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Sergeeva-Olga [200]

3 years ago

8

Write the balanced equation for the reaction of solid potassium chlorate decomposing to form solid potassium chloride and oxygen

gas.

1 answer:

DanielleElmas [232]3 years ago

8 0

2KClO3(s) ----> 2KCl(s) + 3O2(g)

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How many times more hydroxide ions are there in a solution with a ph of 9 than in a solution with a ph of 3?

cestrela7 [59]

The pH unit has 10x as many hydrogens ions as the unit above.
Ex: A pH of 5 would have 10x more hydrogen ions than a pH of 6
and 100x more than if it had a pH of 7.
With a pH of 9 and 3, this is equivalent to 10⁶
So your answer should be:
1,000,000

4 0

3 years ago

reativity of alkali metals increases down the group while reativity decreases down the group in helogens

BigorU [14]

Complete question is;

Chemical reactivity of alkali metals increases down the group while reactivity of halogens decreases down the group. Give reasons

Answer:

Explained below

Explanation:

Alkali metals exhibit reactivity due to their electropositivity. Now, for alkalis, their electro-positivity increases down their group. Since their reactivity increases with increase in electropositivity, it means their reactivity also increases down the group.

Whereas, the reactivity of halogens occurs as a result of their electronegativity. Now, electronegativity for halogens decreases down the group. Since their reactivity decreases with decrease in electronegativity, it means that their reactivity will also decrease down the group.

4 0

3 years ago

We want to calculate the concentrations of all species in a 0.58 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants

NeX [460]

Explanation:

Reaction equation is as follows.

$Na_{2}SO_{3}(s) \rightarrow 2Na^{+}(aq) + SO^{2-}_{3}(aq)$

Here, 1 mole of $Na_{2}SO_{3}$ produces 2 moles of cations.

$[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58$

= 1.16 M

$[SO^{2-}_{3}] = [Na_{2}SO_{3}]$ = 0.58 M

The sulphite anion will act as a base and react with $H_{2}O$ to form $HSO^{-}_{3}$ and $OH^{-}$ .

As, $K_{b} = \frac{K_{w}}{K_{a_{2}}}$

= $\frac{10^{-14}}{6.3 \times 10^{-8}}$

= $1.59 \times 10^{-7}$

According to the ICE table for the given reaction,

$SO^{2-}_{3} + H_{2}O \rightleftharpoons HSO^{-}_{3} + OH^{-}$

Initial: 0.58 0 0

Change: -x +x +x

Equilibrium: 0.58 - x x x

So,

$K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}$

$1.59 \times 10^{-7} = \frac{x^{2}}{0.58 - x}$

$x^{2} = 1.59 \times 10^{-7} \times (0.58 - x)$

x = 0.0003 M

So, x = $[HSO^{-}_{3}] = [OH^{-}]$ = 0.0003 M

$[SO^{2-}_{3}]$ = 0.58 - 0.0003

= 0.579 M

Now, we will use $[HSO^{-}_{3}]$ = 0.0003 M

The reaction will be as follows.

$HSO^{2-}_{3} + H_{2}O \rightleftharpoons H_{2}SO_{3} + OH^{-}$

Initial: 0.0003

Equilibrium: 0.0003 - x x x

$K_{b} = \frac{x^{2}}{0.0003 - x}$

$K_{b} = \frac{K_{w}}{K_{a_{1}}}$

= $\frac{10^{-14}}{1.4 \times 10^{-2}}$

= $7.14 \times 10^{-13}$

Therefore, $7.14 \times 10^{-13} = \frac{x^{2}}{0.0003 - x}$

As, x <<<< 0.0003. So, we can neglect x.

Therefore, $x^{2} = 7.14 \times 10^{-13} \times 0.0003$

= $0.00214 \times 10^{-13}$

x = $0.0146 \times 10^{-6}$

x = $[OH^{-}] = [H_{2}SO_{3}]$ = $1.46 \times 10^{-8}$

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

= $\frac{10^{-14}}{0.0003}$

= $3.33 \times 10^{-11}$ M

Thus, we can conclude that the concentration of spectator ion is $3.33 \times 10^{-11}$ M.

5 0

3 years ago

Pleaseee helpppp!!!!!!!!

Katarina [22]

Answer:

a covalent would be the two that are nonmetals

7 0

3 years ago

45 Three samples of the same solution are tested, each with a different indicator. All three indicators, bromthymol blue, bromcr

ANEK [815]

<u>Answer:</u> The correct answer is Option 4.

<u>Explanation:</u>

Bromothymol blue, Bromocresol green and Thymol blue are the indicators which change their color according to the change in pH of the solution.

The pH range and color change of these indicators are:

Bromothymol Blue: The pH range for this indicator is 6.0 to 7.5 and color change is from yellow to blue. It appears yellow below pH 6.0 and blue above pH 7.5
Bromocresol green: The pH range for this indicator is 3.5 to 6.0 and color change is from yellow to blue. It appears yellow below pH 3.5 and blue above pH 6.0
Thymol Blue: The pH range for this indicator is 8.0 to 9.6 and color change is from yellow to blue. It appears yellow below pH 8.0 and blue above pH 9.6

As, the highest pH of all the indicators is 9.6, so every indicator will appear blue above pH 9.6.

Hence, the correct answer is Option 4.

3 0

3 years ago

Read 2 more answers