Use the Clausius-Clapeyron equation...
<span>Let T1 be the normal boiling point, which will occur at standard pressure (P1), which is 101.3 kPa (aka 760 torr or 1.00 atm). You know the vapour pressure (P2) at a different temperature (T2). And you are given the enthalpy of vaporization. Therefore, we can use the Clausius-Clapeyron equation.
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![ln(P_1/P_2) = \frac{-\delta H_{vap}}{R} \times [\frac{1}{T_1} - \frac{1}{T_2}]](https://tex.z-dn.net/?f=ln%28P_1%2FP_2%29%20%3D%20%5Cfrac%7B-%5Cdelta%20H_%7Bvap%7D%7D%7BR%7D%20%20%5Ctimes%20%5B%5Cfrac%7B1%7D%7BT_1%7D%20-%20%5Cfrac%7B1%7D%7BT_2%7D%5D)
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</span><span>ln(101.3 kPa / 52.7 kPa) = (-29.82 kJ/mol / 8.314x10^{-3} kJ/molK) (1/T - 1/329 K)
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------ some algebra goes here -----
<span>T = 349.99K ...... or ...... 76.8C </span>
Answer:
i think its 2
Explanation:
because the H2 is seperate from the others it is the only one with a number.
When an atomic nucleus emits an alpha particle it decay into an atom with atomic number 2 less and mass number 4 less. Thus Thorium 230 decay as follows.
230 90Th -------> 226 88Th + 4 2 He
thorium is in the atomic number 90 thus it during alpha decay it reduces to atomic number 88 while its 230 mass number reduces to 226
Hey there!:
is the solution <span>saturated , ie :
</span>They are the ones that have reached the exact solubility coefficient.<span>If we mix 64.4 g of KCl at 200.0 g of water at 20 º C, we'll see that the 32.2 g will dissolve and the remainder (32.2 g) will precipitate, forming the bottom body. In this case we will then have a solution saturated with background. However, if we want only the saturated solution, simply perform a simple filtration to separate the precipitate from the saturated solution.
hope this helps!</span>