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3 years ago

What are two sources of salt in ocean water

1 answer:

7 0

Hello there!

The two sources of salt in ocean water are:

Sea salts and mines.

As always, it is my pleasure to help students like you!

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A voltaic cell is constructed in which the anode is a Zn|Zn2+ half cell and the cathode is a Fe2+|Fe3+ half cell. The half-cell

zmey [24]

Answer:

See below explanation

Explanation:

Checking the reduction potencials:

Zn⁺²/Zn° , E° = -0.76 V

Fe⁺²/Fe° , E° = -0.44 V

For which Fe⁺² will be the ion that will reduce, and Zn° will lose electrons

ANODE: Zn° ⇄ Zn⁺² + 2e⁻

CATODE: Fe⁺² + 2e⁻ ⇄ Fe°

NET CELL REACTION: FeSO₄ (ac) + Zn° ⇄ Fe° + ZnSO₄ (ac)

In the external circuit, electrons will migrate from anode (Zn|Zn⁺²) to catode (Fe|Fe⁺²), and in the salt bridge; anions migrate from Fe⁺²|Fe compartment to Zn|Zn⁺² compartment

7 0

3 years ago

¿Qué es un enlace químico?

fredd [130]

Answer:

Un enlace químico es una atracción duradera entre átomos, iones o moléculas que permite la formación de compuestos químicos.

Explanation:

4 0

4 years ago

If the products of a reaction are Co2 and 2H20, what is known about the<br>reactants?

Svetradugi [14.3K]

Answer:

Reactant : A combustion of hydrocarbon.

Explanation:

It is known that when hydrocarbons are involved in combustion, they produce carbon dioxide and water.

CxHy + (x+y/4)O2 ===> xCO2 + y/2H2O

4 0

3 years ago

Solid lead acetate is slowly added to 75.0 mL of a 0.0492 M sodium sulfate solution. What is the concentration of lead ion requi

Firdavs [7]

Answer:

The concentration of lead ion required to just initiate precipitation is - $2.37\times10^-^5 M$

Explanation:

Lets calculate -:

Solubility equilibrium -: $PbI_2(s)$ ⇄ $Pb^2^+ (aq) + 2I^- (aq)$

Solubility product of $PbI_2$ , $Q=[Pb^2^+]_i_n_i_t_i_a_l$ $[I^-]^2_i_n_i_t_i_a_l$ $=9.8\times10^-^9$

Concentration of $I^-$ $=[KI]=0.0492M$

When the ionic product exceeds the solubility product , precipitation of salt takes place .

$Q_s_p\geq K_s_p$

$[Pb^2^+]_i_n_i_t_i_a_l$ $[I^-]^2_i_n_i_t_i_a_l$ $\geq 9.8\times10^-^9$

$[Pb^2^+]_i_n_i_t_i_a_l$ $[0.0492]^2$ $\geq 9.8\times10^-^9$

$[Pb^2^+]_i_n_i_t_i_a_l$ $\geq \frac{9.8\times10^-^9}{[0.0492]^2}$

$[Pb^2^+]_i_n_i_t_i_a_l$ $\geq$ $\frac{9.8\times10^-^9}{2.42\times10^-^3}$

$[Pb^2^+]_i_n_i_t_i_a_l$ $\geq$ $2.37\times10^-^5 M$

Thus , $PbI_2$ will start precipitating when $[Pb^2^+]_i_n_i_t_i_a_l$ $\geq 2.37\times10^-^5 M$ .

3 0

3 years ago

If carbon dioxide is broken down. what element will it give?

katrin2010 [14]

Carbon dioxide is CO2...if it’s broken down it will be C and O which are carbon and oxygen

7 0

3 years ago