Answer:
a) 
b) 1657 €
Explanation:
Hola,
a) En esta cuestión analizaremos el millón de litros de agua anualmente, dado que este dato nos permite calcular el calor necesario para calentar dicha cantidad, considerando que la densidad del agua es de 1 kg/L:
A continuación, utilizamos la entalpía de combustión del metano para determinar la cantidad en kilogramos necesaria, ya que la energía calórica perdida por el metano es equivalente a la energía obtenida por el agua:
b) En este supuesto, tenemos que, bajo condiciones normales de 1 bar y 273 K, el precio de 1 metro cúbico de metano es 0,45 €, lo que nos permite calcular las moles de metano en esas condiciones:
En consecuencia, los kilogramos de metano que se obtienen por 0,45 € son:
Finalmente, usando regla de tres:
0.715 kg ⇒ 0.45 €
2630 kg ⇒ X
X = (2630 kg x 0.45 €) / 0.715 kg
X = 1657 €
Regards.
1) The ionic compound present in solution b is K₂CrO₄ (potassium chromate). This compound contains two potassiums (oxidation state +1), a single chromium (oxidation state +6), and four oxygen atoms. The oxidation state of oxygen is -2, resulting in a neutral compound: 2 · (+1) + 6 + x · (-2) = 0. Hence, x = 4, denoting the count of oxygen atoms. 2) The ionic compound in solution a is AgNO₃ (silver nitrate). ω(N) = 8.246% ÷ 100%. Thus, ω(N) = 0.08246, indicating the mass percentage of nitrogen. M(MNO₃) = M(N) ÷ ω(N). It follows that M(MNO₃) = 14 g/mol ÷ 0.08246, leading to M(MNO₃) = 169.8 g/mol; the molar mass of the metal nitrate. M(M) = M(MNO₃) - M(N) - 3 · M(O). Consequently, M(M) = 169.8 g/mol - 14 g/mol - 3 · 16 g/mol, resulting in M(M) = 107.8 g/mol which is the atomic mass of silver (Ag). 3) The balanced chemical equation is: 2AgNO₃(aq) + K₂CrO₄(aq) → Ag₂CrO₄(s) + 2KNO₃(aq). In ionic form: 2Ag⁺(aq) + 2NO₃⁻(aq) + 2K⁺(aq) + CrO₄²⁻(aq) → Ag₂CrO₄(s) + 2K⁺(aq) + 2NO₃⁻(aq). The net ionic equation is: 2Ag⁺(aq) + CrO₄²⁻(aq) → Ag₂CrO₄(s). Thus, the red precipitate is identified as silver chromate (Ag₂CrO₄). 4) The mass of solid silver chromate created is m(Ag₂CrO₄) = 331.8 g. The amount is determined by n(Ag₂CrO₄) = m(Ag₂CrO₄) ÷ M(Ag₂CrO₄). Therefore, n(Ag₂CrO₄) = 331.8 g ÷ 331.8 g/mol yields n(Ag₂CrO₄) = 1 mol. From the balanced equation, n(Ag₂CrO₄): n(AgNO₃) = 1: 2, it follows n(AgNO₃) = 2 · 1 mol, which means n(AgNO₃) = 2 mol. Then, the mass of silver nitrate is computed as m(AgNO₃) = n(AgNO₃) · M(AgNO₃). Hence, m(AgNO₃) = 2 mol · 169.8 g/mol gives m(AgNO₃) = 339.6 g; thus, m(AgNO₄) equals m(K₂CrO₄). Therefore, m(K₂CrO₄) = 339.6 g; amount of potassium chromate is n(K₂CrO₄) = m(K₂CrO₄) ÷ M(K₂CrO₄). Thus, n(K₂CrO₄) = 339.6 g ÷ 194.2 g/mol thus arrives at n(K₂CrO₄) = 1.75 mol. 5) The dissociation of silver nitrate in water is expressed as: AgNO₃(aq) → Ag⁺(aq) + NO₃⁻(aq). Volume of solution a = 500 mL ÷ 1000 mL/L results in V(solution a) = 0.5 L. Concentration equation c(AgNO₃) = n(AgNO₃) ÷ V(solution a), thus c(AgNO₃) = 2 mol ÷ 0.5 L, yielding c(AgNO₃) = 4 mol/L = 4 M. As a result: c(AgNO₃) = c(Ag⁺) = c(NO₃⁻). Thus, c(Ag⁺) = 4 M; the concentration of silver ions in the initial solution a. 6) The dissociation of potassium chromate in water is represented as: K₂CrO₄(aq) → 2K⁺(aq) + CrO₄²⁻(aq). Volume of solution b = 500 mL ÷ 1000 mL/L results in V(solution b) = 0.5 L. Following, c(K₂CrO₄) is calculated as n(K₂CrO₄) ÷ V(solution b). So c(AgNO₃) = 1.75 mol ÷ 0.5 L gives c(AgNO₃) = 3.5 mol/L = 3.5 M. Consequently: c(K⁺) = 7 M; the concentration of potassium ions in solution b. Therefore, c(CrO₄²⁻) = 3.5 M; the concentration of chromium ions in the same solution. 