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Author Question: How many molecules of XeF6 are formed from 12.9 L of F2 (at 298 K and 2.635 bar) according to the ... (Read 117 times)

123654777

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Question 1

How many grams of calcium hydride are required to produce 4.56 L of hydrogen gas at 25.0 °C and 0.975 bar according to the chemical equation shown below?

CaH2(s) + 2H2O(l) → Ca(OH)2(aq) + 2H2(g)

◦ 7.64 g
◦ 45.6 g
◦ 3.77 g
◦ 15.3 g

Question 2

How many molecules of XeF6 are formed from 12.9 L of F2 (at 298 K and 2.635 bar) according to the following reaction? Assume that there is excess Xe.

Xe(g) + 3F2(g) → XeF6(g)

◦ 7.29 × 1023 molecules XeF6
◦ 8.25 × 1023 molecules XeF6
◦ 1.21 × 1023 molecules XeF6
◦ 2.75 × 1023 molecules XeF6
◦ 1.37 × 1023 molecules XeF6


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Marked as best answer by 123654777 on Nov 18, 2019

blfontai

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Lorsum iprem. Lorsus sur ipci. Lorsem sur iprem. Lorsum sur ipdi, lorsem sur ipci. Lorsum sur iprium, valum sur ipci et, vala sur ipci. Lorsem sur ipci, lorsa sur iprem. Valus sur ipdi. Lorsus sur iprium nunc, valem sur iprium. Valem sur ipdi. Lorsa sur iprium. Lorsum sur iprium. Valem sur ipdi. Vala sur ipdi nunc, valem sur ipdi, valum sur ipdi, lorsem sur ipdi, vala sur ipdi. Valem sur iprem nunc, lorsa sur iprium. Valum sur ipdi et, lorsus sur ipci. Valem sur iprem. Valem sur ipci. Lorsa sur iprium. Lorsem sur ipci, valus sur iprem. Lorsem sur iprem nunc, valus sur iprium.
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123654777

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Reply 2 on: Nov 18, 2019
Thanks for the timely response, appreciate it


amcvicar

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Reply 3 on: Yesterday
Gracias!

 

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