What are the three simplest particles of iron

Why are fixed resistors’ values indicated by color bands rather than printing the numeric value on their exterior?

gladu [14]

Because there are so many different values of numbers, it would be impractical to use 1Ω, 2Ω, 3Ω... etc... Using colored bands helps make reading it a little easier to the trained eye. There are hundreds of thousands, if not tens of millions of different resistors would need to exist to cover every value. So you just use something called "preferred values" with their resistance values posted on them instead.

6 0

3 years ago

Three roots of a fifth degree polynomial function f(x) are –2, 2, and 4 + i. Which statement describes the number and nature of

Daniel [21]

The two roots of this functions are the exact roots and one is an imaginary root. The -2 is the root of a negative number -16 because if you compute the -2 on its 3rd power you come up -16, the 2 is the root of 16 and the 4+i contains a imaginary number that is why it is an imaginary root

3 0

4 years ago

Read 2 more answers

How many miles can a commercial jetliner flying at 253 meters per second travel in 6.0 hours

Goryan [66]

1 mile = 1609.344 meters
1 hour = 3600 seconds 345.48 miles per hour x 1609.344
meters/mile x 1hour/3600seconds = 154.44 meter per second~~ANSWERR

6 0

4 years ago

In the activity, click on the E∘cell and Keq quantities to observe how they are related. Use this relation to calculate Keq for

inessss [21]

Answer:

The standard cell potential of the reaction is 0.78 Volts.

Explanation:

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

Reduction at cathode :

$Cu^+(aq)+2e^-\rightarrow Cu(s)$

Reduction potential of $Cu^{2+}$ to Cu= $E^o_{1}=0.34 V$

Oxidation at anode:

$Fe(s)\rightarrow Fe^{2+}(aq)+2e^-$

Reduction potential of $Fe^{2+}$ to Fe= $E^o_{2}=-0.44 V$

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{red,cathode}-E^o_{red,anode}$

Putting values in above equation, we get:

$E^o_{cell}=0.34V -(-0.44 V)=0.78 V$

The standard cell potential of the reaction is 0.78 Volts.

3 0

3 years ago

Consider the nuclear equation below. Superscript 235 subscript 92 upper U right arrow superscript 4 subscript 2 upper H e. What

Leto [7]

Answer: The nuclide symbol of X is $^{231}_{90}\textrm{Th}$

Explanation:

The given nuclear reaction is a type of alpha decay process. In this process, the nucleus decays by releasing an alpha particle. The mass number of the nucleus is reduced by 4 units and atomic number is also decreased by 2 units. The particle released is a helium nucleus.

The general equation representing alpha decay process is:

$_{Z}^{A}\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_2^4\textrm{He}$

For the given equation :

$^{235}_{92}\textrm{U}\rightarrow ^{A}_{Z}\textrm{X}+^4_2\textrm{He}$

As the atomic number and mass number must be equal on both sides of the nuclear equation:

$^{235}_{92}\textrm{U}\rightarrow ^{231}_{90}\textrm{Th}+^4_2\textrm{He}$

Thus the nuclide symbol of X is $^{231}_{90}\textrm{Th}$

5 0

3 years ago