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chubhunter [2.5K]
3 years ago
14

What tendency is the standard reduction potential of an element based on?

Chemistry
2 answers:
Sedaia [141]3 years ago
7 0

Answer:

the tendency of an element to gain electrons

Explanation:

Electrode potential is considered to be the tendency of a metal to lose or gain electrons when dipped in its own ionic solution.

For example Zn dipped in ZnSO₄.

Now if a metal or element is losing electrons then it is considered as oxidation potential and it an element is gaining electrons then it is considered as reduction potential.

Standard word is related with standard condition of temperature, pressure and concentration.

standard temperature is 298 K

standard pressure is 1 bar

standard concentration is 1M.

Y_Kistochka [10]3 years ago
3 0
I think the answer is B :)
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guapka [62]
<span>Hey! Dependent variable parameter, I believe is the correct answer. (:</span>
3 0
3 years ago
What should be the mole fraction of O2 in the gas mixture the diver breathes in order to have the same partial pressure of oxyge
Nutka1998 [239]

The question is incomplete, here is the complete question:

A scuba diver is at a depth of 355 m, where the pressure is 36.5 atm.

What should be the mole fraction of O_2 in the gas mixture the diver breathes in order to have the same partial pressure of oxygen in his lungs as he would at sea level? Note that the mole fraction of oxygen at sea level is 0.209.

<u>Answer:</u> The mole fraction of oxygen in the gas mixture is 0.00573

<u>Explanation:</u>

To calculate the partial pressure, we use the equation given by Raoult's law, which is:

p_{A}=p_T\times \chi_{A}      ........(1)

where,

p_A = partial pressure of oxygen at sea level = ?

p_T = total pressure at sea level = 1.00 atm

\chi_A = mole fraction of oxygen at sea level = 0.209

Putting values in equation 1, we get:

p_{O_2}=1.00atm\times 0.209\\\\p_{O_2}=0.209atm

As, partial pressure of the oxygen in the diver's lungs is equal to the partial pressure of oxygen at sea level

We are given:

p_T=36.5atm\\p_{O_2}=0.209atm

Putting values in equation 1, we get:

0.209atm=36.5atm\times \chi_{O_2}\\\\\chi_{O_2}=\frac{0.209}{36.5}=0.00573

Hence, the mole fraction of oxygen in the gas mixture is 0.00573

3 0
3 years ago
A solution is prepared by dissolving 25.0 mL ethanol, CH3CH2OH (d=0.789 g/mL), in enough water to produce 250.0 mL solution. Wha
Svetllana [295]

The molarity = 1.716 M

<h3>Further explanation</h3>

Given

25 ml ethanol

d = 0.789 g/ml

250 ml solution

Required

The molarity

Solution

Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution

\large{\boxed {\bold {M ~ = ~ \frac {n} {V}}}

mass of ethanol :

= 25 ml x 0.789 g/ml

= 19.725 g

mol ethanol :

= 19.725 g : 46 g/mol

= 0.429

The molarity :

= 0.429 : 0.25 L

= 1.716 M

4 0
3 years ago
Pure iron has a mass of 5.00 grams what’s the calculation of its volume
mart [117]

Answer:

0.635 cm³

Explanation:

You divide the mass by the density of iron (7.87 g/cm³)

<em>D = m</em>/<em>V</em>                Multiply both sides by V.

<em>DV = m</em>                 Divide both sides by D.

<em>V</em> = <em>m</em>/<em>D</em>                Insert the values.

<em>V </em>= 5.00/7.87      Do the division.

<em>V</em> = 0.635 cm³

5 0
3 years ago
if a solution containing 18.96 g of mercury (II) perchlorate is allowed to react completely with a solution containing 6.256 g o
adell [148]
<span>The balanced chemical reaction would be written as:

</span>Hg(ClO4)2<span> + </span>Na2SO4<span> → 2 NaClO4 + </span>HgSO4(s)<span>

We are given the amounts of both of the reactants to be used for the reaction. We use these values as the starting point of the calculations. 

18.96 g </span>Hg(ClO4)2 ( 1 mol / 399.4912 g ) = 0.0475 mol  Hg(ClO4)2 
6.256 g Na2SO4 ( 1 mol / <span>142.04 g ) = 0.0440 mol Na2SO4

We have a 1 is to 1 ratio of the reactants. Thus, the limiting reactant would be Na2SO4. We use this value for calculating the amount of products in the reaction.

</span>0.0440 mol Na2SO4 ( 1 mol HgSO4 / 1 mol Na2SO4 ) ( 296.65 g / mol ) = 13.06 g HgSO4

mass of excess reactant = (0.0475 - 0.0440) mol ( 399.4912 g / 1 mol ) = 172.180 g Hg(ClO4)2
8 0
4 years ago
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