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Ivanshal [37]
3 years ago
9

What is the [H3O+] in a solution with [OH-] = 1.0 x 10-12 M?

Chemistry
1 answer:
schepotkina [342]3 years ago
8 0

Answer:

The concentration of hydronium ions in the solution is 1.0\times 10^{-2}M.

Explanation:

Concentration of hydroxide ions concentration = [OH^-]=1.0\times 10^{-12} M

The pOH if the solution is given by :

pOH=-\log[OH^-]

pOH=-\log[1.0\times 10^{-12} M}]

pOH = 12

The relation between pH and OH is given by : pH + pOH = 14

pH = 14 - pOH

pH =14 - 12 = 2

The pH of the solution isgiven by :

pH=-\log[H_3O^+]

2=-\log[H_3O^+]

[H_3O^+]=1.0\times 10^{-2} M

The concentration of hydronium ions in the solution is 1.0\times 10^{-2}M.

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slamgirl [31]

Answer: 3173.9cm3

Explanation:Please see attachment for explanation

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3 years ago
How many grams of sodium oxide can be produced when 55.3 g Na react with 64.3 g O2?. . Unbalanced equation: Na + O2 → Na2O. . Sh
GenaCL600 [577]
 <span>this is a limiting reagent problem. 

first, balance the equation 
4Na+ O2 ---> 2Na2O 

use both the mass of Na and mass of O2 to figure out how much possible Na2O you could make. 
start with Na and go to grams of Na2O 

55.3 gNa x (1molNa/23.0gNa) x (2 molNa2O/4 molNa) x (62.0gNa2O/1molNa2O) = 75.5 gNa2O 

do the same with O2 

64.3 gO2 x (1 molO2/32.0gO2) x (2 molNa2O/1 mol O2) x (62.0gNa2O/1molNa2O) = 249.2 g Na2O 

now you must pick the least amount of Na2O for the one that you actually get in the reaction. This is because you have to have both reacts still present for a reaction to occur. So after the Na runs out when it makes 75.5 gNa2O with O2, the reaction stops. 

So, the mass of sodium oxide is 

75.5 g</span>
3 0
3 years ago
Read 2 more answers
Find molar mass of a gas when the density is given as 1.25g / L
adelina 88 [10]
1.25 g x 22.4 l / 1 mol = 28 g/mol
7 0
3 years ago
How many moles of atoms are in 9.00 g of 13c?
Blababa [14]
<span>The atomic weight of 13C should be pretty close to 13.0. (If you have the exact mass, use it in the problem.) So, 9.00 g / 13.0 g/mol = 0.692 moles Therefore, the answer should be 0.692 moles are in 9.00 g of 13C.</span>
8 0
3 years ago
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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

4 0
3 years ago
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