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3 years ago

12

What is the ground-state electron configuration of a neutral atom of manganese?.

1 answer:

Alchen [17]3 years ago

3 0

Explanation:

elctronic configuration of manganese

Mn=1s²2s²2p⁶3s²3p⁶4s²3d⁵

ground state

Mn=Ar3d⁵4s²

note that Ar is argon

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Decrease in predators because then the wild turkeys won't be killed.

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Pick 1 answer, please help!

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2 years ago

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What is the volume of 15.2 grams of SO2 at STP?

kiruha [24]

5.512 litres is the volume of 15.2 grams of sulphur dioxide gas at STP.

Explanation:

Data given:

mass of sulphur dioxide = 15.2 grams

conditions is at STP whech means volume = 22.4 litres

atomic mass of sulphur dioxide = 64.06 grams/mole

Number of moles is calculated as:

number of moles = $\frac{mas}{atomic mass of 1 mole}$

Putting the values in the equation:

number of moles = $\frac{15.2}{64.06}$

= 0.23 moles

Assuming that sulphur dioxide behaves as an ideal gas, we can calculate the volume as:

When 1 mole of sulphur dioxide occupies 22.4 litres at STP

Then 0.23 moles of sulphur dioxide occupies 22.4 x 0.23

= 5.152 litres is the volume.

5 0

3 years ago

Octane (C8H18)reacts oxygen gas (O2) to form carbon dioxide and water

iren [92.7K]

Answer:

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Explanation:

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3 years ago

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Pani-rosa [81]

The question is incomplete, here is the complete question:

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Atmospheric Gas Mole Fraction kH mol/(L*atm)

$N_2$ $7.81\times 10^{-1}$ $6.70\times 10^{-4}$

$O_2$ $2.10\times 10^{-1}$ $1.30\times 10^{-3}$

Ar $9.34\times 10^{-3}$ $1.40\times 10^{-3}$

$CO_2$ $3.33\times 10^{-4}$ $3.50\times 10^{-2}$

$CH_4$ $2.00\times 10^{-6}$ $1.40\times 10^{-3}$

$H_2$ $5.00\times 10^{-7}$ $7.80\times 10^{-4}$

<u>Answer:</u> The solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

<u>Explanation:</u>

To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:

$p_{\text{hydrogen gas}}=p_T\times \chi_{\text{hydrogen gas}}$

where,

$p_A$ = partial pressure of hydrogen gas = ?

$p_T$ = total pressure = 0.380 atm

$\chi_A$ = mole fraction of hydrogen gas = $5.00\times 10^{-7}$

Putting values in above equation, we get:

$p_{\text{hydrogen gas}}=0.380\times 5.00\times 10^{-7}\\\\p_{\text{hydrogen gas}}=1.9\times 10^{-7}atm$

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{H_2}=K_H\times p_{H_2}$

where,

$K_H$ = Henry's constant = $7.80\times 10^{-4}mol/L.atm$

$p_{H_2}$ = partial pressure of hydrogen gas = $1.9\times 10^{-7}atm$

Putting values in above equation, we get:

$C_{H_2}=7.80\times 10^{-4}mol/L.atm\times 1.9\times 10^{-7}atm\\\\C_{CO_2}=1.48\times 10^{-10}M$

Hence, the solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

4 0

3 years ago