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jekas [21]
3 years ago
11

6PbO + O2 → 2Pb304 a. How many grams of Pb304 are produced from 8.50 grams of lead(II) oxide?

Chemistry
1 answer:
AlladinOne [14]3 years ago
7 0

Answer:

8.70g of Pb3O4

Explanation:

6PbO + O2 → 2Pb3O4

Molar Mass of PbO = 207 + 16 = 223g/mol

The mass of the PbO from the balanced equation = 6 x 223 = 1338g

Molar Mass of Pb3O4 = (3x207) +(16x4) = 621 + 64 = 685g/mol

Mass of Pb3O4 from the balanced equation = 2 x 685 = 1370g

From the equation,

1338g oh PbO produced 1370g of Pb3O4.

Therefore, 8.50g of PbO will produce = (8.5 x 1370)/1338 = 8.70g of Pb3O4

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Answer:

1400KJ/mol⁻¹

Explanation:

Amount of heat required can be found by:

Q = m × c × ΔT

<em>Where m is the mass, c is the specific heat capacity (4.2KJ for water) and ΔT is the change in temperature.</em>

Q = 24 × 4.2 × (23 - 9)

= 24 × 4.2 × 14

=   1411.2KJ/mol⁻¹

= <u>1400KJ/mol⁻¹</u>  (to 2 significant figures)

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Explanation:

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How many of each element are present in this compound 4C4HgCL​
frozen [14]

Answer:

Percent Composition of Compounds

The percent composition (by mass) of a compound can be calculated by dividing the mass of each element by the total mass of the compound.

LEARNING OBJECTIVES

Translate between a molecular formula of a compound and its percent composition by mass

  • The atomic composition of chemical compounds can be described in a variety of ways, including molecular formulas and percent composition.
  • The percent composition of a compound is calculated with the molecular formula: divide the mass of each element found in one mole of the compound by the total molar mass of the compound.
  • The percent composition of a compound can be measured experimentally, and these values can be used to determine the empirical formula of a compound.

  • percent by mass: The fraction, by weight, of one element of a compound.
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Explanation:

I hope it's help

5 0
2 years ago
Three gases (8.00 g of methane, ch4, 18.0 g of ethane, c2h6, and an unknown amount of propane, c3h8) were added to the same 10.0
Len [333]
Assume ideal gas behavior, then solve for the total number of moles:

PV = nRT
(5.50 atm)(10 L) = n(0.0821 L-atm/mol-K)(23+273 K)
n = 2.263 mol

Moles methane: 8 g ÷ 16.04 g/mol = 0.499 mol
Moles ethane: 18 g ÷ 30.07 g/mol = 0.599 mol
Moles propane: 2.263 - (0.499+0.599) = 1.165 mol

Applying Raoult's Law:

Partial pressure = Mole fraction * Total Pressure

<em>Partial Pressure of Methane = (0.499/2.263)(5.5 atm) = 1.21 atm</em>
<em>Partial Pressure of Ethane = (0.599/2.263)(5.5 atm) = 1.46 atm</em>
<em>Partial Pressure of Propane = (1.165/2.263)(5.5 atm) = 2.83 atm</em>
4 0
2 years ago
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