1

What is the oh- in a solution with a poh of 5.71

Rudik [331]

Answer:- The hydroxide ion concentration of the solution is $1.95*10^-^6$ .

Solution:- The formula used to calculate pOH from hydroxide ion is:

$pOH=-log[OH^-]$

When pOH is given and we are asked to calculate hydroxide ion concentration then we multiply both sides by negative sign and take antilog and what we get on doing this is:

$[OH^-]=10^-^p^O^H$

pOH is given as 5.71 and we are asked to calculate hydrogen ion concentration. Let's plug in the given value in the formula:

$[OH^-]=10^-^5^.^7^1$

$[OH^-]$ = 0.00000195 or $1.95*10^-^6$

So, the hydroxide ion concentration of the solution is $1.95*10^-^6$ .

3 0

3 years ago

What part

-BARSIC- [3]

Answer:

What arrow????

6 0

3 years ago

Calculate for the following electrochemical cell at 25°C, Pt H2(g) (1.0 atm) H (0.010 M || Ag (0.020 M) Ag if E (H) - +0.000 V a

viva [34]

Answer : The correct option is, (b) +0.799 V

Solution :

The values of standard reduction electrode potential of the cell are:

$E^0_{[H^{+}/H_2]}=+0.00V$

$E^0_{[Ag^{+}/Ag]}=+0.799V$

From the cell representation we conclude that, the hydrogen (H) undergoes oxidation by loss of electrons and thus act as anode. Silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : $H_2\rightarrow 2H^{+}+2e^-$

Reaction at cathode (reduction) : $Ag^{+}+1e^-\rightarrow Ag$

The balanced cell reaction will be,

$H_2+2Ag^{+}\rightarrow 2H^{+}+2Ag$

Now we have to calculate the standard electrode potential of the cell.

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=E^o_{[Ag^{+}/Ag]}-E^o_{[H^{+}/H_2]}$

$E^o_{cell}=(+0.799V)-(+0.00V)=+0.799V$

Therefore, the standard cell potential will be +0.799 V

4 0

3 years ago

Consider the following reaction:

adell [148]

Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

2H₂O₂ ⟶ 2H₂O + O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

d[H₂O]/dt = 2d[O₂]/dt = 2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = 6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂

(a) Initial moles of H₂O₂

$\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }$

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

$\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }$

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

$\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}$

8 0

3 years ago

non metals are generally non conductors of electricity. Graphite is non metal but is is used as electrodes in batteries.

expeople1 [14]

Graphite conduct electricity because it contains delocalize election (free election ) the election move through the graphite

4 0

3 years ago