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3 years ago

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Example 4. Balance the equation below in which methane reacts with oxygen in the atmosphere to release heat and the atoms recomb

ine to form carbon dioxide and water vapour. CH4 + O2------>CO2 + H2O 1 10

2 answers:

bija089 [108]3 years ago

5 0

Answer:

CH4+O2---->CO2 + H2O unbalance

C=1. C=1

H=4. H=2x2

O=2x2. O=2 +(1 x2)

CH4+2O2----->CO2 +2H2O. Balanced

Tems11 [23]3 years ago

4 0

Answer:

CH4 + 2O2 ----> CO2 + 2H2O

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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

De que material estan echos los marcapasos

ZanzabumX [31]

Answer:

Hm creo que es La carcasa es de titanio y la batería está compuesta de acumuladores de litio y yodo, que pueden durar cerca de diez años. Los electrodos que lo componen pueden ser de distintos tipos, pero sus puntas, que no están aisladas, son de platino, y el número de los mismos dependerá de las necesidades de cada paciente

Explanation:

3 0

3 years ago

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How many significant figures are in the mass 3.0047 g?

Rudik [331]

Answer:

5 significant figures

5 0

4 years ago

What was the weight percent of water in the hydrate before heating?

DedPeter [7]

Data:

weight of water before heating = 0.349

weight of hydrate before heateing = 2.107

Formula:

Weight percent of water = [ (weight of water) / (weight of the hydrate) ] * 100

Solution:

Weight percent of water = [ 0.349 / 2.107] * 100 ≈ 16.6 %

Answer: 16.6%

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4 years ago

Lewis structure for 2O2

Mrrafil [7]

the big number describes the number ratio in a chemical equation

so for example,

2H2 + O2 --> 2H2O means

2 moles of hydrogen reacts with one mole of oxygen to form 2 moles of water

and as you know, the small (subscript) number determines the number of atoms of that element in one molecule of a compound

so I believe that drawing a normal lewis structure ( O=O ) should be correct

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3 years ago

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