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3 years ago

1/2 * m * v ^ 2 = mgh

Chemistry

1 answer:

viktelen [127]3 years ago

4 0

C. the square root of 2gh

Explanation:

The square root of 2gh is equivalent to Torricelli's law in the equation.

given expression:

$\frac{1}{2}$ m v² = mgh

to find v, we make it the subject of the expression:

*multiply both sides of the equation by 2:

2 x( $\frac{1}{2}$ m v² )= 2 (mgh)

mv² = 2mgh

*cancel the mass, m appearing on both sides:

v² = 2gh

* find the square root of both side

v = √2gh

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Expression brainly.com/question/11207748

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A galvanic cell consists of a Cu(s)|Cu2+(aq) half-cell and a Cd(s)|Cd2+(aq) half-cell connected by a salt bridge. Oxidation occu

Kryger [21]

Answer:

$Cd(s)|Cd^{2+}(aq) || Cu^{2+}(aq)|Cu(s)$

Explanation:

A galvanic cell is composed of two electrodes immersed in a suitable electrolyte and connected via a salt bridge. One of the electrodes serves as a cathode where reduction or gain of electrons takes place. The other half cell functions as an anode where oxidation or loss of electrons occurs.

The representation is given by writing the anode on left hand side followed by its ion with its molar concentration. It is followed by a salt bridge. Then the cathodic ion with its molar concentration is written and then the cathode.

As it is given that cadmium acts as anode, it must be on the left hand side and copper must be on right hand side.

$Cd(s)|Cd^{2+}(aq) || Cu^{2+}(aq)|Cu(s)$

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4 years ago

What is the lower concentration limit (vol%) at which a mixture of ethanol in air can explode?

liubo4ka [24]

Answer:

Lower explosive limit (LEL) of ethanol = 3.3%

Explanation:

In the case of alcohol, ethanol presents certain fire hazards. Its momentary flash point is 55ºF (12.9ºC), while the momentary flash point of gasoline is -45ºF (-42.8ºC), and the E85 mixture ranges between -20ºF and -4ºF (between -28 , 9ºC and -20ºC), and has a wider range of flammability limits than gasoline. For emergency response teams, this implies that during a release of the typical ethanol / gasoline mixture, the fuel can be expected to behave like gasoline: It is heavier than air - as we mentioned earlier - and can produce vapors and form flammable mixtures in the air, under most environmental conditions.

General properties and comparison with other inflambles products:

Flash point momentary Gasoline = -45 ° F

<u>Ethanol</u> = 55 ° F

E 85 = between -20º and -4º F

<u>Flammability limits </u>

Lower explosive limit (LEL) of ethanol = 3.3%

Upper Explosive Limit (UEL) = 19%

Lower explosive limit (LEL) of the mixture E 85 = 1.4%

Upper Explosive Limit (UEL) 85 = 19%

Lower explosive limit (LEL) of gasoline = 1.4%

Upper Explosive Limit (UEL) = 7.6%

They have a wider range than gasoline

4 0

3 years ago

One mole of an ideal gas with a volume of 1.0 L and a pressure of 5.0 atm is allowed to expand isothermally into an evacuated bu

Deffense [45]

Answer:

w= - 1.7173 kJ, q= 1.7173 kJ, q(rev) = 1717.3 J = 1.7173 kJ.

Explanation:

Okay, from the question we are given the information below;

Number of moles, n= 1 mole; initial volume, v(1) = 1.0 litres (L); pressure (p) = 5atm, final volume(v2) = 2.0 Litres(L) ; the workdone, w= not given; the heat, q and q(rev)= not given and the gas was said to expand isothermally.

So, this question is a question from the part of chemistry known as thermodynamics. Therefore, grip yourself we are delving into thermodynamics 'waters' now.

For expansion isothermally; the workdone, w= -nRT ln v2/v1.

Where T= temperature= 25° C = 298 k and R= gas constant.

Therefore; workdone, w = - 1 × 8.314 × 298 × ln(2/1).

Workdone,w= - 1717.32204643. =

- 1717.3 Joules (J).

==> Workdone,w= - 1.7173 kJ.

Then, we are to find q. q can be solved by using the first law of thermodynamics, which by mathematical representation is:

∆U= q + w. Where ∆U= change in internal enegy. Since the question is dealing with isothermal expansion, there is this rule that says for an isothermal expansion ∆U = 0.

Hence, 0 =q + [- 1717.3 Joules (J)].

q=1717.3 J = 1.7173 kJ.

Finally, the q(rev) which is= nRT ln (v2/V1).

q(rev) = 1 × 8.314 × 298 ln (2/1).

q(rev) = 1717.3 J = 1.7173 kJ.

PS: please note the negative signs in the workdone and the positive sign in the q(rev).

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3 years ago