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1 year ago

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Oxidation state of Nitrogen in N2O5

1 answer:

balandron [24]1 year ago

7 0

Explanation:

Oxidation state of Nitrogen in N2O5 is +5

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CH3COOH mc015-1.jpg CH3COO– (aq) + H+(aq) What will happen to the chemical equilibrium of the solution if CH3COONa is added?

andrey2020 [161]

The equilibrium will shift to the left or the backward reaction since addition of <span>CH3COONa will add more CH3COO- ions to the solution. The formation of reactants are promoted.</span>

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3 years ago

Given the following reaction: \ce{Cu + 2AgNO3 -> 2Ag + Cu(NO3)2}Cu+2AgNO3 2Ag+Cu(NOX 3 )X 2 How many moles of \ce{Ag}

Alexandra [31]

Answer:

0.252 mol

Explanation:

<em>Given the following reaction: </em>

<em>Cu + 2 AgNO₃ → 2 Ag + Cu(NO₃)₂</em>

<em>How many moles of Ag will be produced from 16.0 g Cu, assuming AgNO3 is available in excess.</em>

First, we write the balanced equation.

Cu + 2 AgNO₃ → 2 Ag + Cu(NO₃)₂

We can establish the following relations.

The molar mass of Cu is 63.55 g/mol.
The molar ratio of Cu to Ag is 1:1.

The moles of Ag produced from 16.0 g of Cu are:

$16.0gCu.\frac{1molCu}{63.55gCu} .\frac{1molAg}{1molCu} =0.252 molAg$

6 0

3 years ago

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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.

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3 years ago

Which factors affect a river's load?

Marizza181 [45]

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3 years ago

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Dahasolnce [82]

Of course we would experience them.

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2 years ago