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3 years ago

How is electrolysis most commonly used to produce an energy source?

Chemistry

1 answer:

elena-s [515]3 years ago

4 0

Answer:

Splitting water molecules produces hydrogen gas, which is used to power machines through hydrogen fuel cells. ( B-)

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Will 1 gram of sugar or 1 gram of salt dissolve more quickly which one

Montano1993 [528]

1 gram of sugar because super molecules are bigger then the ions of dissolved salt

6 0

3 years ago

By mistake, a quart of oil was dumped into a swimming pool that measures 25.0 m by 30.0 m. The density of the oil was 0.750 g/cm

kondor19780726 [428]

The oil slick thick = 1.256 x 10⁻⁴ cm

<h3>Further explanation</h3>

Volume is a derivative quantity derived from the length of the principal

The unit of volume can be expressed in liters or milliliters or cubic meters

The conversion is

1 cc = 1 cm3

1 dm = 1 Liter

1 L = 1.06 quart

<em>so for 1 quart = 0.943 L</em>

$\tt 0.943~L=0.943\times 10^{-3}m^3$

Volume of oil dumped = volume of swimming pool

$\tt 0.943\times 10^{-3}~m^3=25\times 30\times h(h=thick)\\\\h=\dfrac{0.943\times 10^{-3}}{750~m^2}=1.257\times 10^{-6}~m=\boxed{\bold{1.256\times 10^{-4}~cm}}$

3 0

3 years ago

For an alloy that consists of 91.0 g copper, 111 g zinc, and 7.51 g lead, what are the concentrations (a) of Cu (in at%), (b) of

viktelen [127]

Answer:

$\% Cu=45.2\%\\\\\% Zn=53.6\%\\\\\% Pb=1.2\%$

Explanation:

Hello!

In this case, given the masses of copper, zinc and lead, it is possible to compute the moles via their atomic masses first:

$n_{Cu}=\frac{91.0gCu}{63.55g/mol}=1.432mol\\\\ n_{Zn}=\frac{111gZn}{65.41g/mol}=1.697mol\\\\n_{Pb}=\frac{7.51gPb}{207.2g/mol}=0.0362mol\\$

Now, we compute the atomic percentages as shown below:

$\% Cu=\frac{1.432}{1.432+1.697+0.0362}*100\% =45.2\%\\\\\% Zn=\frac{1.697}{1.432+1.697+0.0362}*100\%=53.6\%\\\\\% Pb=\frac{0.0362}{1.432+1.697+0.0362}*100\%=1.2\%$

Best regards!

4 0

2 years ago

The most useful ore of aluminum is bauxite, in which Al is present as hydrated oxides, Al2O3 * xH2O.

lesantik [10]

Answer:

44.7 kWh

Explanation:

Let's consider the reduction of Al₂O₃ to Al in the Bayer process.

6 e⁻ + 3 H₂O + Al₂O₃ → 2 Al + 6 OH⁻

We can establish the following relations:

The molar mass of Al is 26.98 g/mol.
2 moles of Al are produced when 6 moles of e⁻ circulate.
1 mol of e⁻ has a charge of 96468 c (Faraday's constant).
1 V = 1 J/c
1 kWh = 3.6 × 10⁶ J

When the applied electromotive force is 5.00 V, the energy required to produce 3.00 kg (3.00 × 10³ g) of aluminum is:

$3.00 \times 10^{3} gAl.\frac{1molAl}{26.98gAl} .\frac{6mole^{-}}{2molAl}.\frac{96468c}{1mole^{-}}.\frac{5.00J}{c}.\frac{1kWh}{3.6 \times 10^{6}J} =44.7kWh$

6 0

3 years ago

A light bulb has resistance 12 12 when the

Mkey [24]

Explanation:

Given :

V=6V

R=12 ohm

I=?

I=V/R

=6/12

=0.5 A

8 0

3 years ago