You must add 75 mL water to 195 mL 90 % alcohol to make 270 mL of 65 % alcohol.
<em>Step 1.</em> Calculate the volume of 90 % alcohol needed
You can use the dilution formula
<em>V</em>1×<em>C</em>1 = <em>V</em>2×<em>C</em>2
where
<em>V</em>1 and<em> V</em>2 are the volumes of the two solutions
<em>C</em>1 and <em>C</em>2 are the concentrations
You can solve the above formula to get
<em>V</em>2 = <em>V</em>1 × <em>C</em>1/<em>C</em>2
<em>V</em>1 = 270 mL; <em>C</em>1 = 65 %
V2 = ?; _____<em>C</em>2 = 90 %
∴<em>V</em>2 = 270 mL × (65 %/90 %) = 195 mL
You need 195 mL of 90 % alcohol to make 270 mL of 65 % RA
<em>Step 2</em>. Calculate the amount of water to add.
Volume of water = 270 mL – 195 mL = 75 mL
Answer:
2
Explanation:
Each orbital can hold two electrons. One spin-up and one spin-down.
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Answers and Explanation:
a)- The chemical equation for the corresponden equilibrium of Ka1 is:
2. HNO2(aq)⇌H+(aq)+NO−2
Because Ka1 correspond to a dissociation equilibrium. Nitrous acid (HNO₂) losses a proton (H⁺) and gives the monovalent anion NO₂⁻.
b)- The relation between Ka and the free energy change (ΔG) is given by the following equation:
ΔG= ΔGº + RT ln Q
Where T is the temperature (T= 25ºc= 298 K) and R is the gases constant (8.314 J/K.mol)
At the equilibrium: ΔG=0 and Q= Ka. So, we can calculate ΔGº by introducing the value of Ka:
⇒ 0 = ΔGº + RT ln Ka
ΔGº= - RT ln Ka
ΔGº= -8.314 J/K.mol x 298 K x ln (4.5 10⁻⁴)
ΔGº= 19092.8 J/mol
c)- According to the previous demonstation, at equilibrium ΔG= 0.
d)- In a non-equilibrium condition, we have Q which is calculated with the concentrations of products and reactions in a non equilibrium state:
ΔG= ΔGº + RT ln Q
Q= ((H⁺) (NO₂⁻))/(HNO₂)
Q= ( (5.9 10⁻² M) x (6.7 10⁻⁴ M) ) / (0.21 M)
Q= 1.88 10⁻⁴
We know that ΔGº= 19092.8 J/mol, so:
ΔG= ΔGº + RT ln Q
ΔG= 19092.8 J/mol + (8.314 J/K.mol x 298 K x ln (1.88 10⁻⁴)
ΔG= -2162.4 J/mol
Notice that ΔG<0, so the process is spontaneous in that direction.
