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3 years ago

What equation explains the relation between amperes, watts and volts?

2 answers:

6 0

W*V i believe because it comes down to this :
I is the current, W is wattage V is volts and the * is the thing that represents the Amperes (i’m not 100% but this is my best)

Ann [662]3 years ago

5 0

I’m pretty sure W*V
i wish you luck :)

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Match the scientist with their achievement:

Tju [1.3M]

Answer:

Frederich Miescher- first person to isolate DNA and RNA

Frederick Griffith- first to demonstrate horizontal transmission of dna using bacteria

Gregor Mendel- documented and demonstrated inheritance patterns

Thomas Hunt Morgan- identified chromosomes as the structures responsible for inheritance

Joachim Hammerling- demonstrated that the hereditary information of of eukaryotes is contained within the nucleus

Alfred Hershey and Martha Chase- demonstrated that dna not protein was the molecule responsible for hereditary

George Beadle and Edward Tatum- used mutants to show the relationship between DNA and proteins

Albrecht Kossel- characterized the structure of adenine, guanine, cytosine, thymine, and uracil

Explanation:

Hope this helps! Pls give brainliest!

5 0

2 years ago

Magma can form when

Setler79 [48]

Hot liquid rock comes in contact with Earth´s cold crust

4 0

3 years ago

Read 2 more answers

SpongeBob & Friends and the Scientific Method

yarga [219]

Answer:

1. Which people are in the control group? The people who received the mint without the secret ingredient

(Group B) would be the control group.

2. What is the independent variable? Secret ingredient in the breath mint

3. What is the dependent variable? Amount of breath odor (or bad breath)

4. What should Mr. Krabs’ conclusion be? The breath mint with the secret ingredient appears to reduce the

amount of breath odor more than half the time, but it is not 100% effective.

5. Why do you think 10 people in group B reported fresher breath? This may be due to the placebo effect.

6 0

3 years ago

If a sample has 25% parent and 75% daughter, how many halflifes have gone by?

Anit [1.1K]

Answer:

Two half lives.

Explanation:

It is known that the decay of isotopes and radioactive material obeys first order kinetics.

Also, it is clear that in first order decay the half-life time is independent of the initial concentration.

That means for a sample 100% to decay to 50 % it will take one half-life, and to decay the remaining 50% to 25% it will take another half-life.

So, for a sample has 25% parent and 75% daughter it will have two half-lives.

6 0

3 years ago

Combine the two half-reactions that give the spontaneous cell reaction with the smallest E∘. Fe2+(aq)+2e−→Fe(s) E∘=−0.45V I2(s)+

Iteru [2.4K]

Answer: The spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

Explanation:

We are given:

$E^o_{(Fe^{2+}/Fe)}=-0.45V\\E^o_{(I_2/I^-)}=0.54V\\E^o_{(Cu^{2+}/Cu)}=0.34V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, iodine will always undergo reduction reaction, then copper and then iron.

The equation used to calculate electrode potential of the cell is:

$E^o_{cell}=E^o_{oxidation}+E^o_{reduction}$

The combination of the cell reactions follows:

Case 1:

Here, iodine is getting reduced and iron is getting oxidized.

The cell equation follows:

$I_2(s)+Fe(s)\rightarrow Fe^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=0.45+0.54=0.99V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

Case 2:

Here, iodine is getting reduced and copper is getting oxidized.

The cell equation follows:

$I_2(s)+Cu(s)\rightarrow Cu^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-$ $E^o_{oxidation}=-0.34V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=-0.34+0.54=0.20V$

Thus, this cell will give the spontaneous cell reaction with smallest $E^o_{cell}$

Case 3:

Here, copper is getting reduced and iron is getting oxidized.

The cell equation follows:

$Cu^{2+}(aq.)+Fe(s)\rightarrow Fe^{2+}(aq.)+Cu(s)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $Cu^{2+}(aq.)+2e^-\rightarrow Cu(s)$ $E^o_{reduction}=0.34V$

$E^o_{cell}=0.45+0.34=0.79V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

Hence, the spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

7 0

3 years ago