N<span>et ionic equation for ammonia and phosphoric acid</span> is
3 NH4OH + H3PO4 >> (NH4)3PO4 + 3 H2O
hope this helps
Answer:
[Cl2] equilibrium = 0.0089 M
Explanation:
<u>Given:</u>
[SbCl5] = 0 M
[SbCl3] = [Cl2] = 0.0546 M
Kc = 1.7*10^-3
<u>To determine:</u>
The equilibrium concentration of Cl2
<u>Calculation:</u>
Set-up an ICE table for the given reaction:
I 0 0.0546 0.0546
C +x -x -x
E x (0.0546-x) (0.0546-x)
![Kc = \frac{[SbCl3][Cl2]}{[SbCl5]}\\\\1.7*10^{-3} =\frac{(0.0546-x)^{2} }{x} \\\\x = 0.0457 M](https://tex.z-dn.net/?f=Kc%20%3D%20%5Cfrac%7B%5BSbCl3%5D%5BCl2%5D%7D%7B%5BSbCl5%5D%7D%5C%5C%5C%5C1.7%2A10%5E%7B-3%7D%20%3D%5Cfrac%7B%280.0546-x%29%5E%7B2%7D%20%7D%7Bx%7D%20%5C%5C%5C%5Cx%20%3D%200.0457%20M)
The equilibrium concentration of Cl2 is:
= 0.0546-x = 0.0546-0.0457 = 0.0089 M
Ammonium is NH₄⁺ and Carbonate is CO₃⁻² => Ammonium Carbonate is (NH₄)₂CO₃
<span>1.75 mol H2O x (6.022x10^23 molecules H2O / 1 mol H2O) = 1.05x10^24 molecules H2O if you need a further example let me know </span>
The sign of the entropy change, ΔS, for the following processes include:
- I2 (s) + ½ Cl2 (g) ← → ICl (g) - positive
- 3 Ag (s) + 4 HNO3 (aq) → 3AgNO3 (aq) + NO (g) + 2 H2O (l) - positive
- Cl2 (g) → Cl2 (l) - negative
- C5H12 (g) + 8 O2 (g) → 5 CO2 (g) + 6 H2O (l) - negative
What is Entropy?
This is referred to the degree of randomness or disorderliness of a system and is denoted as S.
Entropy change is usually positive when solid or liquid reactant is converted into gas product and vice versa.
Read more about Entropy here brainly.com/question/419265
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