Yes. a linear graph does. Pass the vertical line test
Each value in the domain X,results in a unique range ,y, value
Answer:
![[Cu^{2+}]=0.041 M](https://tex.z-dn.net/?f=%5BCu%5E%7B2%2B%7D%5D%3D0.041%20M)
Explanation:
Hello!
In this case, since the molarity of a solution is defined in terms of the moles of the solute and the volume of solution, given that the concentration of Cu(NH₃)₄²⁺ is 0.041 M, and there is only one copper atom per Cu(NH₃)₄²⁺ ion, we can compute the concentration of Cu²⁺ as shown below:
![[Cu^{2+}]=0.041\frac{molCu(NH_3)_4^{2+}}{L}*\frac{1molCu^{2+}}{1molCu(NH_3)_4^{2+}} =0.041 \frac{molCu(NH_3)_4^{2+}}{L}](https://tex.z-dn.net/?f=%5BCu%5E%7B2%2B%7D%5D%3D0.041%5Cfrac%7BmolCu%28NH_3%29_4%5E%7B2%2B%7D%7D%7BL%7D%2A%5Cfrac%7B1molCu%5E%7B2%2B%7D%7D%7B1molCu%28NH_3%29_4%5E%7B2%2B%7D%7D%20%3D0.041%20%5Cfrac%7BmolCu%28NH_3%29_4%5E%7B2%2B%7D%7D%7BL%7D)
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I got a 4 out of 1. not 100 percent sure if this is the right answer but i put 1,2,1,2
Answer:
The most effective buffer at pH 9.25 will be a mixture of 1.0 M NH3 and 1.0 M NH4Cl
Explanation:
Step 1: Data given
pH of a buffer = pKa + log ([A-]/[Ha])
a mixture of 1.0 M HC2H3O2 and 1.0 M NaC2H3O2 (Ka for acetic acid = 1.8 x 10-5)
pH = -log( 1.8 * 10^-5) + log (1/1)
pH = -log( 1.8 * 10^-5)
pH = 4.74
a mixture of 1.0 M NaCN and 1.0 M KCN (Ka for HCN = 4.9 x 10-10)
pH = -log( 4.9 * 10^-10) + log (1/1)
pH = -log( 1.8 * 10^-5)
pH = 9.30
a mixture of 1.0 M HCl and 1.0 M NaCl
The solution made from NaCl and HCl will NOT act as a buffer.
HCl is a strong acid while NaCl is salt of strong acid and strong base which do not from buffer solutions hence due to HCl PH is less than 7.
a mixture of 1.0 M NH3 and 1.0 M NH4Cl (Kb for ammonia = 1.76 x 10^-5)
Ka * Kb = 1*10^-14
Ka = 10^-14 / 1.76*10^-5
Ka = 5.68*10^-10
pH = -log( 5.68*10^-10) + log (1/1)
pH = -log( 5.68*10^-10)
pH = 9.25
The most effective buffer at pH 9.25 will be a mixture of 1.0 M NH3 and 1.0 M NH4Cl
