How many kilograms are in 4 moles of Na2CO3?
1 answer:
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Answer:
Mass percent of nitrogen in the compound is 13,3%
Explanation:
Dumas method is an analytical method to determine nitrogen content in samples, thus:
CₐHₓNₙ + (2a+x/2) CuO → aCO₂ + ˣ/₂ H₂O + ⁿ/₂ N₂ + (2a+x/2) Cu
As the CO₂ is removed with KOH, in the mixture you have H₂O and N₂
At 25°C. the vapor pressure of water is 23,8 torr, that means that the pressure due to N₂ is:
726torr - 23,8torr = 702,2 torr.
Using gas law:
n = PV/RT
Where:
P is pressure (702,2torr≡ 0,924 atm)
V is volume (0,0318L)
R is gas constant (0,082atmL/molK)
And T is temperature (25°C≡298,15K)
Replacing, number of moles of N₂, n, are:
n = 1,20x10⁻³moles of N₂. In grams:
1,20x10⁻³moles of N₂× =<em> 0,0336 g of N₂</em>.
Thus, mass percent of nitrogen in the compound is:
×100= <em>13,3%</em>
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I hope it helps!
Answer:
( About ) 0.03232 M
Explanation:
Based on the units for this reaction it should be a second order reaction, and hence you would apply the integrated rate law equation "1 / [X] = kt + 1 / []"
This formula would be true for the following information -
{ = the initial concentration of X, k = rate constant, [ X ] = the concentration after a certain time ( which is what you need to determine ), and t = time in minutes }
________
Therefore, all we have left to do is plug in the known values. The initial concentration of X is 0.467 at a time of 0 minutes, as you can tell from the given data. This is not relevant to the time needed in the formula, as we need to calculate the concentration of X after 18 minutes ( time = 18 minutes ). And of course k, the rate constant = 1.6
1 / [X] = ( 1.6 )( 18 minutes ) + 1 / ( 0.467 ) - Now let's solve for X
1 / [X] = 28.8 + 1 / ( 0.467 ),
1 / [X] = 28.8 + 2.1413...,
1 / [X] = 31,
[X] = 1 / 31 = ( About ) 0.03232 M
Now for this last bit here you probably are wondering why 1 / 31 is not 0.03232, rather 0.032258... Well, I did approximate one of the numbers along the way ( 2.1413... ) and took the precise value into account on my own and solved a bit more accurately. So that is your solution! The concentration of X after 18 minutes is about 0.03232 M