As per as the Multiplication rules of the significant figures, whenever any numbers in the decimals forms are multiplied or divided then result in mentioned in such a way so that the significant figures after the decimal will be same as that in the given least condition.
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102900/12 = 8575
170 × 1.27 = 215.9
∴ (102,900 ÷ 12) + (170 × 1.27) = 8575 + 215.9
= 8790.9
Now, As per as Above rules, answer in correct significant figures will be = 8791.
There seems to be a mistake. If u mistyped the 'h', then the reaction is single replacement.
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.
<span>4 Al + 3 O2 → 2 Al2O3
(10.0 g Al) / (26.98154 g Al/mol) = 0.37062 mol Al
(19.0 g O2) / (31.99886 g O2/mol) = 0.59377 mol O2
0.37062 mole of Al would react completely with 0.37062 x (3/4) = 0.277965 mole of O2, but there is more O2 present than that, so O2 is in excess.
((0.59377 mol O2 initially) - (0.277965 mol O2 reacted)) x (31.99886 g O2/mol) =
10.1 g O2 left over</span><span>
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