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True [87]
4 years ago
8

Phosphorus trichloride, PCl₃ is a liquid at room temperature.

Chemistry
1 answer:
Simora [160]4 years ago
4 0

Answer:

Phosphorus trichloride, PCl₃ undergoes change in bonding and molecular force of attraction, causing it to be liquid at room temperature.

Explanation:

Unlike other chlorides of Period 3 elements, Phosphorus trichloride, PCl₃ changes the structure of its molecular bonding from ionic to covalent bonds as it transitions to fluids (liquids or gases). The PCl₃ molecule also has the weak Van der Waals dispersion and dipole-dipole attraction, making it a fuming liquid at room temperature,  with no electrical conductivity.

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Determine the AMOUNT OF NO2, LIMITING REACTANT, AND THE AMOUNT AND NAME OF EXCESS REACTANT.
zepelin [54]

Answer:

balanced equation mole ratio 5 2 mol NO/1 mol O2

10.00 g O2 3 1 mol O2/32.00 g O2 5 0.3125 mol O2

20.00 g NO 3 1 mol NO/30.01 g NO 5 0.6664 mol NO

actual mole ratio 5 0.6664 mol NO/0.3125 mol O2 5 2.132 mol NO/1.000 mol O2

Because the actual mole ratio of NO:O2 is larger than the balanced equation mole

ratio of NO:O2, there is an excess of NO; O2 is the limiting reactant.

Mass of NO used 5 0.3125 mol O2 3 2 mol NO/1 mol O2 5 0.6250 mol NO

0.6250 mol NO 3 30.01 g NO/1 mol NO 5 18.76 g NO

Mass of NO2 produced 5 0.6250 mol NO2 3 46.01 g NO2/1 mol NO2 5 28.76 g NO2

Excess NO 5 20.00 g NO 2 18.76 g NO 5 1.24 g N

Explanation:

3 0
3 years ago
Describe the overall enthalpy of the chemical reactants compared to the enthalpy of the chemical products in the combustion of p
Veronika [31]

Answer:

A thermochemical equation for the combustion of propane (C3H8)(C3H8) is written as follows:

C3H8(l)+5O2(g)→3CO2(g)+4H2O(g);ΔH∘rxnC3H8(l)+5O2(g)→3CO2(g)+4H2O(g);ΔHrxn∘ = -2202.0 kJ/mol

The value given for ΔH∘rxnΔHrxn∘ means that:

a. the reaction of one mole of propane absorbs 2202 kJ of energy from the surroundings.

b. the reaction is endothermic.

c. the enthalpy of formation of propane is 2202 kJ/mol.

d. the reaction of one mole of propane releases 2202 kJ of energy to the surroundings.

e. None of these.

3 0
1 year ago
A foce that pushes or pulls is known as​
Radda [10]

Answer:

Newton 3rd Law of Motion or the Law of Force Pairs

(An applied force)

3 0
3 years ago
Which is a component of John Dalton’s atomic theory?
Dmitry [639]
Atoms of different elements can be identical
7 0
2 years ago
A mixture of noble gases [helium (MW 4), argon (MW 40), krypton (MW 83.8), and xenon (MW 131.3)] is at a total pressure of 150 k
kodGreya [7K]

Answer:

a) 1,6%

b) 64,775 g/mol

c) 3,6×10⁻² M

d) 2,3×10⁻³ g/mL

Explanation:

a) The mass fractium of helium is obtained converting the moles of the four gases to grams with molar weight and then caculating of the total of grams how many are of helium, thus:

  • Helium: 0,25 moles ×\frac{4 g}{1 mol} = 1 g of Helium
  • Argon: 0,25 moles ×\frac{40 g}{1 mol} = 10 g of Argon
  • Krypton: 0,25 moles ×\frac{83,8 g}{1 mol} = 20,95 g of krypton
  • Xenon: 0,25 moles ×\frac{131,3 g}{1 mol} = 32,825 g of Xenon

Total grams: 1g+10g+20,85g+30,825g= 62,675 g

Mass fraction of helium: \frac{1 gHelium}{62,675 g} × 100 = <em>1,6%</em>

<em />

<em>The mass fraction of Helium is 1,6%</em>

<em />

<em>b)</em><em>  </em>Because the mole fraction of all gases is the same the average molecular weight of the mixture is:

\frac{4+40+83,8+131,3}{4} = 64,775 g/mol

c) The molar concentration is possible to know ussing ideal gas law, thus:

\frac{P}{R.T} = M

Where:

P is pressure: 150 kPa

R is gas constant: 8,3145\frac{L.kPa}{K.mol}

T is temperature: 500 K

And M is molar concentration. Replacing:

M = 3,6×10⁻² M

d) The mass density is possible to know converting the moles of molarity to grams with average molecular weight and liters to mililiters, thus:

3,6×10⁻² \frac{mol}{L} × \frac{64,775 g}{mol} × \frac{1L}{1000 mL} =

2,3×10⁻³ g/mL

I hope it helps!

7 0
3 years ago
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