What is the property of a substance for ml

Current passes through a solution of sodium chloride. In 1.00 second, 2.68×1016Na+ ions arrive at the negative electrode and 3.9

EleoNora [17]

Answer : The current passing between the electrodes is, $1.056\times 10^{-2}A$

Explanation :

First we have to calculate the charge of sodium ion.

$q=ne$

where,

q = charge of sodium ion

n = number of sodium ion = $2.68\times 10^{16}$

e = charge on electron = $1.6\times 10^{-19}C$

Now put all the given values in the above formula, we get:

$q=(2.68\times 10^{16})\times (1.6\times 10^{-19}C)=4.288\times 10^{-3}C$

Now we have to calculate the charge of chlorine ion.

$q'=ne$

where,

q' = charge of chlorine ion

n = number of chlorine ion = $3.92\times 10^{16}$

e = charge on electron = $1.6\times 10^{-19}C$

Now put all the given values in the above formula, we get:

$q'=(3.92\times 10^{16})\times (1.6\times 10^{-19}C)=6.272\times 10^{-3}C$

Now we have to calculate the current passing between the electrodes.

$I=\frac{q}{t}+\frac{q'}{t}$

$I=\frac{4.288\times 10^{-3}}{1.00}+\frac{6.272\times 10^{-3}}{1.00}$

$I=1.056\times 10^{-2}A$

Thus, the current passing between the electrodes is, $1.056\times 10^{-2}A$

4 0

3 years ago

Why did scientists need to establish an international language for elements

Ulleksa [173]

So it could be used in every country(different languages) yet still understood

5 0

3 years ago

The osmotic pressure of an aqueoussolution of urea at 300 K is

den301095 [7]

<u>Answer:</u> The freezing point of solution is -0.09°C

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=imRT$

where,

$\pi$ = osmotic pressure of the solution = 120 kPa

i = Van't hoff factor = 1 (for non-electrolytes)

m = concentration of solute in terms of molality = ?

R = Gas constant = $8.31\text{L kPa }mol^{-1}K^{-1}$

T = temperature of the solution = 300 K

Putting values in above equation, we get:

$120kPa=1\times m\times 8.31\text{ L kPa }mol^{-1}K^{-1}\times 300K\\\\m=0.05m$

To calculate the depression in freezing point, we use the equation:

$\Delta T=i\times K_f\times m$

where,

i = Vant hoff factor = 1 (for non-electrolytes)

$K_f$ = molal freezing point depression constant = 1.86°C/m

m = molality of solution = 0.05 m

Putting values in above equation, we get:

$\Delta T=1\times 1.86^oC/m\times 0.05m\\\\\Delta T=0.09^oC$

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

The equation used to calculate freezing point of solution is:

$\Delta T=\text{freezing point of water}-\text{freezing point of solution}$

where,

$\Delta T$ = Depression in freezing point = 0.09 °C

Freezing point of water = 0°C

Freezing point of solution = ?

Putting values in above equation, we get:

$0.09^oC=0^oC-\text{Freezing point of solution}\\\\\text{Freezing point of solution}=-0.09^oC$

Hence, the freezing point of solution is -0.09°C

4 0

3 years ago

What is the molarity of a solution made by dissolving 20.0 g of H3PO4 in 50.0 mL of solution

Rom4ik [11]

(20*1000)÷(molecular weight of H3po4*50)

8 0

3 years ago

What is the term for a combination of two or more substances in any proportions in which the substances do not combine chemicall

Evgesh-ka [11]

A) mixture because they are not chemically combined

8 0

3 years ago