8.8 × 10-5 M is the [H3O+] concentration in 0.265 M HClO solution.
Explanation:
HClO is a weak acid and does not completely dissociate in water as ions.
the equation of dissociation can be written and ice table to be formed.
HClO +H2O ⇒ ClO- + H3O+
I 0.265 0 0
C -x +x +x
E 0.265-x +x +x
Now applying the equation of Ka, where Ka is given as 2.9 × 10-8.
Ka = ![\frac{[ClO-][H3O+]}{[HClO]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BClO-%5D%5BH3O%2B%5D%7D%7B%5BHClO%5D%7D)
2.9 × 10^-8 = ![\frac{[x] [x]}{[0.265-x]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Bx%5D%20%5Bx%5D%7D%7B%5B0.265-x%5D%7D)
= 7.698 x
x = 8.8 × 10-5 M
The hydronium ion concentration is 8.8 × 10-5 M in 0.265 M solution of HClO.
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<h3>Answer:</h3>
89.6 L of O₂
<h3>Solution:</h3>
The balanced chemical equation is as,
CH₄ + 2 O₂ → CO₂ + 2 H₂O
As at STP, one mole of any gas (Ideal gas) occupies exactly 22.4 L of Volume. Therefore, According to equation,
44 g ( 1 mol) CO₂ is produced by = 44.8 L (2 mol) of O₂
So,
88 g CO₂ will be produced by = X L of O₂
Solving for X,
X = (88 g × 44.8 L) ÷ 44 g
X = 89.6 L of O₂
Answer:
1120 L.
Explanation:
Hello!
In this case, as no conditions of pressure of temperature are given for this problem, we can assume that the scuba diver dives at STP (1 atm and 273.15 K), which means that 1 mole of air would occupy a volume of 22.4 L.
In such a way, since she needs 50.0 moles of air, the following ratio is useful to compute the size (volume) of the tank she needs:

Thereby, we plug in to obtain:

Best regards!
<span>1. 1 molecule of C6H12O6(dextrose sugar), 2 molecles of c2h6o (ethyl alcohol), 2 molecules of Co2
2. 48 hydrogen atoms </span>