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Liula [17]
3 years ago
15

determine the lone pairs and electron and electron densities in C in CH4, O in CH3OH, F in PH2F, N in NOH

Chemistry
1 answer:
Crazy boy [7]3 years ago
6 0
In CH4, C has no lone pairs, the electron density is zero since it is equally distributed. 
in CH3OH, O has 2 lone pairs.
In PH2F, F has 3 lone pairs
In NOH, N has 1 lone pair, in this case, oxygen has more lone pairs than N as such the electron density of oxygen is higher than that of nitrogen.

*Reference pictures to see how they each connect. 

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A monatomic ideal gas that is initially at 1.50 * 105 Pa and has a volume of 0.0800 m3 is compressed adiabatically to a volume o
lubasha [3.4K]

Answer:

300000Pa or 3×10^5 Pa

Explanation:

Since the problem involves only two parameters of volume and pressure, the formula for Boyle's law is suitably used.

Using Boyle's law

P1V1 = P2V2

P1 is the initial pressure = 1.5×10^5Pa

V1 is the initial volume = 0.08m3

P2 is the final pressure (required)

V2 is the final volume = 0.04 m3

From the formula, P2 = P1V1/V2

P2 = 1.5×10^5 × 0.08 ÷ 0.04

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8 0
3 years ago
How does the mass per nucleon of an element change as the atomic number increases?
alex41 [277]

Answer:

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8 0
2 years ago
The conversion factor that is present in almost all stoichiometry calculations is called the ________ __________ .
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5 0
3 years ago
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Explanation:

Element Symbol Mass Percent

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3 years ago
Vanadium (III) hydroxide reacts with hydrobromic acid according to the following balanced chemical equation: V(OH)3 (s) + 3 HBr
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Answer:

TY and I like the country studies.

Explanation:

ok bye por a mathmu n bye for now and hope you have patience.

3 0
3 years ago
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