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3 years ago

What types of energy can be moved through conductors

1 answer:

8 0

<span>The two types of energy that can be moved through conductors are electrical energy and thermal energy. Conductors are materials that allow electrons to flow freely or transfers heat more easily than other substances. Heat is transfered in conductors when fast moving particles (contains more heat) crash into slow moving particles. conductors allow electrons to flow freely from one object to another in contact. Metals are usually excelent conductors of heat and electricity. </span>

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How many polar bonds are found in this molecule

Law Incorporation [45]

A bond is non polar if it is between same atoms and polar if it is between different atoms.

Same atoms are like two dogs of same strength pulling a bone towards towards each other. But when it’s different atoms it’s like a big dog and small dog then the bone is more towards bigger dog. So it’s the same way in bonds.

Bonds are made up of electrons, when the more stronger pulling atom is present than other the electrons are more towards it and as a result we have polar bond. There is development of a kind of a negative pole and a positive pole.

The stronger atom has electrons towards itself so it has a little more negative charge while the other atom has positive charge. This makes bond polar.

So just look for bond between two different atoms, it would be polar.

Look at the pic below to see the answer.

Marked with green is bond between same atoms... one carbon and another carbon so it is not polar and test marked with blue are polar.

Well the answer should have been 10 but since the bonds at 3 and 8 are two of same type we count only one of them.

The answer is 8... well the answer should be 10 otherwise... discuss it with ur teacher

6 0

3 years ago

Help please lol I’m stuggling

Mazyrski [523]

Answer:

balenced

Explanation:

6 0

3 years ago

Consider the group a1 element sodium atomic number 11, the group 3a element aluminum atomic number 13, and the group 7a element

Akimi4 [234]

According to there chemical properties

7 0

4 years ago

Consider the following reaction: A(g)⇌2B(g). Find the equilibrium partial pressures of A and B for each of the following differe

Illusion [34]

Answer:

a. Kp=1.4

$P_{A}=0.2215 atm$

$P_{B}=0.556 atm$

b.Kp=2.0 * 10^-4

$P_{A}=0.495atm$

$P_{B}=0.00995 atm$

c.Kp=2.0 * 10^5

$P_{A}=5*10^{-6}atm$

$P_{B}=0.9999 atm$

Explanation:

For the reaction

A(g)⇌2B(g)

Kp is defined as:

$Kp=\frac{(P_{B})^{2}}{P_{A}}$

The conditions in the system are:

A B

initial 0 1 atm

equilibrium x 1atm-2x

At the beginning, we don’t have any A in the system, so B starts to react to produce A until the system reaches the equilibrium producing x amount of A. From the stoichiometric relationship in the reaction we get that to produce x amount of A we need to 2x amount of B so in the equilibrium we will have 1 atm – 2x of B, as it is showed in the table.

Replacing these values in the expression for Kp we get:

$Kp=\frac{(1-2x)^{2}}{x}$

Working with this equation:

$x*Kp=(1-2x)^{2} - -> x*Kp=4x^{2}-4x+1- - >4x^{2}-(4+Kp)*x+1=0$

This last expression is quadratic expression with a=4, b=-(4+Kp) and c=1

The general expression to solve these kinds of equations is:

$x=\frac{-b(+-)*\sqrt{(b^{2}-4ac)}}{2a}$ (equation 1)

We just take the positive values from the solution since negative partial pressures don´t make physical sense.

Kp = 1.4

$x_{1}=\frac{(1.4+4)+\sqrt{(-(1.4+4)^{2}-4*4*1)}}{2*4}=1.128$

$x_{1}=\frac{(1.4+4)-\sqrt{(-(1.4+4)^{2}-4*4*1)}}{2*4}=0.2215$

With x1 we get a partial pressure of:

$P_{A}=1.128 atm$

$P_{B}=1-2*1.128 = -1.256 atm$

Since negative partial pressure don´t make physical sense x1 is not the solution for the system.

With x2 we get:

$P_{A}=0.2215 atm$

$P_{B}=1-2*0.2215 = 0.556 atm$

These partial pressures make sense so x2 is the solution for the equation.

We follow the same analysis for the other values of Kp.

Kp=2*10^-4

X1=0.505

X2=0.495

With x1

$P_{A}=0.505atm$

$P_{B}=1-2*0.505 = -0.01005 atm$

Not sense.

With x2

$P_{A}=0.495atm$

$P_{B}=1-2*0.495 = 0.00995 atm$

X2 is the solution for this equation.

Kp=2*10^5

X1=50001

$X2=5*10^{-6}$

With x1

$P_{A}=50001atm$

$P_{B}=1-2*50001=-100001atm$

Not sense.

With x2

$P_{A}= 5*10^{-6}atm$

$P_{B}=1-2*5*10^{-6}= 0.9999 atm$

X2 is the solution for this equation.

8 0

4 years ago

How are mutation affects different?

dsp73

Mutation affects can be different just with changes as small as the substitution of a single DNA building block or nucleotide base with another nucleotide base

6 0

2 years ago