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3 years ago

Write the chemical symbols for three different atoms or atomic cations with 12 electrons.

Chemistry

2 answers:

ladessa [460]3 years ago

8 0

1) Atomic number of magnesium (Mg) is 12, it means that it has 12 protons and 12 electrons.

Electron configuration of magnesium atom: ₁₂Mg 1s² 2s² 2p⁶ 3s².

2) Atomic number of aluminium (Al) is 13, it means that it has 13 protons and 13 electrons., but if it lost one electron, it will have 12 electrons and 13 protons and become aluminium cation Al⁺.

Electron configuration of aluminium cation: ₁₃Al⁺ 1s² 2s² 2p⁶ 3s².

3) Atomic number of silicon (Si) is 14, it means that it has 14 protons and 14 electrons., but if it lost two electrons, it will have 12 electrons and 14 protons and become silicon cation Si²⁺.

Electron configuration of silicon cation: ₁₄Si²⁺ 1s² 2s² 2p⁶ 3s².

Tom [10]3 years ago

5 0

Mg, Al+1, Si+2

I think that's the answer

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4 years ago

Pure chlorobenzene is contained in a flask attached to an open-end mercury manometer. When the flask contents are at 58.3°C, the

Elan Coil [88]

Answer:

The slope is 4661.4K

(B), the intercept is 18.156

The vapor pressure of chlorobenzene is 731.4mmHg

The percentage of chlorobenzene originally in the vapor that condenses is = 99.7%

Explanation:

Two sets of conditions (a and b) are observed for L1 and L2 at two different temperatures.

$T_{a}$ = 58.3°C

$L1_{a}$ = 747mmHg

$L2_{a}$ = 52mmHg

$T_{b}$ = 110°C

$L1_{a}$ = 577mmHg

$L2_{b}$ = 222mmHg

We need to convert the temperatures from Celsius to Kelvin (K)

$T_{a}$ = 58.3°C + 273.2

$T_{a}$ = 331.5k

$T_{b}$ = 110°C + 273.2

$T_{b}$ = 383.2k

We then calculate the vapor pressures of the chlorobenzene at each set of conditions by measuring the difference in the mercury levels.

$P^{0}$ = $P_{atm} - (P_1 -P_2)$

The vapor pressure under the first set of conditions is:

$P^{a}$ = 755mmHg - (747mmHg - 52mmHg)

$P^{a}$ = 60mmHg

The vapor pressure under the second set of conditions is:

$P^{b}$ = 755mmHg - (577mmHg = 222mmHg)

$P^{b}$ = 400mmHg

First question say we should find ΔH(slope) and B(intercept) in the Clausius- Clapeyron equation:

Using the Formula :

$In_p^{0} = \frac{- (delta) H}{RT} + B$

where (delta) H = ΔH

The slope of the $In_p^{0} = \frac{T_1T_2 In (P_2/P_1)}{T_1-T_2}$

Calculating the slope; we have:

$\frac{- (delta) H}{R} = \frac{T_1T_2 In (P_2/P_1)}{T_1-T_2}$

$\frac{- (delta) H}{R}$ = $\frac{331.5 * 383.2 * In (400/60)}{(331.5-383.2)}$

$\frac{- (delta) H}{R}$ = -4661.4k

$\frac{ (delta) H}{R}$ = 4661.4k

The intercept can be derived from the Clausius-Clayperon Equation by making B the subject of the formula. To calculate the intercept using the first set of condition above; we have:

B = $In_p_{1} + \frac{ (delta) H}{RT}$

B = $In 60 + \frac{4661.4}{331.5}$

B = 18.156

Thus the Clausius-Clayperon equation for chlorobenzene can be expressed as:

In P = $\frac{-4661.4k}{T} + 18.156$

In question (b), the air saturated with chlorobenzeneis 130°C, converting he temperature of 130°C to absolute units of kelvin(k) we have;

T = 130°C + 273.2

T = 403.2k

Calculating the vapor pressure using Clausius-Clapyeron equation: we have;

$In_p_{0}$ = $In_p_{0} = \frac{-4661.4}{403.2} + 18.156$

$In_p_{0}$ = 6.595

$p_{0}$ = $e^{6.595}$

$p_{0}$ = 731.mmHg

The vapor pressure of chlorobenzene is 731.4mmHg

The diagram of the flowchart with the seperate vapor and liquid stream can be found in the attached document below.

Afterwards, both the inlet and outlet conditions contain saturated liquid.

From the flowchart, the vapor pressure of chlorobenzene at the inlet and outlet temperatures are known:

$P_1(130^{0}C)$ = 731.44mmHg

$P_1(58.3^{0}C)$ = 60mmHg

To calculate the percentage of the chlorobenzene originally in the vapor pressure that condenses; we must first calculate the mole fractions of chlorobenzene for the vapor inlet and outlet using Raoult's Law:

$y_1 = \frac{P^0 (T)}{P}$