Dimitri mastered in Elements
Answer:
a) heat it from 23.0 to 78.3
q = (50.0 g) (55.3 °C) (2.46 J/g·°C) =
b) boil it at 78.3
(39.3 kJ/mol) (50.0 g / 46.0684 g/mol) =
c) sum up the answers from the two calculations above. Be sure to change the J from the first calc into kJ
Explanation:
<u>Answer:</u> The concentration of solute is 0.503 mol/L
<u>Explanation:</u>
To calculate the concentration of solute, we use the equation for osmotic pressure, which is:
where,
= osmotic pressure of the solution = 24 atm
i = Van't hoff factor = 2 (for NaCl)
c = concentration of solute = ?
R = Gas constant = 
T = temperature of the solution = ![25^oC=[273+25]=298K](https://tex.z-dn.net/?f=25%5EoC%3D%5B273%2B25%5D%3D298K)
Putting values in above equation, we get:
Hence, the concentration of solute is 0.503 mol/L
Answer:
cesium
In particular, cesium (Cs) can give up its valence electron more easily than can lithium (Li). In fact, for the alkali metals (the elements in Group 1), the ease of giving up an electron varies as follows: Cs > Rb > K > Na > Li with Cs the most likely, and Li the least likely, to lose an electron
Explanation:
Answer:
The answer is...
Explanation:
Smaller crystals have more surface area.
