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notka56 [123]
3 years ago
14

A(n) _____ is a natural object that orbits another object in the solar system besides the Sun.

Chemistry
1 answer:
Stolb23 [73]3 years ago
4 0
The answer is the moon

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14.28 explain how you would distinguish between each pair of compounds using high- resolution mass spectrometry
vagabundo [1.1K]

The distinguish between each pair of compounds using high- resolution mass spectrometry by the exact mass rather than nominal mass are utilizes to measure the compound.

The mass spectrometry is involves the following steps :

  • The ionization
  • acceleration
  • deflection
  • detection

Mass spectrometry is the analytical method useful for the calculating the mass to charge ratio ( m / z ). the mass spectrometry is based on the newton's second law and the momentum.

Thus, the mass spectroscopy is method to measure the molecular mass of the compound and indirectly helps examine the isotopes and based on the newton's second law .

To learn more about mass spectroscopy here

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7 0
1 year ago
Draw the orbital notation for the element Boron
Wewaii [24]

Answer:

I can't draw but you could draw 2 electrons in the first orbit and 3 electrons in the second orbit.

Explanation:

6 0
2 years ago
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What is the definition of a Lewis acid?
Sveta_85 [38]

Answer:

A Lewis acid is a chemical species that contains an empty orbital which is capable of accepting an electron pair from a Lewis base to form a Lewis adduct

Explanation:

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8 0
3 years ago
Explain why is sodium less reactive than potassium
leva [86]

Answer:

As potassium is larger than sodium, potassium's valence electron is at a greater distance from the attractive nucleus and is so removed more easily than sodium's valence electron. As it is removed more easily, it requires less energy, and can be said to be more reactive.

Explanation:

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8 0
3 years ago
A city continuously disposes of effluent from a wastewater treatment plant into a river. The minimum flow in the river is 130 m3
Vera_Pavlovna [14]

Answer:

2.54\ \text{mg/L}

Explanation:

C = Allowable concentration = 1.1 mg/L

Q_1 = Flow rate of river = 130\ \text{m}^/\text{s}

Q_2 = Discharge from plant = 37\ \text{m}^3/\text{s}

C_1 = Background concentration = 0.69 mg/L

C_2 = Maximum concentration that of the pollutant

The concentration of the mixture will be

C=\dfrac{Q_1C_1+Q_2C_2}{Q_1+Q_2}\\\Rightarrow C_2=\dfrac{C(Q_1+Q_2)-Q_1C_1}{Q_2}\\\Rightarrow C_2=\dfrac{1.1(130+37)-130\times 0.69}{37}\\\Rightarrow C_2=2.54\ \text{mg/L}

The maximum concentration that of the pollutant (in mg/L) that can be safely discharged from the wastewater treatment plant is 2.54\ \text{mg/L}.

6 0
2 years ago
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