<span> If mass remains the same while the volume of a substance increases and the density of the substance will decrease or, volume decreases and density increase, 'cause density equals mass divided in volume, which means that volume and density are inversely proporcional</span>
Answer:
I would help but the picture will not load for me :(
Explanation:
The question is incomplete, here is the complete question:
Write a balanced chemical equation for each single replacement reaction that takes place in aqueous solution. write no reaction if a reaction does not occur
1.) Zn + PbCl₂
2.) Cu + Fe(NO₃)₂
<u>Answer:</u>
<u>For 1:</u> The reaction does occur.
<u>For 2:</u> The reaction does not occur.
<u>Explanation:</u>
Single displacement reaction is defined as the reaction in which more reactive element displaces a less reactive element.
The reactivity of metal is determined by a series known as reactivity series. The metals lying above in the series are more reactive than the metals which lie below in the series.
![A+BC\rightarrow AC+B](https://tex.z-dn.net/?f=A%2BBC%5Crightarrow%20AC%2BB)
For the given options:
Zinc is more reactive than lead as it lies above in the series. So, it will displace lead from its chemical equation.
The chemical equation for the reaction of zinc and lead chloride follows:
![Zn+PbCl_2\rightarrow ZnCl_2+Pb](https://tex.z-dn.net/?f=Zn%2BPbCl_2%5Crightarrow%20ZnCl_2%2BPb)
Copper is less reactive than iron as it lies below in the series. So, it will not displace iron from its chemical equation.
The chemical equation for the reaction of copper and iron (II) nitrate follows:
![Cu+Fe(NO_3)_2\rightarrow \text{No reaction}](https://tex.z-dn.net/?f=Cu%2BFe%28NO_3%29_2%5Crightarrow%20%5Ctext%7BNo%20reaction%7D)
Answer:
1.1 M
Explanation:
The dissociation of
is as follows:
![AgCl \leftrightharpoons Ag^+_{(aq)}+Cl^-_{(aq)}](https://tex.z-dn.net/?f=AgCl%20%5Cleftrightharpoons%20Ag%5E%2B_%7B%28aq%29%7D%2BCl%5E-_%7B%28aq%29%7D)
Given Value for ![K_{sp} = 1.77*10^{-10}](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%201.77%2A10%5E%7B-10%7D)
The equation for the reaction for the formation of complex ion
is :
![Ag^+_{(aq)} + 2NH_3_{(aq)} \rightleftharpoons Ag(NH_3)_2^+_{(aq)}](https://tex.z-dn.net/?f=Ag%5E%2B_%7B%28aq%29%7D%20%20%20%20%2B%20%20%20%20%202NH_3_%7B%28aq%29%7D%20%5Crightleftharpoons%20Ag%28NH_3%29_2%5E%2B_%7B%28aq%29%7D)
The value of ![K_f = 1.6*10^7](https://tex.z-dn.net/?f=K_f%20%3D%201.6%2A10%5E7)
If we combine both equation and find the overall equilibrium constant will be:
![AgCl \leftrightharpoons Ag^+_{(aq)}+Cl^-_{(aq)}](https://tex.z-dn.net/?f=AgCl%20%5Cleftrightharpoons%20Ag%5E%2B_%7B%28aq%29%7D%2BCl%5E-_%7B%28aq%29%7D)
![Ag^+_{(aq)} + 2NH_3_{(aq)} \rightleftharpoons Ag(NH_3)_2^+_{(aq)}](https://tex.z-dn.net/?f=Ag%5E%2B_%7B%28aq%29%7D%20%20%20%20%2B%20%20%20%20%202NH_3_%7B%28aq%29%7D%20%5Crightleftharpoons%20Ag%28NH_3%29_2%5E%2B_%7B%28aq%29%7D)
<u> </u>
![AgCl_{(s)}+2NH_3_{(aq)} \rightleftharpoons Ag(NH_3)_2^+_{(aq)} + Cl^-_{(aq)}](https://tex.z-dn.net/?f=AgCl_%7B%28s%29%7D%2B2NH_3_%7B%28aq%29%7D%20%5Crightleftharpoons%20Ag%28NH_3%29_2%5E%2B_%7B%28aq%29%7D%20%2B%20Cl%5E-_%7B%28aq%29%7D)
![K = (1.77*10^{-10})(1.6*10^7)](https://tex.z-dn.net/?f=K%20%3D%20%281.77%2A10%5E%7B-10%7D%29%281.6%2A10%5E7%29)
![K = 0.00283](https://tex.z-dn.net/?f=K%20%3D%200.00283)
If
= x M
The solubility of
in the
solution will be:
![x = 743*10^{-3}g * \frac{mol AgCl}{143.32g}*\frac{1}{0.1000L}](https://tex.z-dn.net/?f=x%20%3D%20743%2A10%5E%7B-3%7Dg%20%2A%20%5Cfrac%7Bmol%20AgCl%7D%7B143.32g%7D%2A%5Cfrac%7B1%7D%7B0.1000L%7D)
![x = 0.0518 M](https://tex.z-dn.net/?f=x%20%3D%200.0518%20M)
Constructing an ICE Table; we have :
![AgCl_{(s)} + 2 NH_3{(aq)} \rightleftharpoons Ag(NH_3)_2^+_{(aq)} + Cl^-_{(aq)}](https://tex.z-dn.net/?f=AgCl_%7B%28s%29%7D%20%2B%202%20NH_3%7B%28aq%29%7D%20%5Crightleftharpoons%20Ag%28NH_3%29_2%5E%2B_%7B%28aq%29%7D%20%2B%20Cl%5E-_%7B%28aq%29%7D)
Initial (M) x 0 0
Change (M) -2 (0.0518) + 0.0518 + 0.0518
Equilibrium (M) x - 0.1156 0.0518 0.0518
Equilibrium constant;
(K) = ![\frac{[Ag(NH_3)_2^+][Cl^-]}{[NH_3]^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BAg%28NH_3%29_2%5E%2B%5D%5BCl%5E-%5D%7D%7B%5BNH_3%5D%5E2%7D)
![0.00283 = \frac{(0.0518)^2}{(x-0.1156)^2}](https://tex.z-dn.net/?f=0.00283%20%3D%20%5Cfrac%7B%280.0518%29%5E2%7D%7B%28x-0.1156%29%5E2%7D)
![0.00283 = (\frac{0.0518}{x-0.1156})^2](https://tex.z-dn.net/?f=0.00283%20%3D%20%28%5Cfrac%7B0.0518%7D%7Bx-0.1156%7D%29%5E2)
![x = 0.1156 + \sqrt{\frac{0.0518^2}{0.00283} }](https://tex.z-dn.net/?f=x%20%3D%200.1156%20%2B%20%5Csqrt%7B%5Cfrac%7B0.0518%5E2%7D%7B0.00283%7D%20%7D)
x = [NH₃] = 1.089 M
[NH₃] ≅ 1.1 M
Answer:
C 41.2
Explanation:
I took the test ez
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