Xe + F_2 xrightarrow{573 K, 60.70 bar} AWha | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The reaction of Xenon with Fluorine under specific conditions of temperature and pressure yields different Xenon fluorides.When $Xe$ is reacted wi

Vedclass · Accepted Answer

$XeF_6$

Vedclass · Answer

(A) The reaction of Xenon with Fluorine under specific conditions of temperature and pressure yields different Xenon fluorides.When $Xe$ is reacted with $F_2$ in a $1:5$ molar ratio at $573 \ K$ and $60-70 \ bar$ pressure,the product formed is $XeF_6$.The balanced chemical equation is:$Xe(g) + 3F_2(g) \xrightarrow{573 \ K, 60-70 \ bar} XeF_6(s)$.

$Xe + F_2 \xrightarrow{573 \ K, 60.70 \ bar} A$
What is $A$?

Similar Questions

Argon was discovered by

Which of the following represents a noble gas configuration?

Consider the following transformations:
$I. XeF_6 + NaF \to Na^+[XeF_7]^-$
$II. 2PCl_{5(s)} \to [PCl_4]^+[PCl_6]^-$
$III. [Al(H_2O)_6]^{3+} + H_2O \to [Al(H_2O)_5OH]^{2+} + H_3O^+$
Which of the above transformations are correct?

What is the general electronic configuration of noble gases other than helium?

The shape of $XeF_6$ is,

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$Xe + F_2 \xrightarrow{573 \ K, 60.70 \ bar} A$What is $A$?