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Aleksandr [31]
2 years ago
5

Identify the compound with the smallest dipole moment in the gas phase. identify the compound with the smallest dipole moment in

the gas phase. lif cl2 clf hf
Chemistry
1 answer:
stepladder [879]2 years ago
3 0
Without solving for the dipole moment, we can easily determine which among the common gases has the smallest dipole moment just by determining the differences in their electronegativity. The greater the difference in the electronegativity, the higher is the value of the dipole moment. 

From the given above, there are obvious differences between the electronegativity between the atoms composing LiF, ClF, and HF. For Cl2, since this is the same molecule then, the difference in the electronegativity is zero.

Answer: Cl2. 
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coefficient: they balance the chemical equation you have to make sure the number is as small as it can. It is also used to convert different compounds to compounds or quantities to quantities.

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3 years ago
The standard unit for concentration, Molarity (M) can also be written as?
emmasim [6.3K]

mole is the standardized form of molarity

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Are your knees the elbows of your legs or are your elbows the knees of your
Nimfa-mama [501]

Answer:

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28 grams of Al react with 30 Oxygen gas (O2) to produce Al Oxide (Al2O3). How many grams of Al Oxide will be produced
Kamila [148]

Answer:

58

Explanation:

Due to conservation of energy, the sum of the reactants = the sum of the products. So 28 + 30 = 58g

8 0
2 years ago
Realice las siguientes conversiones: a) 72°F a °C, b) 213.8°C a °F, c)180°C a K, d) 315K a °F, e) 1750°F a K, f) 0K a °F.
bonufazy [111]

Answer:

a) 72 °F= 22.22 °C

b)  213.8 °C=  416.84°F

c) 180 °C= 453.15 °K

d) 315 °K=  107.33 °F

e) 1750 °F= 1227.594 °K

f) 0 °K=  -459.67 °F

Explanation:

Para realizar el intercambio de unidades debes tener en cuenta las siguientes conversiones:

  • Fahrenheit a Celsius: C=\frac{F-32}{1.8}
  • Celsius a Fahrenheit: °F= °C*1.8 + 32
  • Celsius a Kelvin: °K= °C + 273.15
  • Kelvin a Fahrenheit: F= (K -273.15)*1.8 + 32
  • Fahrenheit a Kelvin:K=\frac{F-32}{1.8} + 273.15

Entonces se obtiene:

a) 72 °F= \frac{72-32}{1.8}=22.22 °C

b)  213.8 °C= 213.8*1.8 + 32= 416.84°F

c) 180 °C= 180°C + 273.15= 453.15 °K

d) 315 °K= (315 -273.15)*1.8 + 32= 107.33 °F

e) 1750 °F= \frac{1750-32}{1.8} + 273.15= 1227.594 °K

f) 0 °K= (0 -273.15)*1.8 + 32= -459.67 °F

7 0
2 years ago
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