Answer:
0.036 M of 
Explanation:
It is an example of acid-base neutralization reaction.
KOH + ----> 
+ 
Base Acid Salt
When two component react then the number of moles of both the component should be same, therefore the number of moles and acids and bases should be the same in the following .
Molarity= 
No.of moles= Molarity × Volume of the Particular Solution
Therefore,
------------------------------(1)
where
= Molarity of Acid
= Volume of Acid
= Molarity of Base
= Volume of Base
=0.3330 M
=10.62 mL
=98.2 mL
=??(in M)
Plugging in Equation 1,
0.3330 × 10.62 = × 98.2
=
=0.036 M
Answer:
Two
Explanation:
Elements in group 16 wants to bond with elements in group IIA, the group of alkaline earth metals.
- The bonding will make it easier for them complete their octet.
- Elements in group 16 has 6 valence electrons.
- To have a complete octet, they require 2 more electrons.
- Group II elements are willing donors as they are metals.
- For Group II elements to fill their octets, they must lose two electrons.
- So the willingness of group II elements to lose two electrons and the readiness for group 16 elements to gain the electrons makes the desire one another.
Answer:
Step 1;
q = w = -0.52571 kJ, ΔS = 0.876 J/K
Step 2
q = 0, w = ΔU = -7.5 kJ, ΔH = -5.00574 kJ
Explanation:
The given parameters are;
= 100 N·m
= 327 K
= 90 N·m
Step 1
For isothermal expansion, we have;
ΔU = ΔH = 0
w = n·R·T·ln(/) = 1 × 8.314 × 600.15 × ln(90/100) = -525.71
w ≈<em> -0.52571</em> kJ
At state 1, q = w = -0.52571 kJ
ΔS = -n·R·ln(/) = -1 × 8.314 × ln(90/100) ≈ 0.876
ΔS ≈ 0.876 J/K
Step 2
q = 0 for adiabatic process
ΔU = 25×(27 - 327) = -7,500
w = ΔU = <em>-7.5 kJ</em>
ΔH = ΔU + n·R·ΔT
ΔH = -7,500 + 8.3142 × 300 = -5,005.74
ΔH = ΔU = <em>-5.00574 kJ</em>
The subscript 2 in the formula o2 indicates that there are 2 atoms of the element oxygen, represented by the symbol "O." The subscript to the right of the symbol tells how many atoms are present of that element.
I'd say the answer is D because they need to intake and keep as much water as possible.
