Answer:
[OH-]= 1.995×10^-11M
Explanation:
pOH= 14- pH = 14-3.3 =10.7
pOH= -log[OH-]
[OH-]=antilog (-10.7)
[OH-]= 1.995×10^-11M
Explanation:
The given data is as follows.
Enthalpy of vaporization, 
= 38.56 kJ/mol
Temperature (T) = 
= (78.4 + 273) K
= 351.4 K
Now, we will calculate the change in entropy using the formula as follows.
= 
= 0.109 kJ/mol K
Thus, w ecan conclude that the entropy change for vaporization is 0.109 kJ/mol K.
Answer:
c
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cc
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AMOUNG US AND THE ANSWER IS C
Explanation:
From the periodic table you get the atomic masses of each element.
These are the values that I have in my periodic table (use those numbers of your periodic table if your teache gave you a specific one)
Na: 23 g/mol
O: 16 g/mol
H: 1 g/mol
C: 12 g/mol
Ca: 40 g/mol
S: 32 g/mol
Mg: 24 g/mol
P: 31 g/mol
Now I will do some examples and you do the others:
1) NaOH: 1 atom of Na * 23 g/mol + 1 atom of O * 16 g/mol + 1 atom of H * 1 g/mol
=> 1*23g/mol + 1*16g/mol + 1*1g/mol = 40 g/mol
2) H2O
=> 2 atoms of H * 1 g/mol + 1 atom of O * 16 g/mol = 2*1g/mol + 1*16g/mol = 18 g/mol
3) Glucose: C6H12O6
6*12 g/mol + 12 * 1g/mol + 6*16 g/mol = 72g/mol + 12g/mol + 96 g/mol = 180 g/mol
4) CaSO4:
1*40 g/mol + 1*32g/mol + 4*16g/mol = 136 g/mol
Now you only have to do the last one by your own.
Answer:
It says in the question its up to you to decide, so you have to graph the weather changes then explain why you graphed it in that way.
