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.
</span>
![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)
<span>
</span><span>ln(101.3 kPa / 52.7 kPa) = (-29.82 kJ/mol / 8.314x10^{-3} kJ/molK) (1/T - 1/329 K)
</span>
------ some algebra goes here -----
<span>T = 349.99K ...... or ...... 76.8C </span>
Limestone is calcium carbonate Ca(Co3) it's dissociation equation is as follows:
Ca(CO3) <----> Ca+2 + CO3-2
but the solubility is very small
in the presence of acid by LeChatlier's Principle you can shift to the right by adding acid
H+ + CO3-2 <-----> H+CO3- (the bicarbonate ion which is really soluble)
The system will pull more and more limestone out into solution as acid is added and neutralize the acid coming from the rain.
In 1803AD, Osmium was discovered by Smithson Tennant when dissolving an impure sample of platinum in aqua regia, a mixture of hydrochloric and nitric acids. He was able to determine that the black powder left behind after dissolving the platinum was actually a mixture of two new elements, indium and osmium.
•Osmium is a bluish-white and shiny metal.
•Osmium is very hard and is brittle even at very high temperatures.
•Osmium has the lowast vapor pressure and the highest melting point among the platinum group of metals.
•Osmium's density is slightly more than iridium hence is credited as the heaviest element.
Answer:
There will be produced 1.71 moles of B which contain 1.03×10²⁴ molecules
Explanation:
The example reaction is:
2A → 3B
2 moles of A produce 3 moles of B
If we have the mass of A, we convert it to moles and then, we make the rule of three: 29.2 g / 25.6g/mol = 1.14 moles
Therefore 2 moles of A produce 3 moles of B
1.14 moles of A will produce (1.14 . 3) / 2 = 1.71 moles of B are produced
Now we can determine, the number of molecules
1 mol has NA molecules (6.02×10²³)
1.71 moles have (1.71 . NA) = 1.03×10²⁴ molecules
C. Oxidized and reduced are the same.
