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evablogger [386]

2 years ago

3

Incoming waves force air upward, which spins a turbine, which in turn spins a generator to produce electricity. What factor woul

d increase the electricity production of this apparatus by the greatest amount?

1 answer:

schepotkina [342]2 years ago

4 0

Answer:

Wind speed largely determines the amount of electricity generated by a turbine. Higher wind speeds generate more power because stronger winds allow the blades to rotate faster. Faster rotation translates to more mechanical power and more electrical power from the generator.

Explanation:

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HIPVs can cause what two more severe illnesses?

Nat2105 [25]

One of them are Cancer

4 0

2 years ago

Calculate the daily aluminum production of a 150,000 [A] aluminum cell that operates at a faradaic efficiency of 89%. The cell r

Gala2k [10]

Explanation:

It is known that in one day there are 24 hours. Hence, number of seconds in 24 hours are as follows.

$24 \times 3600 sec$

Hence, total charge passed daily is calculated as follows.

$150,000 \times 24 \times 3600 sec$

And, number of Faraday of charge is as follows.

$\frac{150,000 \times 24 \times 3600 sec}{96500}$

= 134300.52 F

The oxidation state of aluminium in $Al_{2}O_{3}$ is +3.

$Al^{3+} + 3e^{-} \rightarrow Al(s)$

So, if we have to produce 1 mole of Al(s) we need 3 Faraday of charge.

Therefore, from 134300.52 F the moles of Al obtained with 89% efficiency is calculated as follows.

$\frac{134300.52 F}{3} \times \frac{89}{100}$

= 39842.487 mol

or, = $3.9842 \times 10^{4} mol$

Molar mass of Al = 27 g/mol

Therefore, mass in gram will be calculated as follows.

Mass in grams = $3.9842 \times 10^{4} mol \times 27$

= $107.57 \times 10^{4} g$

= 1075.7 kg/day

Thus, we can conclude that the daily aluminum production of given aluminium is 1075.7 kg/day.

8 0

2 years ago

How long would it take for 1.50 mol of water at 100.0 ∘c to be converted completely into steam if heat were added at a constant

Aleksandr-060686 [28]

To determine the time it takes to completely vaporize the given amount of water, we first determine the total heat that is being absorbed from the process. To do this, we need information on the latent heat of vaporization of water. This heat is being absorbed by the process of phase change without any change in the temperature of the system. For water, it is equal to 40.8 kJ / mol.

Total heat = 40.8 kJ / mol ( 1.50 mol ) = 61.2 kJ of heat is to be absorbed

Given the constant rate of 19.0 J/s supply of energy to the system, we determine the time as follows:

Time = 61.2 kJ ( 1000 J / 1 kJ ) / 19.0 J/s = 3221.05 s

5 0

2 years ago

Read 2 more answers

The decomposition of nitramide, O2NNH2, in water has the chemical equation and rate law O2NNH2(aq)⟶N2O(g)+H2O(l)rate=k[O2NNH2][H

FinnZ [79.3K]

Answer:

It can be concluded that the third step of the reaction is very fast, in this way, it does not contribute to the rate law

Explanation:

Please, observe the solution in the attached Word document.

7 0

2 years ago

Coal gasification is a multistep process to convert coal into cleaner-burning fuels. In one step, a coal sample reacts with supe

ddd [48]

Answer :

The enthalpy of reaction is, -187.6 kJ/mol

The total heat will be, -2251 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

(a) The formation of $CH_4$ will be,

$2C(coal)+2H_2O(g)\rightarrow CH_4(g)+CO_2(g)$ $\Delta H_{rxn}=?$

The intermediate balanced chemical reaction will be,

(1) $C(coal)+H_2O(g)\rightarrow CO(g)+H_2(g)$ $\Delta H_1=29.7kJ$

(2) $CO(g)+H_2O(g)\rightarrow CO_2(g)+H_2(g)$ $\Delta H_2=-41kJ$

(3) $CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)$ $\Delta H_3=-206kJ$

We are multiplying equation 1 by 2 and then adding all the equations, we get :

(b) The expression for enthalpy of reaction will be,

$\Delta H_{rxn}=2\times \Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H_{rxn}=(2\times 29.7)+(-41)+(-206)$

$\Delta H_{rxn}=-187.6kJ/mol$

Therefore, the enthalpy of reaction is, -187.6 kJ/mol

(c) Now we have to calculate the total heat.

$\Delta H=\frac{q}{n}$

or,

$q=\Delta H\times n$

where,

$\Delta H$ = enthalpy change = -187.6 kJ/mol

q = heat = ?

n = number of moles of coal = $\frac{1.00\times 1000g}{12.00g/mol}=83.33mol$

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

$q=(-187.6kJ/mol)\times (83.33mol)=-2.251kJ$

Thus, the total heat will be, -2251 kJ

4 0

2 years ago