Van der Waals equation is as follows:
(P + a(n/V)²) * (V -nb) = n RT
Moles of H₂ is calculated by dividing 25 g by 2 (molecular weight of H₂) = 12.5 moles
Values of a and b are:
a = 0.02444 atm L² / mol
b = 0.0266 L / mol
(P + 0.02444 (12.5/1)²) * (1 -(12.5 *0.0266)) = 12.5 * 0.0821 * 293 K
If we solve this equation we get pressure of 412.29 atm
With ideal gas equation we get:
P V = n R T
P = n * R * T / V = (12.5 * 0.0821 * 293 / 1) = 300.69 atm
The given substance combusts following the reaction:
C2H2 + (5/2)O2 -> 2CO2 + H2O
Assume C2H2 is an ideal gas. At STP, 1 mol of an ideal gas
occupies 22.4 L. Given 100.50 mL of C2H2, this means that there is 4.4866 x
10^(-3) mol. Combusting 1 mol of C2H2 consumes (5/2) mol of O2, then combusting
the given amount of C2H2 consumes 0.01121 mol of O2. At STP, this amount of O2
occupies 251.25 mL.
Body cells and tissues
They help give oxygen and other nutrients to keep you healthy
Hope this helps, Godbless :)
Answer:That all noble gases will have 8 valence electrons which makes a atom stable
Explanation:
Answer:
0.00735°C
Explanation:
By seeing the question, we can see the elevation in boiling point with addition of BaCl₂ in water
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<u>The</u><u> </u><u>elevation</u><u> </u><u>in</u><u> </u><u>boiling</u><u> </u><u>point</u><u> </u><u>is</u><u> </u><u>a</u><u> </u><u>phenomenon</u><u> </u><u>in</u><u> </u><u>which</u><u> </u><u>there</u><u> </u><u>is</u><u> </u><u>increase</u><u> </u><u>in</u><u> </u><u>boiling</u><u> </u><u>point</u><u> </u><u>in</u><u> </u><u>solution</u><u>,</u><u> </u><u>when</u><u> </u><u>the</u><u> </u><u>particular</u><u> </u><u>type</u><u> </u><u>of</u><u> </u><u>solute</u><u> </u><u>is</u><u> </u><u>added</u><u> </u><u>to</u><u> </u><u>pure</u><u> </u><u>solvent</u><u>.</u>
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Where 'i' is van't hoff factor which represents the ratio of observed osmotic pressure and the value to be expected.
and 'i' is 3 (as given in the question)
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'Kb' is molal boiling point constant. And it's value is 0.51°C/mol(given in question)
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'm' represent the molality of solution. Molatity is no. of moles of solution present in 1kg of solution.
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<u>To</u><u> </u><u>find</u><u> </u><u>molality</u><u>,</u><u> </u><u>we</u><u> </u><u>have</u><u> </u><u>to</u><u> </u><u>divide</u><u> </u><u>no</u><u>.</u><u> </u><u>of</u><u> </u><u>moles</u><u> </u><u>of</u><u> </u><u>solute</u><u> </u><u>by</u><u> </u><u>weight</u><u> </u><u>of</u><u> </u><u>solution</u>
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While first we need to no. of moles
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<u>Now</u><u>,</u><u> </u><u>we</u><u> </u><u>will</u><u> </u><u>find</u><u> </u><u>molality</u>
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<u>Henceforth</u><u>,</u><u> </u><u>the</u><u> </u><u>change</u><u> </u><u>in</u><u> </u><u>boiling</u><u> </u><u>point</u><u> </u><u>is</u><u> </u><u>0</u><u>.</u><u>0</u><u>0</u><u>7</u><u>3</u><u>5</u><u>°</u><u>C</u><u>.</u>