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3 years ago

If an atom had 2 electrons in its third shell, it would need to ____ electron(s) to become stable.

Chemistry

1 answer:

Anni [7]3 years ago

7 0

6 electron to be stable

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Hey can someone please help me with this

vivado [14]

Answer:

This is not a redox reaction.

Explanation:

A redox reaction is a reaction that involves reduction and oxidation processes. The best way to identify a redox reaction. Changes in the oxidation number f the species indicates that it is a redox reaction.

Na+

Reactant side = +1, Product Side = +1; No change

Cl-

Reactant side = -1, Product Side = -1; No change

Ag+

Reactant side = +1, Product Side = +1; No change

Reactant side = +5, Product Side = +5; No change

Reactant side = -2, Product Side = -2; No change

There is no change in any of the oxidation numbers, Hence this is not a redox reaction.

6 0

3 years ago

When a magnet spins within a coil of wires, what is produced?

Andrei [34K]

Decreasing the turns of wire

7 0

2 years ago

How many moles of silver chloride are produced from 15.0 moles of silver nitrate

Ghella [55]

15.0 moles of silver chloride are produced.

7 0

3 years ago

In the sodium-potassium pump, sodium ion release causes two K ions to be released into the cell. __________________ True False

Solnce55 [7]

Answer:

false

Explanation:

As we know that in sodium-potassium pump .

sodium potassium move 3Na+ outside the cells

and moving 2k+ inside the cells

so that we can say that given statement is false

Answer FALSE

6 0

3 years ago

A 18.08-g sample of the ionic compound , where is the anion of a weak acid, was dissolved in enough water to make 116.0 mL of so

Oksi-84 [34.3K]

Answer:

a) 129.14 g/mol

b) 8.87

Explanation:

Given that:

mass of the ionic compound [NaA] = 18.08 g

Volume of water = 116.0 mL = 0.116 L

Let the mole of the acid HCl = 0.140 M

Volume of the acid = 500.0 mL = 0.500 L

pH = 4.63

$V_{equivalence}_{acid}$ = 1.00 L

Equation for the reaction can be represented as:

$NaA_{(aq)} + HCl_{(aq)} -----> HA_{(aq)} + NaCl_{(aq)$

From above; 1 mole of an ionic compound reacts with 1 mole of an acid to reach equivalence point = 0.140 M × 1.00 L

= 0.140 mol

Thus, 0.140 mol of HCl neutralize 0.140 mol of ionic compound at equilibrium

Thus, the molar mass of the sample = $\frac{18.08g}{0.140 mole}$

= 129.14 g/mol

b) since pH = pKa

Then pKa of HA = 4.63

Ka = $10^{-4.63]$

= $2.3*10^{-5}$

$[A^-]equ = \frac{0.140M*1.00L}{1.00L+0.116L}$

$= \frac{0.140 mol}{1.116 L}$

= 0.1255 M

$K_a$ of HA = $2.3*10^{-5}$

$K_b = \frac{1.0*10^{-14}}{2.3*10^{-5}}$

$= 4.35*10^{-10}$

$A_{(aq)}$ + $H_2O_{(l)}$ $\rightleftharpoons$ $HA_{(aq)}$ + $OH^-_{(aq)}$

Initial 0.1255 0 0

Change - x + x + x

Equilibrium 0.1255 - x x x

$K_b = \frac{[HA][OH^-]}{[A^-]}$

$4.35*10^{-10} = \frac{[x][x]}{[0.1255-x]}$

As $K_b$ is very small, (o.1255 - x) = 0.1255

$x = \sqrt{0.1255*4.35*10^{-10}}$

[OH⁻] = $x = 7.4 *10^{-6}$

But pOH = - log [OH⁻]

= - log [ $7.4*10^{-6}$ ]

= 5.13

pH = 14.00 = 5.13

pH = 8.87

6 0

3 years ago