Featured 1 year ago
· Astrophysics

The layer chromosphere is **km** thick.

The radius of the sun is **km**.

To calculate this you have to just do same as we do in calculation of our exam's marks percentage.

*Just imagine,*

Your exam is of total marks of

out of this you got

To calculate the percentage you have to do something like this:

Right?

Just do same calculation for this question,

Total radius of sun is **km**.

out of this **km** is chromosphere.

I have added image for your help by which you can know where is chromosphere of the sun.

Here's what I got.

First of all, you're dealing with **freezing-point depression**, not freezing-point *elevation*.

Simply put, the freezing point of a solution will be **lower** than the freezing point of the *pure solvent*. The *freezing-point depression* tells you **how low** the freezing point of the solution will actually be compared with that of the pure solvent.

Now, the equation that allows you to calculate freezing-point depression looks like this

#color(blue)(|bar(ul(color(white)(a/a)DeltaT_f = i * K_f * bcolor(white)(a/a)|)))#

Here

*van't Hoff factor*

*cryoscopic constant* of the solvent;

As you can see, the freezing-point depression **cannot** be negative because the van't Hoff factor, the cryoscopic constant, and the *molality* of the solution are all **positive values**.

In your case, aluminium nitrate, **soluble** in aqueous solution, which means that it dissociates completely to form aluminium cations,

#"Al"("NO"_ 3)_ (3(aq)) -> "Al"_ ((aq))^(3+) + 3"NO"_ (3(aq))^(-)#

Since one mole of solute produces **moles** of particles of solute in solution, the van't Hoff factor is equal to

The cryoscopic constant of water is equal to

#K_f = 1.86^@"C kg mol"^(-1)#

The freezing-point depression will thus be

#DeltaT_f = 4 * 1.86^@"C" color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 0.5 color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1))))#

#DeltaT_f = 3.72^@"C"#

This is the correct value for the freezing-point depression of a **water is the solvent**.

Since freezing-point depression tells you by how many degrees the freezing point of the solution **decreased** compared to that of the pure solvent, which is **the solution** will be

#T_"f sol" = 0^@"C" - 3.72^@"C" = -3.72^@"C"#

Now, the question asks you to find the freezing point of a solution in which **ethanol** is the solvent, * not* water.

The cryoscopic constant for ethanol is

#K_f = 1.99^@"C kg mol"^(-1)#

http://www.vaxasoftware.com/doc_eduen/qui/tcriosebu.pdf

Assuming that you have complete dissociation, the freezing-point depression will be

#DeltaT_f = 4 * 1.99^@"C" color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 0.5 color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1))))#

#DeltaT_f = 3.98^@"C"#

This means that the freezing point of the solution will be **lower** than the freezing point of the pure solvent, which for ethanol is

#T_"f sol" = -18^@"C" - 3.98^@"C" = -21.98^@"C"#

So remember, the freezing-point depression is **always positive** and it must be subtracted from the freezing point of the *pure solvent* in order to get the freezing point of the **solution**.

#"the initial statement is "ypropx#

#"to convert to an equation multiply by k the constant"#

#"of variation"#

#rArry=kx#

#"to find k use the given condition"#

#x=6" when "y=102#

#y=kxrArrk=y/x=102/6=17#

#"equation is " color(red)(bar(ul(|color(white)(2/2)color(black)(y=17x)color(white)(2/2)|)))#

Featured 9 hours ago
· Trigonometry

We know

Inserting these values in the given relation.

This means the triangle is right angled.

Featured 9 hours ago
· Trigonometry

1st part of LHS

2nd part of LHS

Adding we get

LHS

Intramolecular forces/ bonds: covalent

Intermolecular forces: dipole-dipole, London dispersion forces, hydrogen bonding

Ibuprofen:

**Intramolecular bonds**

Organic compounds mostly have **covalent** bonds.

The covalent bond is formed by non-metallic bonding, in which two or more atoms share up to three valence electrons.

Here below are some examples of covalent bonds between identical atoms and various atoms in ibuprofen:

**Intermolecular forces**

- Dipole-dipole forces

The dipole-dipole forces occur with the polar molecules of the compound

e.g.: - London dispersion forces

The London forces are present among all chemical groups and usually represent a major part of the total force of interaction in the condensed matter, although they are usually weaker than ionic bonds and hydrogen bonds.

e.g.: - Hydrogen bonding

The hydrogen bonding occurs with the oxygen atoms to the hydrogen atoms of the molecule.

