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Nimfa-mama [501]
3 years ago
11

The atom below with the greatest number of valence electrons is _______. A. Cl: 1S2 2S2 2P6 3S2 3P5 B. Br: 1S2 2S2 2P6 3S2 3P6 4

S2 3D10 4P5 C. I: 1S2 2S2 2P6 3S2 3P6 4S2 3D10 4P6 5S2 4D10 5P5 D. Ne: 1S2 2S2 2P6 E. None of the Above
Chemistry
1 answer:
allochka39001 [22]3 years ago
6 0
Me. It has 6 valence electrons. Everything else has 5
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2 years ago
A gas has a volume of 1.75L at -23°C and 150.0 kPa. At what temperature would the gas occupy 1.30L at 210.0 kPa?
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As amount of gas remains constant in both state therefore in accordance with combined gas law for an ideal gas-

                                          \frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}

where P_{1} and P_{2} are initial and final pressure respectively.

           V_{1}  and V_{2} are initial and final volume respectively.

           T_{1} and T_{2} are initial and final temperature in kelvin scale respectively.

Here P_{1}=150.0kPa , V_{1}=1.75L , T_{1}=(273-23)K=250K, P_{2}=210.0kPa and V_{2}=1.30L

Hence    T_{2}=\frac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}

            \Rightarrow T_{2}=\frac{(210.0kPa)\times (1.30L)\times (250K)}{(150.0kPa)\times (1.75L)}

            \Rightarrow T_{2}=260K

            \Rightarrow T_{2}=(260-273)^{0}\textrm{C}=-13^{0}\textrm{C}

So at -13 ^{0}\textrm{C} , the gas would occupy 1.30L at 210.0 kPa.

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3 years ago
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Quantitative data is numerical.
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