The metalloids are mostly concentrated in groups 14, 15, and 16. (Some simpler charts will show them as 4A, 5A, and 6A - take a look at the top of the periodic table your class uses to double-check).
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Answer:
The pressure inside the flask is 0,09 atm
Explanation:
We calculate the number of moles of O2, we convert the Temperature into unit KELVIN and use the formula of the ideal gases to calculate the pressure:
Weight 1 mol of 02= 15,999g x2= 32g
32g---1mol 02
0,60g---x=(0,60gx1mol O2)/32g= 0,019mol O2
T(K)=273+22=295K
PV=nRT
Px5.0l= 0,019mol x 0,082 lxatm/Kxmolx295 K
P=(0,019mol x 0,082 lxatm/Kxmolx295 K)/5.0l
P=0,09atm
<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
Answer:
a) Can be cut into tiny pieces - Physical Change
b) Fizzling occurs when placed water -Chemical Change
c) Light is emitted when burned -Chemical Change
d) Turns to ash -Chemical Change
Explanation:
Yes, this is a balanced chemical equation, and it obeys the Law of Conservation of Mass.
A balanced chemical equation obeys the Law of Conservation of Mass if there are the <em>same numbers of each type of atom on each side of the reaction arro</em>w.
2H₂ + O₂ ⟶ 2H₂O
Let’s check the numbers.
<u>Atom</u> <u>On the left</u> <u>On the right</u><u>
</u>
H 4 4
O 2 2
We have the same numbers of H and O atoms on each side of the reaction arrow.
The equation is<em> </em>balanced and it obeys the Law of Conservation of Mass.
