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2 years ago

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The development of a nuclear power plant at an ocean site is expected to produce an enormous amount of electric power for a larg

e population. Ocean water will be used to cool the plant and then returned to the ocean. What unintended consequences should be analyzed before starting construction?
A. The effects of nutrients on algae
B. The effects of warmed water on aquatic life
C. The impacts of other energy-producing resources
D. The impacts of human populations on coastal areas

2 answers:

Oksanka [162]2 years ago

8 0

Answer:

B. The effects of warmed water on aquatic life

Explanation:

quizizz

agasfer [191]2 years ago

4 0

I think the answer is B

PS: Sorry if I’m wrong

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2 years ago

What is the last step in the scientific method?

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2 years ago

How many total atoms are in 0.680 g of P₂O5?

Ludmilka [50]

Answer: 2.88× $10^{24}$ atoms $P_{2}O_{5}$

Explanation: First, using stoichiometry, we must convert this from grams to moles, then from moles to atoms.

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$0.680g P_{2} O_{5} *\frac{1molP_{2} O_{5} }{141.948gP_{2} O_{5} } =4.79mol P_{2} O_{5}$

2. Now that it is converted to moles, we must convert it to atoms by multiplying it by Avogadro's number.

$4.79molP_{2} O_{5} *\frac{6.022X10^{23} }{1mol} =2.88*10^{24}$

With this information, we know that there are $2.88X10^{24}$ total atoms in 0.680 grams $P_{2} O_{5}$ .

I hope this helps! Pls give brainliest!! :)

8 0

2 years ago

For the reaction 2Co3+(aq)+2Cl−(aq)→2Co2+(aq)+Cl2(g). E∘=0.483 V what is the cell potential at 25 ∘C if the concentrations are [

sveta [45]

Explanation:

The given data is as follows.

$E^{o}$ = 0.483, $[Co^{3+}]$ = 0.173 M,

$[Co^{2+}]$ = 0.433 M, $[Cl^{-}]$ = 0.306 M,

$P_{Cl_{2}}$ = 9.0 atm

According to the ideal gas equation, PV = nRT

or, P = $\frac{n}{V}RT$

Also, we know that

Density = $\frac{mass}{volume}$

So, P = MRT

and, M = $\frac{P}{RT}$

= $\frac{9.0 atm}{0.0820 L atm/mol K \times 298 K}$

= $\frac{9.0}{24.436}$

= 0.368 mol/L

Now, we will calculate the cell potential as follows.

E = $E^{o} - \frac{0.0591}{n} log \frac{[Co^{2+}]^{2}[Cl_{2}]}{[Co^{3+}][Cl^{-}]^{2}}$

= $0.483 - \frac{0.0591}{2} log \frac{(0.433)^{2}(0.368)}{(0.173)(0.306)^{2}}$

= $0.483 - 0.02955 log \frac{0.0689}{0.0162}$

= $0.483 - 0.02955 \times 0.628$

= 0.483 - 0.0185

= 0.4645 V

Thus, we can conclude that the cell potential of given cell at $25^{o}C$ is 0.4645 V.

4 0

3 years ago