Answer:
a. ![K_c = \dfrac{[ FeSCN^{3+}_{(aq)}] }{[Fe^{3+}_{(aq)}] [SCN^-_{(aq)}]}](https://tex.z-dn.net/?f=K_c%20%3D%20%5Cdfrac%7B%5B%20FeSCN%5E%7B3%2B%7D_%7B%28aq%29%7D%5D%20%7D%7B%5BFe%5E%7B3%2B%7D_%7B%28aq%29%7D%5D%20%5BSCN%5E-_%7B%28aq%29%7D%5D%7D)
b. ![K_p = \dfrac{[H_2]^4}{[H_2O]^4}](https://tex.z-dn.net/?f=K_p%20%3D%20%5Cdfrac%7B%5BH_2%5D%5E4%7D%7B%5BH_2O%5D%5E4%7D)
Explanation:
Untuk semua jenis reaksi umum:

Konstanta kesetimbangan ![K_c = \dfrac{[C]^c [D]^d}{[A]^a[B]^b}](https://tex.z-dn.net/?f=K_c%20%3D%20%5Cdfrac%7B%5BC%5D%5Ec%20%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
Dari pertanyaan yang diberikan:
a. 
Konstanta kesetimbangan:
![K_c = \dfrac{[ FeSCN^{3+}_{(aq)}] }{[Fe^{3+}_{(aq)}] [SCN^-_{(aq)}]}](https://tex.z-dn.net/?f=K_c%20%3D%20%5Cdfrac%7B%5B%20FeSCN%5E%7B3%2B%7D_%7B%28aq%29%7D%5D%20%7D%7B%5BFe%5E%7B3%2B%7D_%7B%28aq%29%7D%5D%20%5BSCN%5E-_%7B%28aq%29%7D%5D%7D)
b. 
Konstanta kesetimbangan untuk tekanan parsial 
![K_p = \dfrac{[H_2]^4}{[H_2O]^4}](https://tex.z-dn.net/?f=K_p%20%3D%20%5Cdfrac%7B%5BH_2%5D%5E4%7D%7B%5BH_2O%5D%5E4%7D)
Karena Fe3O4 (s) hadir sebagai padatan.
Answer: The correct answer is D. 273 Kelvin, 0 degrees Celsius, 32 degrees Fahrenheit.
Explanation:
Conversion of degree Celsius to Kelvin :
K=^oC+273
Conversion of degree Celsius to degrees Fahrenheit :
^oF=(\frac{9}{5}\times ^oC)+32
By using these two conversion factors, we get the three temperature readings all mean the same thing.
For option A :
K=^oC+273=100+273=373K
^oF=(\frac{9}{5}\times ^oC)+32=(\frac{9}{5}\times 100)+32=212^oF
For option B :
K=^oC+273=100+273=373K
^oF=(\frac{9}{5}\times ^oC)+32=(\frac{9}{5}\times 100)+32=212^oF
For option C :
K=^oC+273=0+273=273K
^oF=(\frac{9}{5}\times ^oC)+32=(\frac{9}{5}\times 0)+32=32^oF
For option D :
K=^oC+273=0+273=273K
^oF=(\frac{9}{5}\times ^oC)+32=(\frac{9}{5}\times 0)+32=32^oF
From the given options, only option (D) is correct.
Hence, the correct option is, (D) 273 Kelvin, 0 degrees Celsius, 32 degrees Fahrenheit
Hope this helps!
Answer:
2.9 grams.
Explanation:
- From the balanced reaction:
<em>Mg + 1/2O₂ → MgO,</em>
1.0 mole of Mg reacts with 0.5 mole of oxygen to produce 1.0 mole of MgO.
- We need to calculate the no. of moles of (1.8 g) of Mg and (6.0 g) of oxygen:
no. of moles of Mg = mass/molar mass = (1.8 g)/(24.3 g/mol) = 0.074 mol.
no. of moles of O₂ = mass/molar mass = (6.0 g)/(16.0 g/mol) = 0.375 mol.
<em>So. 0.074 mol of Mg reacts completely with (0.074/2 = 0.037 mol) of O₂ which be in excess.</em>
<em></em>
<em><u>Using cross multiplication:</u></em>
1.0 mole of Mg produce → 1.0 mol of MgO.
∴ 0.074 mol of Mg produce → 0.074 mol of MgO.
<em>∴ The amount of MgO produced = no. of moles x molar mass </em>= (0.074 mol)(40.3 g/mol) = <em>2.98 g.</em>
Answer:
0.2024 M
Explanation:
For the decomposition reactio given, let's do an equilibrium chart. Let's call the initial concentration of NH₃ as C:
2NH₃(g) ⇄ N₂(g) + 3H₂(g)
C 0 0 Initial
-2x +x +3x Reacts (stoichiometry is 1:1:3)
C - 2x x 3x Equilibrium
3x = 0.252
x = 0.084 M
The equilibrium constant (Kc) is the multiplication of the concentrations of the products elevated by their coefficients, divided by the multiplication of reactants concentrations elevated by their coefficients.
Kc = ([H₂]³*[N₂])/([NH₃]²)
4.50 = [(0.252)³*(0.084)]/(C - 2*0.084)²
4.50 = 0.00533/(C - 0.168)²
4.50 = 0.00533/(C² - 0.336C + 0.028224)
4.50C² - 1.512C + 0.127008 = 0.00533
4.50C² - 1.512C + 0.121678 = 0
Solving the equation by a graphic calculator, for C > 0.168
C = 0.2024 M
The answer that would best complete the given statement above is the term ATOM. <span>The smallest unit which maintains the physical properties of a compound is an ATOM. But for a compound, it would be molecule. Hope this answers your question.</span>