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2 years ago

Which best describes a difference between energy transformations in power plants and dams?

2 answers:

5 0

I think the answer would be the last option. The difference between the energy transformations in power plants and dams would be that dams are the ones who uses mechanical energy to produce electricity. It uses turbines to convert mechanical to electrical.

Fed [463]2 years ago

5 0

Answer:

The answer is A

Explanation:

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Four different solutions (I, II, III, and IV) are labeled on the pH scale below. A p H scale is shown, fading from red at the le

andreyandreev [35.5K]

Answer: It has the highest number of hydroxide ions

Explanation: I took the review

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2 years ago

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Gina wants to use models to better understand how the types of bonds in a molecule relate to the presence of geometric isomers.

vazorg [7]

The answer to this question is D! The ball and stick model! Hope this helps :)

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2 years ago

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A 40.0 mL sample of 0.25 M KOH is added to 60.0 mL of 0.15 M Ba(OH)2. What is the molar concentration of OH-(aq) in the resultin

solmaris [256]

Answer:

C) 0.28 M

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Potassium hydroxide will furnish hydroxide ions as:

$KOH\rightarrow K^{+}+OH^-$

Given :

For Potassium hydroxide :

Molarity = 0.25 M

Volume = 40.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 40.0×10⁻³ L

Thus, moles of hydroxide ions furnished by Potassium hydroxide is same as the moles of Potassium hydroxide as shown below:

$Moles =0.25 \times {40.0\times 10^{-3}}\ moles$

Moles of hydroxide ions by Potassium hydroxide = 0.01 moles

Barium hydroxide will furnish hydroxide ions as:

$Ba(OH)_2\rightarrow Ba^{2+}+2OH^-$

Given :

For Barium hydroxide :

Molarity = 0.15 M

Volume = 60.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 60.0×10⁻³ L

Thus, moles of hydroxide ions furnished by Barium hydroxide is twice the moles of Barium hydroxide as shown below:

$Moles =2\times 0.15 \times {60.0\times 10^{-3}}\ moles$

Moles of hydroxide ions by Barium hydroxide = 0.018 moles

Total moles = 0.01 moles + 0.018 moles = 0.028 moles

Total volume = 40.0×10⁻³ L + 60.0×10⁻³ L = 100×10⁻³ L

Concentration of hydroxide ions is:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity_{OH^-}=\frac{0.028 }{100\times 10^{-3}}$

The final concentration of hydroxide ion = 0.28 M

5 0

2 years ago

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Consider a 20.0 % (m/v) solution. how can this be written as a conversion factor?

White raven [17]

A conversion factor is a fraction or a ratio representing a relationship of two different measurement values. To write 20% m/v to a conversion factor, we need to remember that a percent is a value that represents the amount of a part per 100 units of the whole. M/v in the given value represents that the percentage is by mass per volume. So, to write it as a conversion factor, we do as follows:

20% m/v = 20 mass units / 100 volume units = 1 mass units / 5 volume units

Usually units of this are in g per L. So, it is equivalent to 1 g / 5 L

4 0

2 years ago

How many grams of (nh4)3po4 are needed to make 0.250 l of 0.150 m (nh4)3po4? hints how many grams of (nh4)3po4 are needed to mak

NeX [460]

There are a number of ways to express concentration of a solution. This includes molarity. Molarity is expressed as the number of moles of solute per volume of the solution. We calculate the mass of the solute by first determining the number of moles needed. And by using the molar mass, we can convert it to units of mass.

Moles (nh4)3po4 = 0.250 L (0.150 M) = 0.0375 moles (nh4)3po4
Mass = 0.0375 mol (nh4)3po4 (149.0867 g / mol) = 5.59 g (nh4)3po4

5 0

2 years ago