Hello.
balance the equation first
<span>Cu + 2AgNO3 ----> [2Ag] + Cu(NO3)2
</span>
<span>Now set up the equation </span>
<span>350gCu*(1molCu/ 63.546g Cu) * (2molAg/1molCu) * (107.8682g Ag/1molAg)
</span>
<span>Cancel out 1molCu with 1molCu, and cancel out 2molAg with 1molAg (it still remains 2mol Ag) </span>
<span>You are left with </span>
<span>350gCu*(1/63.546g Cu)*(2Ag/1)* (107.8682g Ag/1)
</span>
<span>350gCu*1*2*107.8682= 75507.74g
</span>
<span>now divide 75507.74g by 63.546g
</span>
<span>you get [1188.2374972460896g Ag]
</span>
Have a nice day
<span>3.68 liters
First, determine the number of moles of butane you have. Start with the atomic weights of the involved elements:
Atomic weight carbon = 12.0107
Atomic weight hydrogen = 1.00794
Atomic weight oxygen = 15.999
Molar mass butane = 4*12.0107 + 10*1.00794 = 58.1222 g/mol
Moles butane = 2.20 g / 58.1222 g/mol = 0.037851286
Looking at the balanced equation for the reaction which is
2 C4H10(g)+13 O2(g)→8 CO2(g)+10 H2O(l)
It indicates that for every 2 moles of butane used, 8 moles of carbon dioxide is produced. Simplified, for each mole of butane, 4 moles of CO2 are produced. So let's calculate how many moles of CO2 we have:
0.037851286 mol * 4 = 0.151405143 mol
The ideal gas law is
PV = nRT
where
P = Pressure
V = Volume
n = number of moles
R = Ideal gas constant ( 0.082057338 L*atm/(K*mol) )
T = absolute temperature (23C + 273.15K = 296.15K)
So let's solve the formula for V and the calculate using known values:
PV = nRT
V = nRT/P
V = (0.151405143 mol) (0.082057338 L*atm/(K*mol))(296.15K)/(1 atm)
V = (3.679338871 L*atm)/(1 atm)
V = 3.679338871 L
So the volume of CO2 produced will occupy 3.68 liters.</span>
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Answer:
Aluminium is malleable. It means it can be hammered or changed into other shapes easily. In this case, the shape of a soft drink can.
Explanation:
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