • NUCLEAR CHEMISTRY

    AAKASH SERIES|Exercise PRACTICE SHEET - 2 (Integer answer type Questions )|8 Videos

Similar Questions

Explore conceptually related problems

(5x + 6y) (3x - 2y) = ____

CaCO_(3)undersetrarr(Delta)CaO + X . CaO+H_2O to Y . X+Y to Z . X , Y , Z are:

Let find the radical and centre of the circles x^2 + y^2 - 2x + 6y = 0 " ___"(1) x^2 + y^2 - 4x - 2y + 6 = 0 " ____"(2) and x^2 + y^2 -12x + 2y + 3 = 0 "___"(3)

The hydrolysis of XeF_6 is not a redox reaction (R) XeF_6 on complete hydrolysis will give XeO_3

Among the following molecules, a) XeO_3, b) XeOF_4 " and " c) XeF_6 , Those having same number of lone pairs on Xe are

Number of following compounds in which central atom has "+6" oxidation state XeF_2, XeF_6, XeO_3, XeO_4, XeF_4,XeOF_4, XeO_2F_2

Passage: When all the coefficients in a balanced chemical equation are multiplied by a constant factor X the equilibrium constant (originally K) becomes K^J . Similarly, when balanced equations are added together, the equilibrium constant for the combined process is equal to the product of the equilibrium constants for each step. Equilibrium constant of the reversed reaction is numerically equal to the reciprocal of the equilibrium constant of the original equation. Unit of K_p = ("atm")^(Deltan) , Unit of K_c =("mol" L^(-1))^(Deltan) Consider the two reactions: XeF_(6(g))+H_(2)O_((g)) harr XeOF_(4(g))+2HF_((g)), K_(1)" "XeO_(4(g))+XeF_(6(g)) harr XeOF_(4(g))+XeO_(3)F_(2(g)), K_(2) Then the equilibrium constant for the following reaction will be XeO_(4(g))+2HF_((g)) harr XeO_(3)F_(2(g))+H_(2)O_((g))

Sucrose underset ( H^(-)) overset (H_2O) to Y and X. The total no of optical isomers possible for Y and X are