Answer:
B)3
Explanation:
I'm assuming the 2 is meant to be smaller. If it were a regular size 2 the answer would be 2 and that isn't an option
Answer:
The final volume in mL is 7.14 mL or 7.1 mL.
Explanation:
1.Use Boyle's Law(
). Re-arrange to solve for 
<em> for the final volume.</em>
<em />
<em>2. Plug in values. </em>
Answer:
The value of Q must be less than that of K.
Explanation:
The difference of K and Q can be understood with the help of an example as follows
A ⇄ B
In this reaction A is converted into B but after some A is converted , forward reaction stops At this point , let equilibrium concentration of B be [B] and let equilibrium concentration of A be [A]
In this case ratio of [B] and [A] that is
K = [B] / [A] which is called equilibrium constant.
But if we measure the concentration of A and B ,before equilibrium is reached , then the ratio of the concentration of A and B will be called Q. As reaction continues concentration of A increases and concentration of B decreases. Hence Q tends to be equal to K.
Q = [B] / [A] . It is clear that Q < K before equilibrium.
If Q < K , reaction will proceed towards equilibrium or forward reaction will
proceed .
1 gram of sugar because super molecules are bigger then the ions of dissolved salt
The number of mole of HCl needed for the solution is 1.035×10¯³ mole
<h3>How to determine the pKa</h3>
We'll begin by calculating the pKa of the solution. This can be obtained as follow:
- Equilibrium constant (Ka) = 2.3×10¯⁵
- pKa =?
pKa = –Log Ka
pKa = –Log 2.3×10¯⁵
pKa = 4.64
<h3>How to determine the molarity of HCl </h3>
- pKa = 4.64
- pH = 6.5
- Molarity of salt [NaZ] = 0.5 M
- Molarity of HCl [HCl] =?
pH = pKa + Log[salt]/[acid]
6.5 = 4.64 + Log[0.5]/[HCl]
Collect like terms
6.5 – 4.64 = Log[0.5]/[HCl]
1.86 = Log[0.5]/[HCl]
Take the anti-log
0.5 / [HCl] = anti-log 1.86
0.5 / [HCl] = 72.44
Cross multiply
0.5 = [HCl] × 72.44
Divide both side by 72.44
[HCl] = 0.5 / 72.4
[HCl] = 0.0069 M
<h3>How to determine the mole of HCl </h3>
- Molarity of HCl = 0.0069 M
- Volume = 150 mL = 150 / 1000 = 0.15 L
Mole = Molarity x Volume
Mole of HCl = 0.0069 × 0.15
Mole of HCl = 1.035×10¯³ mole
<h3>Complete question</h3>
How many moles of HCl need to be added to 150.0 mL of 0.50 M NaZ to have a solution with a pH of 6.50? (Ka of HZ is 2.3 x 10 -5 .) Assume negligible volume of the HCl
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