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marshall27 [118]
3 years ago
15

Why does hydrogen have a -1 oxidation state in the compound NaH?

Chemistry
1 answer:
ruslelena [56]3 years ago
6 0

your answer will be :

B. <u>Na has a lower</u> <u>electronegativity than H</u>

because Na belongs to alkali metals which are least electronegative (most electro positive) but hydrogen is a non metal, it has higher electronegativity as compared to metals like Sodium (Na).

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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
Am i the only one getting a red text saying my question is mean?
Triss [41]

Answer:

I dont know

Explanation:

8 0
3 years ago
In what two ways can an object possess energy?
-BARSIC- [3]

Explanation:

An object can possess energy in tow ways by it's motion or position

5 0
3 years ago
Which pH indicates a substance that is more acidic than a substance with a pH of 4
dexar [7]
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8 0
4 years ago
The ΔHcomb value for anethole is -5539 kJ/mol. Assume 0.840 g of anethole is combusted in a calorimeter whose heat capacity (Cal
bonufazy [111]

Answer:

Final temperature of calorimeter is 25.36^{0}\textrm{C}

Explanation:

Molar mass of anethole = 148.2 g/mol

So, 0.840 g of anethole = \frac{0.840}{148.2}moles of anethole = 0.00567 moles of anethole

1 mol of anethole releases 5539 kJ of heat upon combustion

So, 0.00567 moles of anethole release (5539\times 0.00567)kJ of heat or 31.41 kJ of heat

6.60 kJ of heat increases 1^{0}\textrm{C} temperature of calorimeter.

So, 31.41 kJ of heat increases (\frac{1}{6.60}\times 31.41)^{0}\textrm{C} or 4.76^{0}\textrm{C} temperature of calorimeter

So, the final temperature of calorimeter = (20.6+4.76)^{0}\textrm{C}=25.36^{0}\textrm{C}

3 0
4 years ago
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