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Anna [14]
2 years ago
5

chemistry moles PLEASE HELP!! A package of foil weighs 360g. How many moles of aluminium are there in this package??

Chemistry
1 answer:
Ksenya-84 [330]2 years ago
6 0

Answer:

I am not 100% sure but I think it is 26.981538

Explanation:

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.00 mol of gas occupy a volume of 112 L at a pressure of 1.00 atm.  What is the kelvin temperature of the gas
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A gas cylinder contains 2.5 atm of ne, 0.8 atm of co2, and 3.4 atm of ar. what is the total pressure in the cylinder? g
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To answer this problem, we must make assumptions for simplicity. The first assumption is that, the system only consist of these 3 gases. The second assumption is that, these gases behave ideally. Thus, from Dalton's Law of Partial Pressure, the total pressure is simply the sum of their individual partial pressures.

Total pressure = 2.5 + 0.8 + 3.4 = <em>6.7 atm</em>
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Name the type of equation shown here: 2NaBr+Ca(OH)2 —&gt;CaBr2+2NaOH
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The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If t
11111nata11111 [884]

Answer:

The catalyzed reaction will take 2.85 seconds to occur.

Explanation:

The activation energy of a reaction is given by:                                                        

k = Ae^{-\frac{E_{a}}{RT}}

For the reaction without catalyst we have:

k_{1} = Ae^{-\frac{E_{a_{1}}}{RT}}   (1)

And for the reaction with the catalyst:

k_{2} = Ae^{-\frac{E_{a_{2}}}{RT}}   (2)

Assuming that frequency factor (A) and the temperature (T) are constant, by dividing equation (1) with equation (2) we have:                      

\frac{k_{1}}{k_{2}} = \frac{Ae^{-\frac{E_{a_{1}}}{RT}}}{Ae^{-\frac{E_{a_{2}}}{RT}}}

\frac{k_{1}}{k_{2}} = e^{\frac{E_{a_{2}} - E_{a_{1}}}{RT}    

\frac{k_{1}}{k_{2}} = e^{\frac{59.0 \cdot 10^{3}J/mol - 184 \cdot 10^{3} J/mol}{8.314 J/Kmol*600 K} = 1.31 \cdot 10^{-11}    

Since the reaction rate is related to the time as follow:

k = \frac{\Delta [R]}{t}

And assuming that the initial concentrations ([R]) are the same, we have:

\frac{k_{1}}{k_{2}} = \frac{\Delta [R]/t_{1}}{\Delta [R]/t_{2}}

\frac{k_{1}}{k_{2}} = \frac{t_{2}}{t_{1}}

t_{2} = t_{1}\frac{k_{1}}{k_{2}} = 6900 y*1.31 \cdot 10^{-11} = 9.04 \cdot 10^{-8} y*\frac{365 d}{1 y}*\frac{24 h}{1 d}*\frac{3600 s}{1 h} = 2.85 s

Therefore, the catalyzed reaction will take 2.85 seconds to occur.

I hope it helps you!                            

4 0
3 years ago
The arrows represent tge movement of which substances
11111nata11111 [884]
The arrows represent the movement of starting substances
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