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3 years ago

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 3d⁶

zepelin [54]3 years ago

6 0

Fe iron atomic number 26 i took the test

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3 years ago

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Methane molecule is depicted here

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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]

Answer: The new volume of the gas is 0.11 L

Explanation:

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

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3 years ago

Explain why reaction speed increases with temperature.

Dahasolnce [82]

As the reaction speed involves the overcoming of the activation energy (due to a required amount of kinetic energy between the particles). When temperature is increased it will provide more energy for the movement of the particles to gain more kinetic energy meaning the rate at which the particles interact increase

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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 $350 \mathrm{~K}$

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

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 $\mathrm{~s}^{-1}$

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3 years ago