Linear (a) and cyclic (b) hydrogen bonds:

Featured 9 hours ago
· Trigonometry

We can find the angle between vectors using the **Dot Product**

The dot product states that for vectors **a** and **b**:

The dot product is sometimes called the inner product, because of the way the vectors **a** and **b** are multiplied and summed.

In algebra we are used to multiplying brackets in the following way.

In the case of the dot product we multiply the vectors in the following way.

So we are just multiplying corresponding components and then adding them together.

Let

Magnitude of

Now to our example:

First find the product of:

Now find the magnitudes of A and C:

So we have for:

**( 2 .d.p.)**

So the angle between vectors A and C is

Notice from the diagram that the angle found by the dot product, is the angle between the vectors where they are pointing in the same relative direction.

Lets just visualise this on a graph.

Since our x value is the same all the way through, the distance is the difference in the y-values.

OK, we need to understand a bit of quantum theory and the standard model.

The universe is best understood by reducing everything to the smallest possible number of entities (the goal of physics.) The “standard model” is our current best attempt and is based on quantum theory (more specifically, quantum electrodynamics or Q.E.D or it’s rather more bohemian cousin quantum chromodynamics, Q.C.D.)

The standard model is much like a family tree the way I teach it ... and the first split down from “everything” is into two groups called ‘fermions’ and ‘bosons’. Fermions include all matter. That means solids, liquids, gases, ceramics, polymers, you, me etc. Every single particle we know of including electrons (and there are hundreds) is a fermion. They all have one fundamental thing in common - they obey something called the “Pauli exclusion principle.”

The other group, the bosons, concerns us less in the context of your question, but includes all the known forces (called strong, weak, electromagnetic and gravity, though this last one is troublesome.)

All the known particles can be described as having a set of “quantum numbers” that determine all the properties that exist in quantum theory. Effectively this states that energy, momentum, **spin** and possibly even space and time exist as **quantised states,** meaning they cannot take any value, but only discrete ones (i.e. they are like an uneven ladder where you **have to** be on one rung or another, you cannot exist in between, so we would say your height is then “quantised”). The opposite would be something like an escalator, where your height could vary continuously.

The Pauli exclusion principle simply states that any fermion (particle to us, including the electrons in your question) **cannot** exist in the same state (complete set of quantum numbers) as another fermion. In other words (rather loosely stated, but it helps) they can be in the same place, but not at the same time, or they can be in the same time, but not at the same energy etc. It is forbidden for two or more to occupy the *same complete set*.

This explains why two electrons can be in the same orbit (same energy, possibly even the same position at the same time) but then *cannot* have the same spin. One **must** be spin “up” and the other spin “down” to prevent violation of this fundamental principle.

Finally, no you don’t get anything like a black hole, it is not even dangerous, just forbidden by the laws of physics.

Featured 8 hours ago
· Algebra

We can graph this equation using a sign chart.

First, let's identify the holes, vertical and horizontal asymptotes and x and y intercepts.

**Horizontal Asymptotes**

Observing the formula, we can see the degree of the denominator is higher than the numerator (The degree is the largest exponent in a polynomial). This means there will be an asymptote at

**Vertical Asymptotes**

In order to find vertical asymptotes, we muct factor the denominator. Factoring out an

**Holes**

Holes occur when there is a zero in both the numerator and denominator that will cancel. There are none in this formula.

**X-Intercepts**

To find x-intercepts, substitute 0 for

There are no x-intercepts as the equation above would result in imaginary numbers as answers (

**Y-Intercepts**

To find y-intercepts substitute 0 for

As the denominator equals zero the equation is undefined. Therefore there are no y-intercepts.

**Sign Chart**

Using the values above, also known as critical values, we will create a sign chart. We do this by sorting the critical values from least to greatest, and surrounding them with

We will then select a value for

We will test the

We don't necessarily care about the value, just whether it is positive or negative. Since it is positive, we will mark a

We are now able to graph now that we know where the function is positive and negative and we have the asymptotes.

graph{(2x^2+1)/(x^3-x) [-10, 10, -5, 5]}

Notice how the graph is positive where we noted positive, and negative where we noted negative.

Yes, we do multiply by the conjugate, but it is easy to be mistaken about what the conjugate is.

The conjugate of

So if either of

The analog for cube roots is based on the fact that

So the conjugate of

# = 1/((x+h)^(2/3)+(x(x+h))^(1/3) + x^(2/3))#

And, now we can evaluate the limit