A lot of molecules will be in 2.570 moles of H2
Answer:
The molarity of the HCl solution should be 4.04 M
Explanation:
<u>Step 1:</u> Data given
volume of HCl solution = 10.00 mL = 0.01 L
volume of a 1.6 M NaOH solution = 25.24 mL = 0.02524 L
<u>Step 2:</u> The balanced equation
HCl + NaOH → NaCL + H2O
Step 3: Calculate molarity of HCl
n1*C1*V1 = n2*C2*V2
Since the mole ratio for HCl and NaOH is 1:1 we can just write:
C1*V1 =C2*V2
⇒ with C1 : the molarity of HCl = TO BE DETERMINED
⇒ with V1 = the volume og HCl = 10 mL = 0.01 L
⇒ with C2 = The molarity of NaOH = 1.6 M
⇒ with V2 = volume of NaOH = 25.24 mL = 0.02524 L
C1 * 0.01 = 1.6 * 0.02524
C1 = (1.6*0.02524)/0.01
C1 = 4.04M
The molarity of the HCl solution should be 4.04 M
Answer:
1. 0.97 V
2. 
Explanation:
In this case, we can start with the <u>half-reactions</u>:
With this in mind we can <u>add the electrons</u>:
<u>Reduction</u>
<u>Oxidation</u>
The reduction potential values for each half-reaction are:
- 0.69 V
-1.66 V
In the aluminum half-reaction, we have an oxidation reaction, therefore we have to <u>flip</u> the reduction potential value:
+1.66 V
Finally, to calculate the overall potential we have to <u>add</u> the two values:
1.66 V - 0.69 V = <u>0.97 V</u>
For the second question, we have to keep in mind that in the cell notation we put the anode (the oxidation half-reaction) in the left and the cathode (the reduction half-reaction) in the right. Additionally, we have to use "//" for the salt bridge, therefore:
I hope it helps!
Answer:
The right to refuse work that could affect their health and safety and that of others.
Answer:
<h2>All Group 1 metals form halides that are white solids at room temperature. The melting point is correlated to the strength of intermolecular</h2>
