The concentration of the drug stock solution is 1.5*10^-9 M i.e. 1.5 * 10^-9 moles of the drug per Liter of the solution
Therefore, the number of moles present in 1 ml i.e. 1*10^-3 L of the solution would be = 1 *10^-3 L * 1.5 * 10^-9 moles/1 L = 1.5 * 10^-12 moles
1 mole of the drug will contain 6.023*10^23 drug molecules
Therefore, 1.5*10^-12 moles of the drug will correspond to :
1.5 * 10^-12 moles * 6.023*10^23 molecules/1 mole = 9.035 * 10^11 molecules
The number of cancer cells = 2.0 * 10^5
Hence the ratio = drug molecules/cancer cells
= 9.035 *10^11/2.0 *10^5
= 4.5 * 10^6
Answer:
Avogadro's number or Avogadro's constant.
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!
Explanation:
Reaction:
Cu + 2AgC₂H₃O₂ → Cu(C₂H₃O₂)₂ + 2Ag
The problem is to split the reaction into oxidation and reduction halves:
The oxidation half is the sub-reaction that undergoes oxidation
The reduction half is the one that undergoes reduction:
The ionic equation:
Cu + 2Ag⁺ + 2C₂H₃O₂⁻ → Cu²⁺ + 2C₂H₃O₂⁻ + 2Ag
Oxidation half:
Cu → Cu²⁺ + 2e⁻
Reduction half:
2Ag⁺ + 2e⁻ → 2Ag
C₂H₃O₂⁻ is neither oxidized nor reduced in the reaction.
learn more:
Oxidation state brainly.com/question/10017129
#learnwithBrainly
Answer:
Mass = 0.158 g
Explanation:
Formula used,
P V = n R T
Or,
n = P V / R T
Putting values,
n = 0.948 atm . 0.025 L / 0.0821 L.atm.K⁻¹.mol⁻¹ . 291.45
n = 0.00099 mol
Note: we have changed pressure from mmHg to atm, volume from mL to L and temperature from C to K)
Also,
Mass = n . Molecular Mass
Mass = 0.00099 mol × 159.808 g/mol
Mass = 0.158 g
