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NeTakaya
3 years ago
12

will mark brainlest 1) Now that you have studied the facts about the Severn River, you are ready to make a decision. What questi

ons would you want answered if you were to determine whether to build a tidal power plant at the Severn River?
Chemistry
1 answer:
chubhunter [2.5K]3 years ago
5 0

Answer:

The questions I would like to answer are:

Will the productivity of the river severn offset the cost of installing the plant?

Is it possible to build a dam on site?

Is it possible to establish submerged turbines in this location?

How would human life and nature be impacted by the plant?

Explanation:

Tidal power plants are signals that use tidal energy to generate electricity. The installation of this type of plant is quite expensive and involves the establishment of dams on site to take advantage of the differences between high and low tides and the establishment of submerged turbines to take advantage of sea currents.

To decide whether I would decide to build a tidal plant on the Severn River, I would need to evaluate a few things to make sure that the project is viable. For that I would ask myself some questions like:

Will the productivity of the river severn offset the cost of installing the plant? (If the installation costs are greater than the capacity of this river to promote high generation of electricity, it would be better to abandon the project)

Is it possible to build a dam on site? Is it possible to establish submerged turbines in this location? (It is necessary to consider the physical and chemical characteristics of the place to know if it is possible to install these two equipment and in a way that they work efficiently)

How would human life and nature be impacted by the plant? (Projects like this include major environmental and social impacts. These impacts must be analyzed and ways to mitigate them must be considered and included in the project.)

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5400 L to KL<br> With work??
Gnesinka [82]

Answer:

5.4KL

Explanation:

Divide by 1000

3 0
2 years ago
Very urgent/easy chem question easy points
aev [14]
Sodium chloride is the answer
8 0
3 years ago
Read 2 more answers
Consider the reversible reaction.
Dmitry [639]

Answer:

Option D is correct.

The concentrations of both PCl₅ and PCl₃ are changing at equilibrium

Explanation:

Chemical equilibrium during a reversible chemical reaction, is characterised by an equal rate of forward reaction and backward reaction. It is better described as dynamic equilibrium.

This is because, the concentration of the elements and compounds involved in the reversible chemical reaction at equilibrium changes, but the rate of change of the reactants is always equal to the rate of change of products.

Hence, the concentration of reactants and products, such as PCl₅ and PCl₃ are allowed to change at equilibrium, but alas, the rate of forward reaction must always match the rate of backward reaction for the process to remain in a state of Chemical equilibrium.

Hope this Helps!!!

3 0
3 years ago
Using the data, which of the following is the rate constant for the rearrangement of methyl isonitrile at 320 °C? (HINT: the act
algol13

Answer:

D) 2.3 x 10⁻¹ s⁻¹

Explanation:

The rate constant is related to the activation energy through the formula:

k= Ae^(-Eₐ /RT)

where A is the collision factor, Eₐ the activation energy, R is the gas constant ( 8.314 J/Kmol ) , and T is the temperature (K)

So a plot of lnk versus 1/T ( Arrehenius plot ) gives us a straight line with slope equal -Eₐ/R and intercept lnA

lnk = -(Eₐ/T)(1/T) + lnA

which has the form y= mx + b

In this problem, we can use the data provided to:

a) Using a calculator determine the slope and intercept and then calculate the value of rate constant at 320 ºC, or

b) Plot the data and determine the equation of the best line , and answer the question for k @ 320 ºC by reading the value from the plot.

Once you do the plot, the resulting equation is:

y = - 19 x 10³ x + 30,582 ( R² = 0.999 )

So for T = 320 + 273 K = 593 K

Y = 19 x 10³ X + 30.58

So for T = (320 + 273)K = 593 K

Y =  -19 x 10³ ( 1/593) + 30.58 = -32.04 +30.58 = - 1.46

and then since

y = lnk ⇒ e^y = k

k= e^-1.46 = 2.3 x 10⁻¹ s⁻¹

Note: there is an error of transcription in the value for T = 472.1 ( 1/T = 2.118 x 10⁻³  and  not 2.228 x 10⁻³). You can  recognize this mistake if you plot the data and notice it produces an outlier.

5 0
3 years ago
How many moles of Cu(OH)2 are soluble in 1L of sodium hydroxide (NaOH) when the pH is 8.23?
Morgarella [4.7K]

Answer:

4.96E-8 moles of Cu(OH)2

Explanation:

Kps es the constant referring to how much a substance can be dissolved in water. Using Kps, it is possible to know the concentration of weak electrolytes. Then, pKps is the minus logarithm of Kps.

Now, we know that sodium hydroxide (NaOH) is a strong electrolyte, who is completely dissolved in water. Therefore the pH depends only on OH concentration originating from NaOH. Let us to figure out how much is that OH concentration.

pH= -log[H]\\pH= -log (\frac{kw}{[OH]})

8.23 = - log(\frac{Kw}{[OH]} \\10^{-8.23} = Kw/[OH]\\ [OH] = Kw/10^{-8.23}

[OH]=1.69E-6

This concentration of OH affects the disociation of Cu(OH)2. Let us see the dissociation reaction:

Cu(OH)_2 -> Cu^{2+} + 2OH^-

In the equilibrum, exist a concentration of OH already, that we knew, and it will be added that from dissociation, called "s":

The expression for Kps is:

Kps= [Cu^{2+}] [OH]^2

The moles of (CuOH)2 soluble are limitated for the concentration of OH present, according to the next equation.

Kps= s*(2s+1.69E-6)^2

"s" is the soluble quantity of Cu(OH)2.

The solution for this third grade equation is s=4.96E-8 mol/L

Now, let us calculate the moles in 1 L:

moles Cu(OH)_2 = 4.96E-8 mol/L * 1 L = 4.96E-8 moles

7 0
3 years ago
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