<u>Answer:</u> The vapor pressure of the liquid is 0.293 atm
<u>Explanation:</u>
To calculate the vapor pressure of the liquid, we use the Clausius-Clayperon equation, which is:
![\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BP_2%7D%7BP_1%7D%29%3D%5Cfrac%7B%5CDelta%20H_%7Bvap%7D%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= initial pressure which is the pressure at normal boiling point = 1 atm
= pressure of the liquid = ?
= Heat of vaporization = 28.9 kJ/mol = 28900 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
= initial temperature = 341.88 K
= final temperature = 305.03 K
Putting values in above equation, we get:
![\ln(\frac{P_2}{1})=\frac{28900J/mol}{8.314J/mol.K}[\frac{1}{341.88}-\frac{1}{305.03}]\\\\\ln P_2=-1.228atm\\\\P_2=e^{-1.228}=0.293atm](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BP_2%7D%7B1%7D%29%3D%5Cfrac%7B28900J%2Fmol%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B341.88%7D-%5Cfrac%7B1%7D%7B305.03%7D%5D%5C%5C%5C%5C%5Cln%20P_2%3D-1.228atm%5C%5C%5C%5CP_2%3De%5E%7B-1.228%7D%3D0.293atm)
Hence, the vapor pressure of the liquid is 0.293 atm
First we need to find the number of moles that 43.9g of gallium metal is. We can do this by finding the molar weight of gallium and cross-multiplying to cancel out units:
So we are dealing with 0.63 moles of gallium metal.
We can take from the balanced equation that 4 moles of gallium metal will react completely with 3 moles of oxygen gas. We can take this ratio and make a proportion to find the amount of oxygen gas, in moles, that will react completely with 0.63 moles of gallium metal:
Cross multiply and solve for x:
So now we know that 0.47 moles of oxygen gas will react with 43.9g of gallium metal.
Answer:
Mass = 2.12 g
Explanation:
Given data:
Volume of KMnO₄ = 255 mL (255/1000 = 0.255 L)
Molarity = 0.0525 M
Mass in gram = ?
Solution:
First of all we will calculate the number of moles.
<em>Molarity = number of moles of solute / volume in litter</em>
0.0525 M = number of moles of solute / 0.255 L
Number of moles of solute = 0.0525 M ×0.255 L
Number of moles of solute = 0.0134 mol
Mass in gram:
<em>Number of moles = mass/ molar mass</em>
Mass = moles × molar mass
Mass = 0.0134 mol × 158.04 g/mol
Mass = 2.12 g
Answer:
Number of grams of mineral salt in one liter of water
.
Explanation:
Salinity :
It is the concentration of dissolved salts .
Salinity means the saltiness or the dissolved salt component in the water body . It is defined as mass (in gram) of the dissolved matter in the water bodies .
salts found in ocean water are :
Salts of chlorides, sulfates , sodium , magnesium, potassium and calcium
For example : world's oceans salinity is 3.5% parts per thousand (g/L) . It means 3.5 g of salt is present in 1 L of seawater .
Note : Sodium chloride is major salt found in ocean . But it is not the only salt present (other salts are also found) . So D can't be the answer
