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WINSTONCH [101]

3 years ago

12

What quantity of aluminium is deposited when a current of 10A is passed through a solution of aluminium salt for 1930s? (All=27,

F=96500cmol)

1 answer:

Arlecino [84]3 years ago

6 0

Answer:

AL*3+ + 3e- =AL.

（10A×1930）÷96500=0.2mole e-

0.2÷3×27=1.8g（AL）

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Picric acid has been used in the leather industry and in etching copper. However, its laboratory use has been restricted because

olchik [2.2K]

Answer:

0.3023 M

Explanation:

Let Picric acid = $H_{picric}$

So, $H_{picric}$ + $H_2}O$ ⇄ $H_3}O^+$ + $Picric^-$

The ICE table can be given as:

$H_{picric}$ + $H_2}O$ ⇄ $H_3}O^+$ + $Picric^-$

Initial: 0.52 0 0

Change: - x + x + x

Equilibrium: 0.52 - x + x + x

Given that;

acid dissociation constant ( $K_a$ ) = 0.42

$K_a = \frac{[H_3O^+][Picric^-]}{H_{picric}}$

$0.42 = \frac{[x][x]}{0.52-x}}$

$0.42 = \frac{[x]^2}{0.52-x}}$

0.42(0.52-x) = x²

0.2184 - 0.42x = x²

x² + 0.42x - 0.2184 = 0 -------------------- (quadratic equation)

Using the quadratic formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$ ; ( where +/- represent ± )

= $\frac{-0.42+/-\sqrt{(0.42)^2-4(1)(-0.2184)} }{2*1}$

= $\frac{-0.42+/-\sqrt {0.1764+0.8736} }{2}$

= $\frac{-0.42+\sqrt {1.0496} }{2}$ OR $\frac{-0.42-\sqrt {1.0496} }{2}$

= $\frac{-0.42+1.0245}{2}$ OR $\frac{-0.42-1.0245}{2}$

= $\frac{0.6045}{2}$ OR $-\frac{1.4445}{2}$

= 0.30225 OR - 0.72225

So, we go by the +ve integer that says:

x = 0.30225

x = [ $H_3}O^+$ ] = [ $Picric^-$ ] = 0.3023 M

∴ the value of [H3O+] for an 0.52 M solution of picric acid = 0.3023 M (to 4 decimal places).

6 0

3 years ago

A sample of copper absorbs 4.31E+1 kJ of heat, resulting in a temperature rise of 6.71E+1 °C. Determine the mass (in kg) of the

olga2289 [7]

Answer: 1.67 kg

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

$Q=m\times c\times \Delta T$

Q = Heat absorbed= $4.31\times 10^1kJ$ = $43100J$ (1kJ=1000J)

m= mass of substance = ?

c = specific heat capacity = $0.385J/g^0C$

Change in temperature , $\Delta T=T_f-T_i=6.71\times 10^1^0C=67.1^0C$

Putting in the values, we get:

$43100J=m\times 0.385J/g^0C\times 67.1^0C$

$m=1670g=1.67kg$ (1kg=1000g)

Thus the mass (in kg) of the copper sample is 1.67

3 0

3 years ago

A water bath in a physical chemistry lab is 1.75 m long, 0.730 m wide, and 0.650 m deep. If it is filled to within 2.27 inches f

german

It is 0.720 meters cause if the manufacturers of liters contain 2.27 inches it would make a deeply filled of 0.660

7 0

3 years ago

Read 2 more answers

In a sample of oxygen gas at room temperature, the

valina [46]

Answer: The Kinetic Molecular Theory of Matter

Explanation: The answer is: B. Collisions between gas particles are elastic; there is no net gain or loss of kinetic energy.

The Kinetic Molecular Theory of Matter

7 0

2 years ago

12.A piece of magnesium is in the shape of a cylinder with a height of 5.62 cm

yaroslaw [1]

Answer:

Density, d = 1.779 g/cm³

Explanation:

The density of a material is given by its mass per unit volume.

Here, height of a piece of magnesium cylinder, h = 5.62 cm

Its diameter, d = 1.34 cm

Radius = 0.67 cm

Volume of he cylinder,

$V=\pi r^2 h\\\\\text{Putting the value of r and h, we get :}\\\\V=(\pi \times (0.67)^2\times 5.62)\ cm^3$

$d=\dfrac{m}{V}\\\\d=\dfrac{14.1\ g}{(\pi \times (0.67)^2\times 5.62)\ cm^3}\\\\d=1.779\ g/cm^3$

So, the density of the sample is 1.779 g/cm³.

4 0

3 years ago