Answer:
Be the anode
Explanation:
The standard hydrogen electrode is regarded as the standard reference electrode and it has been assigned an electrode potential of 0.0V.
If any substance has an electrode potential that is more negative than hydrogen, then that half cell will function as the anode when connected to the standard hydrogen electrode.
Similarly, any substance that has a more positive electrode potential than hydrogen will serve as the cathode when its half cell is connected to the standard hydrogen electrode.
Answer: A.Light travels in a straight line., B.Light behaves in a predictable way. , D.Light curves around corners or obstructions
-Hope this helps<3
Answer:
Q= 245 =2.5 * 10^2
Explanation:
ΔG = ΔGº + RTLnQ, so also ΔGº= - RTLnK
R= 8,314 J/molK, T=298K
ΔGº= - RTLnK = - 6659.3 J/mol = - 6.7 KJ/mol
ΔG = ΔGº + RTLnQ → -20.5KJ/mol = - 6.7 KJ/mol + 2.5KJ/mol* LnQ
→ 5.5 = LnQ → Q= 245 =2.5 * 10^2
<u>Answer:</u> The concentration of
solution will be 0.094 M.
<u>Explanation:</u>
To calculate the concentration of acid, we use the equation given by neutralization reaction:
![n_1M_1V_1=n_2M_2V_2](https://tex.z-dn.net/?f=n_1M_1V_1%3Dn_2M_2V_2)
where,
are the n-factor, molarity and volume of acid which is ![HNO_3](https://tex.z-dn.net/?f=HNO_3)
are the n-factor, molarity and volume of base which is ![Ba(OH)_2](https://tex.z-dn.net/?f=Ba%28OH%29_2)
We are given:
Conversion factor: 1L = 1000 mL
![n_1=1\\M_1=?M\\V_1=0.110L=110mL\\n_2=2\\M_2=0.100M\\V_2=51.9mL](https://tex.z-dn.net/?f=n_1%3D1%5C%5CM_1%3D%3FM%5C%5CV_1%3D0.110L%3D110mL%5C%5Cn_2%3D2%5C%5CM_2%3D0.100M%5C%5CV_2%3D51.9mL)
Putting values in above equation, we get:
![1\times M_1\times 110=2\times 0.100\times 51.9\\\\M_1=0.094M](https://tex.z-dn.net/?f=1%5Ctimes%20M_1%5Ctimes%20110%3D2%5Ctimes%200.100%5Ctimes%2051.9%5C%5C%5C%5CM_1%3D0.094M)
Hence, the concentration of
solution will be 0.094 M.
Answer:
the change in energy of the gas mixture during the reaction is 227Kj
Explanation:
THIS IS THE COMPLETE QUESTION BELOW
Measurements show that the enthalpy of a mixture of gaseous reactants increases by 319kJ during a certain chemical reaction, which is carried out at a constant pressure. Furthermore, by carefully monitoring the volume change it is determined that -92kJ of work is done on the mixture during the reaction. Calculate the change of energy of the gas mixture during the reaction in kJ.
From thermodynamics
ΔE= q + w
Where w= workdone on the system or by the system
q= heat added or remove
ΔE= change in the internal energy
q=+ 319kJ ( absorbed heat is + ve
w= -92kJ
If we substitute the given values,
ΔE= 319 + (-92)= 227 Kj
With the increase in enthalpy and there is absorbed heat, hence the reaction is an endothermic reaction.