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maksim [4K]
3 years ago
11

If 280cm3 of hydrogen diffuses in 40secs, how long will it take for 490cm3 of a gas x, whose vapor density is 25 to diffuse unde

r the same condition (RMM hydrogen =2)​
Chemistry
1 answer:
BaLLatris [955]3 years ago
6 0

The time taken by gas x : 350 s

<h3>Further explanation  </h3>

Graham's law: the rate of effusion of a gas is inversely proportional to the square root of its molar masses or  

the effusion rates of two gases = the square root of the inverse of their molar masses:  

\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }

Molar mass = 2 x vapour density(VD)

r₁=rate Hydrogen = 280 cm³ / 40 s = 7 cm³/s

M₁=molar mass Hydrogen=2 g/mol

M₂=molar mass gas x=

\tt 2\times VD=2\times 25=50~g/mol

\tt \dfrac{7}{r_2}=\sqrt{\dfrac{50}{2} }\\\\\dfrac{7}{r_2}=5\\\\r_2=1.4~cm^3/s

\tt r_2=\dfrac{V_2}{t_2}\\\\t_2=\dfrac{V_2}{r_2}\\\\t_2=\dfrac{490~cm^3}{1.4~cm^3/s}\\\\t_2=350~s

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The density of Ca will be between that of Mg and Sr

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3 years ago
Determine the amount of heat(in Joules) needed to boil 5.25 grams of ice. (Assume standard conditions - the ice exists at zero d
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Following are important constant that used in present calculations
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Now, energy required for melting of ICE = <span>  334 X 5.25 = 1753.5 J .......(1)
Energy required for raising </span><span>the temperature water from 0 oC to 100 oC =  4.18 X 5.25 X 100 = 2195.18 J .............. (2)
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The answer to your question is:

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b) O2, 15.13 g of CO2

c) It's not posible to know which is the limiting reactant.

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b. Given 15.00 g. of CH4 and 22.00 g. of O2, identify the limiting reactant and calculate the grams of CO2 that can be produced. LR _________ grams CO2 _________ .  

                                CH4   +   2O2   ⇒   CO2   +   2H2O

                                15 g         22 g

                                16 g of CH4 ----------------  64 g of O2

                                15 g of CH4  ---------------   x

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The Limiting reactant is O2 because it is necessary 60g of O2 for 16 g of CH4 and there are only 22.

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