4

## The radius of the sun is 0.7 million km. What percentage of the radius is taken up by the chromosphere?

Hunaid L. Hanfee
Featured 1 year ago · Astrophysics

0.28%

#### Explanation:

The layer chromosphere is $2000$ km thick.

The radius of the sun is $0.7 \cdot {10}^{6}$ 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 $0.7 \cdot {10}^{6}$(not possible but you have $\textcolor{w h i t e}{0000000000000000000000000000000}$to just imagine this )
out of this you got $2000$ marks.

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

$= \frac{2000}{0.7 \cdot {10}^{6}} \cdot 100$

Right?

Just do same calculation for this question,
Total radius of sun is $0.7 \cdot {10}^{6}$ km.
out of this $2000$ km is chromosphere.

$= \frac{2000}{0.7 \cdot {10}^{6}} \cdot 100$

$= 0.002857 \cdot 100$

=0.28%

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

# Make the internet a better place to learn

3

## Find the freezing point elevation for 0.5 m Al(NO3)3? To solve this I got -3.72C as the change in freezing point however, the question asks "WHAT IF THE ORIGINAL FREEZING POINT OF ETHANOL IS -18...." What is the new freezing temperature?

Stefan V.
Featured 20 hours ago · Chemistry

Here's what I got.

#### Explanation:

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

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \Delta {T}_{f} = i \cdot {K}_{f} \cdot b \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Here

$\Delta {T}_{f}$ - the freezing-point depression;
$i$ - the van't Hoff factor
${K}_{f}$ - the cryoscopic constant of the solvent;
$b$ - the molality of the solution

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, "Al"("NO"_3)_3, is soluble in aqueous solution, which means that it dissociates completely to form aluminium cations, ${\text{Al}}^{3 +}$, and nitrate anions, ${\text{NO}}_{3}^{-}$

${\text{Al"("NO"_ 3)_ (3(aq)) -> "Al"_ ((aq))^(3+) + 3"NO}}_{3 \left(a q\right)}^{-}$

Since one mole of solute produces $4$ moles of particles of solute in solution, the van't Hoff factor is equal to $4$.

The cryoscopic constant of water is equal to

${K}_{f} = {1.86}^{\circ} {\text{C kg mol}}^{- 1}$

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))))

$\Delta {T}_{f} = {3.72}^{\circ} \text{C}$

This is the correct value for the freezing-point depression of a $\text{0.5 molal}$ aluminium nitrate solution for which 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 ${0}^{\circ} \text{C}$ for water, you can say that the freezing point of the solution will be

${T}_{\text{f sol" = 0^@"C" - 3.72^@"C" = -3.72^@"C}}$

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

${K}_{f} = {1.99}^{\circ} {\text{C kg mol}}^{- 1}$

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))))

$\Delta {T}_{f} = {3.98}^{\circ} \text{C}$

This means that the freezing point of the solution will be ${3.98}^{\circ} \text{C}$ lower than the freezing point of the pure solvent, which for ethanol is $- {18}^{\circ} \text{C}$

${T}_{\text{f sol" = -18^@"C" - 3.98^@"C" = -21.98^@"C}}$

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.

1

## Given that y varies directly with x, and x = 6, y = 102, how do you write a direct variation equation that relates x and y?

Jim G.
Featured 9 hours ago · Algebra

$y = 17 x$

#### Explanation:

$\text{the initial statement is } y \propto x$

$\text{to convert to an equation multiply by k the constant}$
$\text{of variation}$

$\Rightarrow y = k x$

$\text{to find k use the given condition}$

$x = 6 \text{ when } y = 102$

$y = k x \Rightarrow k = \frac{y}{x} = \frac{102}{6} = 17$

$\text{equation is } \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = 17 x} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

2

## If r_1=r_2+r_3+r prove that the triangle is right angled.?

dk_ch
Featured 9 hours ago · Trigonometry

We know

${r}_{1} = \frac{\Delta}{s - a}$

${r}_{2} = \frac{\Delta}{s - b}$

${r}_{3} = \frac{\Delta}{s - c}$

$r = \frac{\Delta}{s}$

Inserting these values in the given relation.

