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nadya68 [22]
3 years ago
10

Cost of production reports can also be used to compare the materials output quantity to the a.administrative staff input quantit

y. b.materials input quantity. c.conversion output quantity. d.total conversion for the period.
Chemistry
1 answer:
valentinak56 [21]3 years ago
7 0

Answer:

b.materials input quantity

Explanation:

A cost of production report shows the following:

  • total and unit costs transferred from preceding department
  • materials, labor , factory Overhead added by the department
  • unit costs added by the department
  • total and unit costs accumulated to the end of operations
  • the cost of beginning and ending work in process inventories
  • the cost transferred to the next department

A cost of production report determines periodic total and unit costs.

Either in the cost of production report or in the supporting schedule each item of material used is listed, every labor operation is shown separately,factory overhead components are noted individually, and a unit cost is derived for each item.

A Quantity Schedule Of Cost of Production report  shows the following

  • units started in process
  • units transferred to next department
  • units still in process
  • units lost in process

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Answer:

Part 1

The mass of the NaCl that reacted with F₂ at 290.K and 1.5 atm is approximately 132.6 gams

Part 2

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Explanation:

Part 1

The volume of F₂ gas in the reaction, V = 18.0 liters

The ideal gas equation is P·V = n·R·T

∴ n = P·V/(R·T)

The pressure, P = 1.5 atm

The temperature, T = 290 K

The universal gas constant, R = 0.0820573 L·atm/(mol·K)

∴ n = 1.5×18/(0.0820573 × 290) ≈ 1.134615

The number of moles of F₂ in the reaction n ≈ 1.134615 moles

The chemical reaction is given as follows;

F₂ + 2NaCl → Cl₂ + 2NaF

1 mole of F₂ reacts with 2 moles of NaCl

Therefore;

1.134615 moles of F₂ reacted with 2 × 1.134615 moles ≈ 2.26923 moles of NaCl

1 mole of NaCl = The molar mass of NaCl, MM = 58.44 g/mol

The mass, of 2.26923 moles of NaCl, m = Number of moles × MM

∴ m ≈ 2.26923 moles × 58.44 g/mol ≈ 132.6 grams

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Part 2

The volume occupied by 1 mole of all gases at STP = 22.4 l/mole

Therefore, the number of moles of F₂ in 18.0 L of F₂ = 18.0 L/(22.4 L/mole) ≈ 0.804 moles

Therefore;

The number of moles of NaCl, in the reaction n = 2 × The number of moles of F₂ ≈ 2×0.804 moles = 1.608 moles

The number of moles of NaCl, in the reaction n ≈ 1.608 moles

The mass of NaCl in the reaction, m = n × MM

∴ m ≈ 1.608 moles × 58.44 g/mol ≈ 93.97 grams

The mass of NaCl that can react with the same volume of gas at STP ≈ 93.77 grams

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