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Luba_88 [7]
3 years ago
14

Calculate the mass of CaCl2 ( formula wt. = 110.99) thatshould

Chemistry
1 answer:
NNADVOKAT [17]3 years ago
7 0

Answer: 5622.6g

Explanation:

Note: Kf for water is 1.86°C/m.

The simple calculation is in the attachment below.

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How many grams of the excess reactant are left over according to the reaction below given that you start with 10.0 g of Al and 1
valentinak56 [21]
<span>4 Al + 3 O2 → 2 Al2O3 

(10.0 g Al) / (26.98154 g Al/mol) = 0.37062 mol Al 
(19.0 g O2) / (31.99886 g O2/mol) = 0.59377 mol O2 

0.37062 mole of Al would react completely with 0.37062 x (3/4) = 0.277965 mole of O2, but there is more O2 present than that, so O2 is in excess. 

((0.59377 mol O2 initially) - (0.277965 mol O2 reacted)) x (31.99886 g O2/mol) = 
10.1 g O2 left over</span><span>
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What adaptations allow a camel and a cactus to survive in warm environments?
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A camel stores fat in its hump, while the cactus stores water in its thick stem.

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8 0
3 years ago
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How many significant figures are in 0.0233?
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4 years ago
Calculate the number of liters in 3.25 g of ammonia
il63 [147K]

 The liters in   3.25 g   of  ammonia  4.28 L


  <u><em>calculation</em></u>

 Step 1: find moles of ammonia

 moles = mass÷ molar  mass

 From  periodic    table  the molar mass  of ammonia (NH₃)  =  14 +(1×3 ) = 17  g/mol

3.25 g÷ 17 g/mol = 0.191   moles

Step 2: find the number of liters of ammonia

 that is at STP  1  moles = 22.4 L

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<em>by cross  multiplication</em>

 ={( 0.191   moles  ×22.4 L) / 1 mole}  = 4.28 L



8 0
3 years ago
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