At STP (standard temperature and pressure conditions), 1 mol of any gas occupies 22.4 L
This rule is applied to O₂
22.4 L volume occupied by 1 mol
Therefore 83.4 L occupied by - 1/ 22.4 x 83.4 = 3.72 mol
stoichiometry of O₂ to H₂O is 1:2
then the number of moles of water produced - 3.72 mol x 2= 7.44 mol
mass of water produced - 7.44 mol x 18.01 g/mol = 134.1 g
correct answer is D
Answer:
[H3O+] = 1.0*10^-12 M
[OH-] = 0.01 M
Explanation:
We can use the following equation to find the hydronium ion concentration. Plug in the pH and solve for H3O+.
pH = -log[H3O+]
<u>[H3O+] = 1.0*10^-12 M</u>
Now, to find the hydroxide ion concentration we will use the two following equations.
14 = pH + pOH
pOH = -log[OH-]
14 = 12 + pOH
pOH = 2
2 = -log[OH-]
<u>[OH-] = 0.01 M</u>
Lol ok ill help the answer is true and she SHOULDNT be mad at you
Answer:
<h2>The Alkali metal halide may precipitate or there may be no change at all</h2>
Explanation:
Alkali metal cations are positively charged. Halogen anions are negatively charged. When a solution of Alkali metal cations is added to a solution of Halogen anions, there are two possibilities :
- The alkali metal halide( salt formed from reation of the two ions) may precipitate if the Ionic product is higher than the Solubility product.
- However, if it can remain in the solution, it will remain so. No chemical changes happen with respect to these both ions. Nothing willl happen.
There is no reaction happening in either of the cases because both species are already in ionic form before addition, hence they continue to be in this form.
Boyle’s Law P1V1 = P2V2
P1 = 0.80 atm V1 = 1.8 L
P2 = 1.0 atm V2 = ??
(.8 atm)(1.8 L) = (1.0 atm)(V2)
1.44 atm x L = 1 atm V2
