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:
b. 
Konstanta kesetimbangan untuk tekanan parsial 
Karena Fe3O4 (s) hadir sebagai padatan.
Answer:
genetic engineering !! :)
Explanation:
also, i love the pfp <3
Answer:
1.33 L.
Explanation:
- We can use the general law of ideal gas: PV = nRT.
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n and T are constant, and have different values of P and V:
<em>(P₁V₁) = (P₂V₂)</em>
<em></em>
Knowing that:
V₁ = 4.0 L, P₁ = 2.0 atm,
V₂ = ??? L, P₂ = 6.0 atm.
- Applying in the above equation
(P ₁V₁) = (P₂V₂)
<em>∴ V₂ = P ₁V₁/P₂</em> = (2.0 atm)(4.0 L)/(6.0 atm) =<em> 1.33 L.</em>
LMBO, for science.
Answer:
<em>The increase in kinetic energy leads to leakage of water from the syringe. When the outside temperature is more than the liquid temperature, say the syringe is out in sunshine, then the liquid becomes slightly warmer.</em>
Answer:
CH₂ ; 67.1 %
Explanation:
To determine the empirical formula we need to find what the mole ratio is in whole numbers of the atoms in the compound. To do that we will first need the atomic weights of C and H and then perform our calculation
Assume 100 grams of the compound.
# mol C = 85.7 g / 12.01 g/mol = 7.14 mol
# mol H = 14.3 g / 1.008 g/mol = 14.19 mol
The proportion is 14.9 mol H/ 7.14 mol C = 2 mol H/ 1 mol C
So the empirical formula is CH₂
For the second part we will need to first calculate the theoretical yield for the 12.03 g NaBH₄ reacted and then calculate the percent yield given the 0.295 g B₂H₆ produced.
We need to calculate the moles of NaBH₄ ( M.W = 37.83 g/mol )
1.203 g NaBH₄ / 37.83 g/mol = 0.0318 mol
Theoretical yield from balanced chemical equation:
0.0318 mol NaBH₄ x 1 mol B₂H₆ / mol NaBH₄ = 0.0159 mol B₂H₆
Theoretical mass yield B₂H₆ = 0.0159 mol x 27.66 g/ mol = 0.440 g
% yield = 0.295 g/ 0.440 g x 100 = 67.1 %
