Question #28d02

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Question #28d02

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Answer 1 · 2016-01-27T05:51:12

$1.99 x 10^{7} g r a m s$ $1.99x10^7 grams$

Explanation:

$U s e$ $Use$ $Avogadro's$ $n$ $umber$ , $6.022*10^23$ , $which$ $is$ $the$ $n$ $umber$ $o$ $f$ $at$ $oms$ $present$ $i$ $n$ $on$ $e$ $mol$ $e$ $o$ $f$ $anythi$ $ng$ .

$Mol$ $es=$ $At$ $oms$ $o$ $f$ $s$ $ubsta$ $n$ $c$ $e$ $/$ $A$ $vogadro's$

$So$ , $take$ $3.01*10^29$ $and$ $di$ $vide$ $it$ $by$ $6.022*10^23$ , $which$ $gives$ $you$ $4.99*10^5$ , $which$ $is$ $the$ $n$ $umber$ $o$ $f$ $mol$ $es$ $of$ $argon$ .

$Now$ ,
$Mass = Mol$ $ecu$ $l$ $ar$ $Weight$ $*$ $Mol$ $es$

$The$ $mol$ $ecu$ $lar$ $weight$ $of$ $Argon,$ $which$ $can$ $be$ $found$ $on$ the periodic table, $is$ $39.9g$ $/mol$

$So, mass =$ $(39.9)(4.99*10^5)$
$= 1.99*10^7$ $grams$

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Question #28d02

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