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Ivanshal [37]
3 years ago
7

The solid XY decomposes into gaseous X and Y: XY(s) m X(g) + Y(g) Kp = 4.1 (at 0 °C) If the reaction is carried out in a 22.4 L

container, which initial amounts of X and Y will result in the formation of solid XY?a) 5 mol X; 0.5 mol Yb) 2.0 mol X; 2.0 mol Yc) 1 mol X; 1 mol Yd) none of the above

Chemistry
1 answer:
tia_tia [17]3 years ago
4 0

Answer:

None of the conditions will favor either the forward reaction or backward reaction , hence the answer is D

Explanation:

  • The principle of chemical Equilibrium is applied here, where the concentration of the reactants or the forward reaction is same as the concentration of the products or the backward reaction.

  • The equilibrium constants is also involved here, K can be in terms of pressure (Kp) or concentration (Kc) hence equilibrium constant is the ration of the concentration of the products to the concentration of the reactants raised to the power of the coefficient of the reactants and products.
  • Partial pressure , total pressure and the mole fraction relationship is also applied
  • The step by step explanation is as shown in the attachment below.

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Omg pls help i dunno what the frick frack this is
snow_lady [41]

Answer:

1. Mass of KCl produced = 774.8 g of KCl

2. Mass of KNO₃ produced = 13.837g

3. Moles of NaOH made = 0.846 moles

4. Moles of LiCl produced = 0.846 moles

5. Moles of CO₂ produced = 207.6 moles

Explanation:

1. From the equation of reaction, 1 mole of ZnCl₂ produces, 2 moles of KCl.

5.02 moles of ZnCl₂ will produce, 2 × 5.02 moles of KCl = 10.4 moles of KCl

Molar mass of KCl = (39 + 35.5) g/mol = 74.5 g/mol

10.4 moles of KCl = 10.4 × 74.5 g

Mass of KCl produced = 774.8 g of KCl

2. Mole ratio of KNO₃ and KOH = 1:1

O.137 moles of KOH will produce 0.137 moles of KNO₃

Molar mass of KNO₃ = 101 g/mol

Mass of KNO₃ produced = 0.137 × 101 g = 13.837g

3. Molar mas of Ca(OH)₂ = 74.0 g

Moles of Ca(OH)₂ in 31.3 g = 31.3/74.0 = 0.423 moles of Ca(OH)₂

Mole ratio of NaOH and Ca(OH)₂ in the reaction = 2 : 1

Moles of NaOH made = 2 × 0.423 = 0.846 moles

4. Molar mass of MgCl₂ = 95.0 g

Moles of MgCl₂ in 40.2 g = 40.2/95.0 = 0.423 moles

From the reaction equation, mole ratio of MgCl₂ and LiCl = 1:2

Moles of LiCl produced = 2 × 0.423 = 0.846 moles

5. From the equation of reaction, 1 mole of C₆H₁₀O₅ produces 6 moles of cO₂

34.6 moles of C₆H₁₀O₅ will produce 34.6 × 6 moles of CO₂

Moles of CO₂ produced = 207.6 moles

4 0
3 years ago
1. The sodium-iodide symporter plays a role in the accumulation of iodide in the thyroid gland. Here, one iodide gets converted
Serggg [28]

Answer:(1) 6 sodium ions

(2)The movement of iodide ions occurs in the same direction as the sodium ions.

Explanation: Sodium-iodide symporter actively transports 2 sodium ions together with one iodide ions across the basement membrane into the thyroid follicular cells.(therefore for production of a single molecule of T3 3×2=6 sodium ions)This system utilises the concentration of sodium ions so that iodide ions can move against its concentration gradient.

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3 years ago
Calculate the number of moles of iron (III) oxide (Fe 2 O 3 ) produced from 112 L of oxygen
RideAnS [48]
112 L/ 16 L = 7 mol O2

7 mol O2 x 2 mol Fe2O3/ 3 mol O2 = 4.67 mol Fe2O3

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False... Honey is greater in velosity then water
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