All categories

3 years ago

In large generators in power plants, ___________ rotate inside a coil of wire to produce an electric current

Chemistry

1 answer:

lapo4ka [179]3 years ago

8 0

<h3><u>Answer;</u></h3>

Magnets

In large generators in power plants, <u>magnets</u> rotate inside a coil of wire to produce an electric current.

<h3><u>Explanation;</u></h3>

Generators are devices that use electromagnetic induction to convert mechanical energy to electrical energy.
In power plants, large wheel known as turbine that rotates when pushed by steam,wind, or water and provides mechanical energy to a generator.
<u><em>Generators are used in power plants and have very large quantities of copper wire spinning around inside very large magnets, at very high speeds. A turbine spins the shaft that runs the generator. Eventually, electric current is generated.</em></u>

You might be interested in

During photosynthesis,blank energy is transformed into blank energy

11Alexandr11 [23.1K]

Answer:

I am also beginners I don't know I am only completing this answer and sorry!!!!!!

8 0

2 years ago

Read 2 more answers

Which best describes the process that occurs when liquid water becomes<br> ice?

garri49 [273]

Answer:

Freezing

Explanation:

When a liquid goes to a solid, this process is called freezing.

7 0

3 years ago

Read 2 more answers

Which processes of the water cycle involve the movement of water in the liquid state?

dangina [55]

Answer:

Surface runoff and condensation

Explanation:

Let's define each of the given processes in order to understand them better:

evaporation is a process in which liquid phase transforms into a gas phase;
precipitation is a process in which we produce a solid phase, usually this is the case when we precipitate a salt out of a solution, analogy of precipitation for water would be transformation from a liquid to a solid phase, such as freezing;
surface runoff is a process in which water flows over the surface of a land without any change in its phase;
condensation is a process in which a gas transforms into a liquid.

All in all, notice that surface runoff keeps water in its liquid state, while all the other three options consider phase change. The only phase change of interest is condensation: we produce liquid water from water vapor and then we can analyze its movement in the liquid state.

6 0

3 years ago

A man walked 1560 meters in 16000 seconds what is the speed at which he walked

Maslowich

In what meters? miles an hour? i can’t answer unless i know that

5 0

2 years ago

Consider the titration of 100.0 mL of 0.280 M propanoic acid (Ka = 1.3 ✕ 10−5) with 0.140 M NaOH. Calculate the pH of the result

Murljashka [212]

Answer:

(a) 2.7

(b) 4.44

(d) 5.363

(e) 5.570

(f) 12.30

Explanation:

Here we have the titration of a weak acid with the strong base NaOH. So in part (a) simply calculate the pH of a weak acid ; in the other parts we have to consider that a buffer solution will be present after some of the weak acid reacts completely the strong base producing the conjugate base. We may even arrive to the situation in which all of the acid will be just consumed and have only the weak base present in the solution treating it as the pOH and the pH = 14 -pOH. There is also the possibility that all of the weak base will be consumed and then the NaOH will drive the pH.

Lets call HA propanoic acid and A⁻ its conjugate base,

(a) pH = -log √ (HA) Ka =-log √(0.28 x 1.3 x 10⁻⁵) = 2.7

(b) moles reacted HA = 50 x 10⁻³ L x 0.14 mol/L = 0.007 mol

mol left HA = 0.28 - 0.007 = 0.021

mol A⁻ produced = 0.007

Using the Hasselbalch-Henderson equation for buffer solutions:

pH = pKa + log ((A⁻/)/(HA)) = -log (1.3 x 10⁻⁵) + log (0.007/0.021)= 4.89 + (-0.48) = 4.44

mol HA left = 0.028 -0.014 = 0.014 mol

mol A⁻ produced = 0.014

pH = -log (1.3 x 10⁻⁵) + log (0.014/0.014) = 4.886

(d) mol HA reacted = 150 x 10⁻³ L x x 0.14 mol/L = 0.021 mol

mol HA left = 0.028 - 0.021 = 0.007

mol A⁻ produced = 0.021

pH = -log (1.3 x 10⁻⁵) + log (0.021/0.007) = 5.363

(e) mol HA reacted = 200 x 10⁻³ L x 0.14 mol/L = 0.028 mol

mol HA left = 0

Now we only a weak base present and its pH is given by:

pH = √(kb x (A⁻) where Kb= Kw/Ka

Notice that here we will have to calculate the concentration of A⁻ because we have dilution effects the moment we added to the 100 mL of HA, 200 mL of NaOH 0.14 M. (we did not need to concern ourselves before with this since the volumes cancelled each other in the previous formulas)

mol A⁻ = 0.028 mOl

Vol solution = 100 mL + 200 mL = 300 mL

(A⁻) = 0.028 mol /0.3 L = 0.0093 M

and we also need to calculate the Kb for the weak base:

Kw = 10⁻¹⁴ = ka Kb ⇒ Kb = 10⁻¹⁴/1.3x 10⁻⁵ = 7.7 x 10⁻ ¹⁰

pH = -log (√( 7.7 x 10⁻ ¹⁰ x 0.0093) = 5.570

(f) Treat this part as a calculation of the pH of a strong base

moles of OH = 0.250 L x 0.14 mol = 0.0350 mol

mol OH remaining = 0.035 mol - 0.028 reacted with HA

= 0.007 mol

(OH⁻) = 0.007 mol / 0.350 L = 2.00 x 10 ⁻²

pOH = - log (2.00 x 10⁻²) = 1.70

pH = 14 - 1.70 = 12.30

4 0

3 years ago