Answer:
a. fluorine
Explanation:
Fluorine is the element of group 17 and period 2. The electronic configuration of the element is
.
Stable oxidation state = -1 of fluorine as it gains one electron to gain noble gas configuration.
With alkali metals, which have oxidation state of +1 form ionic compound of the form, MX where X is F.
Among the halogens, fluorine forms the most stable halide because of the comparable size of the hydrogen and fluorine. Thus, it is the weakest acid when compared with other hydrogen halides.
Fluorine is the most reactive in the halogen series and thus, combines with most of the elements.
Fluorine forms inter-halogen compounds of form XA only. Example - ClF.
Hence, option a is correct.
Answer:
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LEAD is the element that has this electron configuration
Answer:
78.2 g/mol
Step-by-step explanation:
We can use the <em>Ideal Gas Law</em> to solve this problem:
pV = nRT
Since n = m/M, the equation becomes
pV = (m/M)RT Multiply each side by M
pVM = mRT Divide each side by pV
M = (mRT)/(pV)
Data:
ρ = 2.50 g/L
R = 0.082 16 L·atm·K⁻¹mol⁻¹
T =98 °C
p = 740 mmHg
Calculation:
(a)<em> Convert temperature to kelvins
</em>
T = (98 + 273.15) = 371.15 K
(b) <em>Convert pressure to atmospheres
</em>
p = 740 × 1/760 =0.9737 atm
(c) <em>Calculate the molar mass
</em>
Assume V = 1 L.
Then m = 2.50 g
M = (2.50 × 0.082 06 × 371.15)/(0.9737 × 1)
= 76.14/0.9737
= 78.2 g/mol