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stealth61 [152]
3 years ago
7

How can I balance this equation? ____ KClO3 ---> ____ KCl + ____ O2

Chemistry
1 answer:
tensa zangetsu [6.8K]3 years ago
8 0
__ KClO₃ → __ KCl + __ O₂

Left Side:
1 K
1 Cl
3 O

Right Side:
1 K
1 Cl
2 O

Since the least common multiple of 3 and 2 is 6, we need to multiply the compound with 2 oxygen by 3 and the compound with 3 oxygen by 2.

This gives us 2KClO₃ → __ KCl + 3O₂.

However, this equation is still not balanced.

Left Side:
2 K
2 Cl
6 O

Right Side:
1 K
1 Cl
6 O

In order to balance the K and Cl, we need to multiply the KCl compound on the right side by 2.

2KClO₃ → 2KCl + 3O₂
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At 1 atm of pressure and a temperature of 0 degrees celsius, which phases of H2O can exist?
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Sodium is an inactive metal why​
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7 0
3 years ago
Calculate the pH of: (a) 0.1M HCl; (b) 0.1M NaOH; (c) 3 X 10% M HNO3; (d) 5 X 10-10 M HCIO.; and (e) 2 x 10-8 M KOH.
Shtirlitz [24]

Answer:

(a) pH = -Log (0.1M) = 1

(b) pH = -Log (10^{-13}M) = 13

(c) pH = -Log (3x10^{-3}M) = 2.5

(d) pH = -Log (4.93x10^{-10}M) = 9.3

(e) pH = -Log (5^{-7}M) = 6.3

Explanation:

To calculate de pH of an acid solution the formula is:

pH = -Log ([H^{+}]) = 1

were [H^{+}] is the concentration of protons of the solution. Therefore it is necessary to know the concentration of the protons for every solution in order to solve the problem.

(a) and (c) are strong acids so they dissociate completely in aqueous solution. Thus, the concentration of the acid is the same as the protons.

(b) and (e) are strong bases so they dissociate completely in aqueous solution too. Thus, the concentration of the base is the same as the oxydriles. But in this case it is necessary to consider the water autoionization to calculate the protons concentration:

K_{w} =[H^{+} ][OH^{-}]=10^{-14}

clearing the [H^{+} ]

[H^{+} ]=\frac{10^{-14}}{[OH^{-}]}

(d) is a weak base so it is necessary to solve the equilibrium first, knowing Ka=3.24x10^{-8}

The reaction is HClO  →  H^{+} + CO^{-} so the equilibrium is

Ka=3.24x10^{-8}=\frac{x^{2}}{5x10^{-8}-x}

clearing the <em>x</em>

{x^{2}={1.62x10^{-17}-3.24x10^{-8}x}

x=[H^{+}]=4.93x10^{-10}

6 0
3 years ago
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