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2 years ago

5

In the reaction Mg(s) + 2H2O(l) --> Mg(OH)2(s) + H2(g), what is the oxidation state of each H in H2(g)? A. +2 and –2 B. –1 an

d +1 C. +1 D. 0 no idea how to do this. I know its not D because that's for magnesium

1 answer:

chubhunter [2.5K]2 years ago

6 0

The oxidation state of the atoms of any molecule in the element state equals zero. H2 is is the element state, therefore the oxidation state of the whole molecule equals zero and the oxidation states of the atoms also equals zero. Answer letter D is the correct answer.
Magnesium has an oxidation state = zero before it undergoes the reaction and is oxidized. The oxidations state of the magnesium in Mg(OH)2 is +2.

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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 $1.48\times 10^{-10}M$

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3 years ago

Dissolved hydrofluoric acid reacts with dissolved sodium hydroxide to form water and aqueous sodium fluoride

GalinKa [24]

Answer:

HF(aq)+NaOH(aq)→NaF(aq)+H2O(l)

Explanation:

Complete question

Dissolved hydrofluoric acid reacts with dissolved sodium hydroxide to form water and aqueous sodium fluoride. What is the net ionic equation

Equilibrium equation between the undissociated acid and the dissociated ions

HF(aq)⇌H+(aq)+F−(aq)

Sodium hydroxide will dissociate aqueous solution to produce sodium cations, Na+, and hydroxide anions, OH−

NaOH(aq)→Na+(aq)+OH−(aq)

Hydroxide anions and the hydrogen cations will neutralize each other to produce water.

H+(aq)+OH−(aq)→H2O(l)

On combining both the equation, we get –

HF(aq)+Na+(aq)+OH−(aq)→Na+(aq)+F−(aq)+H2O(l)

The Final equation is

HF(aq)+NaOH(aq)→NaF(aq)+H2O(l)

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