Xe + 2F_2 xrightarrow{573 K, 60.70 bar} AW | vedclass

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

Question

(D) The reaction of Xenon $(Xe)$ with Fluorine $(F_2)$ depends on the molar ratio and reaction conditions.When $Xe$ reacts with $F_2$ in a $1:2$ molar

Vedclass · Accepted Answer

$XeF_4$

Vedclass · Answer

(D) The reaction of Xenon $(Xe)$ with Fluorine $(F_2)$ depends on the molar ratio and reaction conditions.When $Xe$ reacts with $F_2$ in a $1:2$ molar ratio at $573 \ K$ and $60-70 \ bar$ pressure,it produces Xenon tetrafluoride $(XeF_4)$.The balanced chemical equation is: $Xe(g) + 2F_2(g) \xrightarrow{573 \ K, \ 60-70 \ bar} XeF_4(s)$.Therefore,$A$ is $XeF_4$.

$Xe + 2F_2 \xrightarrow{573 \ K, \ 60.70 \ bar} A$
What is $A$?

Similar Questions

Which of the following is primarily contained in the colored discharge tubes used for advertising?

Which elements were discovered by the scientists Ramsay and Glenn $T$. Seaborg?

Which is not correctly matched $:-$

Argon was discovered by

Assertion $(A)$: Noble gases have very low boiling points.
Reason $(R)$: All noble gases have general electronic configuration of $ns^2 np^6$ (except $He$).

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$Xe + 2F_2 \xrightarrow{573 \ K, \ 60.70 \ bar} A$What is $A$?