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muminat
3 years ago
6

Carbon monoxide is a gas at 0 °c and a pressure of 1.58 × 105 pa. it is a diatomic gas, each of its molecules consisting of one

carbon atom (atomic mass = 12.0 u) and one oxygen atom (atomic mass = 16.0 u). assuming that carbon monoxide is an ideal gas, calculate its density ρ.
Chemistry
1 answer:
Hoochie [10]3 years ago
5 0

Given temperature = 0^{0}C = (0^{0} + 273 )K  = 273 K

Pressure = 1.58 ×10^{5} Pa ×\frac{1 atm}{1.01325*10^{5}Pa}

= 1.56 atm

Density =\frac{PM}{RT}

ρ = \frac{(1.56 atm)(28.01 g/mol)}{(0.08206 L.atm/mol.K)(273 K)}

= 1.95 g/L

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Answer:

no.

Density (mass / volume) determines whether an object floats or sinks. If the object is less dense than the medium in which it has been submerged, it floats. if it is more dense, it sinks. Volume will definitely determine if a steel ship floats, as steel is far more dense than water.

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3 years ago
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2C 2 H 6 +7O 2 ***>4CO 2 +6H 2 O if 7.0 g of C 2 H 6 react with 18 g of O 2 , how many grams of water will be produced
Alex787 [66]

Answer:

grams H₂O produced = 8.7 grams

Explanation:

Given 2C₂H₆(g) + 7O₂(g) => 4CO₂(g) + 6H₂O(l)

               7g           18g                             ?g

Plan => Convert gms to moles => determine Limiting reactant => solve for moles water => convert moles water to grams water

Moles Reactants

moles C₂H₆ = 7g/30g/mol = 0.233mol

moles O₂ = 18g/32g/mol = 0.563mol

Limiting Reactant => (Test for Limiting Reactant)  Divide mole value by respective coefficient of balanced equation; the smaller number is the limiting reactant.

moles C₂H₆/2 = 0.233/2 = 0.12

moles O₂/7 = 0.08

<u><em>Limiting Reactant is O₂</em></u>

Moles and Grams of H₂O:

Use Limiting Reactant moles (not division value) to calculate moles of H₂O.

moles H₂O = 6/7(moles O₂) = 6/7(0.562) moles H₂O = 0.482 mole H₂O yield

grams H₂O = (0.482mol)(18g·mol⁻¹) = 8.7 grams H₂O

3 0
3 years ago
Convert 19 kcal into joules
il63 [147K]

19~ \text{kcal} = 19 \times 1000 ~\text{cal } = 19000~ \text{cal} = 19000 \times 4.184~ \text{J} = 79496~ \text{J}

7 0
2 years ago
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