Ba stays as Ba+2 and Cl stays as Cl-
Hey there:
Correct answer is :
(b) NaNH₂
Sodium azanide NaNH₂ is the conjugate base of ammonia NH₃
Correct answer is :
(b) NaNH₂
I hope this will help !
Answer:
1.00 M
Explanation:
Sn^2+ reacts with KMNO4 as follows;
5Sn^2+(aq) + 2MnO4^-(aq) + 16H^+(aq) ----> 5Sn^4+(aq) + 2Mn^+(aq) + 8H2O(l)
The number of moles of MnO4^- reacted = 42.1/1000 L × 0.145 mol/L
= 0.0061 moles
If 5 moles of Sn^2+ reacts with 2 moles of MnO4^-
x moles of Sn^2+ reacts with 0.0061 moles of MnO4^-
x= 5 × 0.0061/2
x= 0.015 moles
Since the volume of the Sn^2+ solution is 15.00mL or 0.015 L
number of moles = concentration × volume
Concentration = number of moles/volume
Concentration= 0.015 moles/0.015 L
Concentration = 1 M
Answer:
11.9 is the pOH of a 0.150 M solution of potassium nitrite.
Explanation:
Solution : Given,
Concentration (c) = 0.150 M
Acid dissociation constant = 
The equilibrium reaction for dissociation of 
(weak acid) is,
initially conc. c 0 0
At eqm. 
First we have to calculate the concentration of value of dissociation constant 
.
Formula used :
Now put all the given values in this formula ,we get the value of dissociation constant .
By solving the terms, we get
No we have to calculate the concentration of hydronium ion or hydrogen ion.
![[H^+]=c\alpha=0.150\times 0.0533=0.007995 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dc%5Calpha%3D0.150%5Ctimes%200.0533%3D0.007995%20M)
Now we have to calculate the pH.
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)
pH + pOH = 14
pOH =14 -2.1 = 11.9
Therefore, the pOH of the solution is 11.9
The balanced chemical equation would be as follows:
<span>K2PtCl4(aq) + 2NH3(aq) --> Pt(NH3)2Cl2(s) + 2KCl(aq)
We are given the amount of </span>K2PtCl4 to be used in the reaction. This will be the starting point for our calculations. We do as follows:
65 g K2PtCl4 ( 1 mol / 415.09 g ) ( 1 mol Pt(NH3)2Cl2 / 1 mol K2PtCl ) ( 300.051 g / 1 mol ) = 46.99 g Pt(NH3)2Cl produced
