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Irina-Kira [14]
3 years ago
5

An element that contains an electron in a d sublevel is A. Mg. B. O. C. Fe. D. Ne.

Chemistry
2 answers:
Dimas [21]3 years ago
8 0
C) Fe

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² <u>3d⁶</u>
zepelin [54]3 years ago
6 0

Fe iron atomic number 26 i took the test

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A sample of a gas in a rigid cylinder with a movable piston has a volume of a lamp then .2 L at STP. What is the volume of this
Gnoma [55]

<u>Answer:</u> The new volume of the gas is 0.11 L

<u>Explanation:</u>

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.

The equation follows:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1,V_1\text{ and }T_1 are the initial pressure, volume and temperature of the gas

P_2,V_2\text{ and }T_2 are the final pressure, volume and temperature of the gas

At STP:

The temperature at this condition is taken as 273 K and the pressure at this condition is taken as 1 atm or 101.3 kPa.

We are given:

P_1=101.3kPa\\V_1=0.2L\\T_1=273K\\P_2=202.6kPa\\V_2=?\\T_2=300K

Putting values in above equation, we get:

\frac{101.3kPa\times 0.2L}{273K}=\frac{202.6kPa\times V_2}{300K}\\\\V_2=\frac{101.3\times 0.2\times 300}{273\times 202.6}=0.11L

Hence, the new volume of the gas is 0.11 L

3 0
3 years ago
Explain why reaction speed increases with temperature.
Dahasolnce [82]
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7 0
2 years ago
At 298 K, the rate constant for a reaction is 0.0346 s-1. What is the rate constant at 350K if the Ea = 50.2kJ/mol
frutty [35]

Answer:

0.702 /s

Explanation:

Rate constant at [298 \mathrm{~K}, \mathrm{~K}_{1}=3.46 \times 10^{-2} \mathrm{~s}^{-1}

Rate constant at 350 \mathrm{~K}, \mathrm{~K}_{2}=?

T_{1}=298 \mathrm{~K}

T_{2}=350 \mathrm{~K}

Activation energy, \mathrm{Ea}=50.2 \times 10^{3} \mathrm{~J} / \mathrm{mol}

Use the following equation to calculate K_{2}$ at $350 \mathrm{~K}

Use the following equation to calculate K_{2}$ at $350 \mathrm{~K}

\ln \frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}=\frac{\mathrm{Ea}}{\mathrm{R}}\left[\frac{1}{\mathrm{~T}_{1}}-\frac{1}{\mathrm{~T}_{2}}\right]

Therefore,

 \ln \left(\frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}}\right) &=\frac{50.2 \times 10^{3} \mathrm{~J} / \mathrm{mol}}{8.314 \mathrm{JK}^{-1} \mathrm{~mole}^{-1}}\left[\frac{1}{298 \mathrm{~K}}-\frac{1}{350 \mathrm{~K}}\right]

\ln \left(\frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}}\right) &=\frac{50.2 \times 10^{3} \mathrm{~J} / \mathrm{mol}}{8.314 \mathrm{JK}^{-1} \mathrm{~mole}^{-1}}\left[\frac{52 \mathrm{~K}}{298 \mathrm{~K} \times 350 \mathrm{~K}}\right]

\frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}} &=\mathrm{e}^{3.01}

\frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}} &=20.3

K_{2} &=20.3 \times 3.46 \times 10^{-2} \mathrm{~s}^{-1}

&=0.702 \mathrm{~s}^{-1}

hence, the rate constant at 350 \mathrm{~K} is 0.702\mathrm{~s}^{-1}

5 0
3 years ago
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