7) The total final volume is V(final solution) = V(solution a) + V(solution b). Thus, V(final solution) = 500.0 mL + 500.0 mL leads to V(final solution) = 1000 mL ÷ 1000 mL/L results in V(final solution) = 1 L. Then n(NO₃⁻) = 2 mol. Therefore, c(NO₃⁻) = n(NO₃⁻) ÷ V(final solution) finds c(NO₃⁻) = 2 mol ÷ 1 L and results in c(NO₃⁻) = 2 M; the concentration of nitrate anions in the final solution. 8) In solution b, there are 3.5 mol of potassium cations while part of that combines with 2 moles of nitrate anions: K⁺(aq) + NO₃⁻(aq) → KNO₃(aq). From the reaction: n(K⁺): n(NO₃⁻) = 1: 1. Thus, Δn(K⁺) = 3.5 mol - 2 mol results in Δn(K⁺) = 1.5 mol, signifying the remaining potassium anions in the final solution. Thus, c(K⁺) = Δn(K⁺) ÷ V(final solution) yields c(K⁺) = 1.5 mol ÷ 1 L, leading to c(K⁺) = 1.5 M; the final concentration of potassium cations.
Answer:
a. H₂O (conjugate acid); b. OH⁻ (conjugate base), H₃O⁺ (conjugate acid); c. H₂CO₃ (conjugate acid), CO₃⁻² (conjugate base); d. NH₄⁺ (conjugate strong acid) e. H₂SO₄ (conjugate acid), SO₄⁻² (conjugate base); f. No conjugate acid or base exists; g. H₂S (conjugate acid), S⁻² (conjugate base);
h. H₄N₂ (conjugate base)
Explanation:
a. OH⁻ + H⁺ ⇄ H₂O
The hydroxide functions as a Bronsted-Lowry base, allowing it to capture a proton, thus water serves as the conjugate acid.
b. H₂O is amphoteric, capable of acting as either an acid or a base. As a base, its conjugate acid is H₃O⁺, whereas as an acid, its conjugate base is OH⁻.
c. HCO₃⁻ + H⁺ ⇄ H₂CO₃
HCO₃⁻ + H₂O ⇄ CO₃⁻² + H₃O⁺
Bicarbonate is also amphoteric. When it captures a proton, it forms carbonic acid as the conjugate acid when acting as a base. When HCO₃⁻ acts as an acid and releases a proton, carbonate becomes the conjugate base.
d. Ammonia functions as a weak base, with ammonium being the conjugate strong acid.
NH₃ + H₂O ⇄ NH₄⁺ + OH⁻
e. Another amphoteric compound. Acid sulfate can function as both an acid and a base.
(similar to bicarbonate). Acting as a base yields sulfuric acid as the conjugate acid, while acting as an acid leads to sulfate as the conjugate base.
HSO₄⁻ + H₂O ⇄ SO₄⁻² + H₃O⁺
HSO₄⁻ + H⁺ ⇄ H₂SO₄
f. H₂O₂ does not accept H⁺ or OH⁻ nor does it expel H⁺. It’s neutral and does not function as an acid or base.
g. HS⁻ is amphoteric.
HS⁻ + H⁺ ⇄ H₂S
HS⁻ + H₂O ⇄ S⁻² + H₃O⁺
This is similar to the case of bicarbonate or acid sulfate.
h. H₅N₂⁺ + H₂O ⇄ H₄N₂ + H₃O⁺
Hydrazinium acts as an acid, making hydrazine its conjugate base.
84.34 grams of iron (III) chloride is the maximum produced since iron is the limiting reagent, and chlorine gas is in excess.
Explanation:
Balanced equation:
2 Fe + 3 Cl2 → 2 FeCl3
DATA PROVIDED:
iron = atoms
mass of chlorine = 67.2 liters
mass of FeCl3 =?
The number of moles of iron will be calculated as
number of moles = 
number of moles = 
number of moles = 0.52 mol of iron
moles of chlorine gas
number of moles = 
Substituting the values into the equation:
n = 
(molar mass of chlorine gas = 70.96 g/mol)
= 947.01 moles
As iron is the limiting reagent therefore
2 moles of Fe lead to 2 moles of FeCl3
0.52 moles of Fe will yield
= 
0.52 moles of FeCl3 is produced.
To express this in grams:
mass = n x molar mass
= 0.52 x 162.2 (molar mass of FeCl3 is 162.2g/mol)
= 84.34 grams