${r}_{1} = {r}_{2} + {r}_{3} + r$

$\implies {r}_{1} - r = {r}_{2} + {r}_{3}$

$\implies \frac{\Delta}{s - a} - \frac{\Delta}{s} = \frac{\Delta}{s - b} + \frac{\Delta}{s - c}$

$\implies \frac{1}{s - a} - \frac{1}{s} = \frac{1}{s - b} + \frac{1}{s - c}$

$\implies \frac{2}{2 s - 2 a} - \frac{2}{2 s} = \frac{2}{2 s - 2 b} + \frac{2}{2 s - 2 c}$

$\implies \frac{1}{b + c - a} - \frac{1}{b + c + a} = \frac{1}{c + a - b} + \frac{1}{a + b - c}$

$\implies \frac{\left(b + c + a\right) - \left(b + c - a\right)}{{\left(b + c\right)}^{2} - {a}^{2}} = \frac{c + a - b + a + b - c}{{a}^{2} - {\left(b - c\right)}^{2}}$

$\implies \frac{2 a}{{\left(b + c\right)}^{2} - {a}^{2}} = \frac{2 a}{{a}^{2} - {\left(b - c\right)}^{2}}$

$\implies {\left(b + c\right)}^{2} - {a}^{2} = {a}^{2} - {\left(b - c\right)}^{2}$

$\implies {\left(b + c\right)}^{2} + {\left(b - c\right)}^{2} = {a}^{2} + {a}^{2}$

$\implies 2 {b}^{2} + 2 {c}^{2} = 2 {a}^{2}$

$\implies {b}^{2} + {c}^{2} = {a}^{2}$

This means the triangle is right angled.

2

## Prove  (1+sinβ-cosβ)/(1+sinβ+cosβ) + (1+sinβ+cosβ)/(1+sinβ-cosβ) = 2cosecβ?

dk_ch
Featured 9 hours ago · Trigonometry

1st part of LHS

 =(1+sinβ-cosβ)/(1+sinβ+cosβ)

 =(sinbeta(1+sinβ-cosβ))/(sinbeta(1+sinβ+cosβ))

 =(sinbeta+sin^2β-sinbetacosβ)/(sinbeta(1+sinβ+cosβ))

 =(sinbeta+1-cos^2β-sinbetacosβ)/(sinbeta(1+sinβ+cosβ))

 =(sinbeta(1-cosbeta)+(1-cosβ)(1+cosβ))/(sinbeta(1+sinβ+cosβ))

 =((1-cosbeta)(sinbeta+1+cosβ))/(sinbeta(1+sinβ+cosβ))

$= \frac{1 - \cos \beta}{\sin} \beta$

$= \frac{1}{\sin} \beta - \cos \frac{\beta}{\sin} \beta$

$= \csc \beta - \cot \beta$

2nd part of LHS
=(1+sinβ+cosβ)/(1+sinβ-cosβ)

 =(sinbeta(1+sinβ+cosβ))/(sinbeta(1+sinβ-cosβ))

 =(sinbeta+sin^2β+sinbetacosβ)/(sinbeta(1+sinβ-cosβ))

 =(sinbeta+1-cos^2β+sinbetacosβ)/(sinbeta(1+sinβ-cosβ))

 =(sinbeta(1+cosbeta)+(1-cosβ)(1+cosβ))/(sinbeta(1+sinβ-cosβ))

 =((1+cosbeta)(sinbeta+1-cosβ))/(sinbeta(1+sinβ-cosβ))

$= \frac{1 + \cos \beta}{\sin} \beta$

$= \frac{1}{\sin} \beta + \cos \frac{\beta}{\sin} \beta$

$= \csc \beta + \cot \beta$

LHS
 = cscβ-cotbeta+cscbeta+cotbeta

$= 2 \csc \beta = R H S$

2

## What bonds does ibuprofen have? What intermolecular forces are present in those bonds?

Marta O.
Featured an hour ago · Chemistry

Intramolecular forces/ bonds: covalent
Intermolecular forces: dipole-dipole, London dispersion forces, hydrogen bonding

Ibuprofen:

#### Explanation:

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:
1

## If A = <3 ,8 ,-1 >, B = <4 ,-3 ,-1 > and C=A-B, what is the angle between A and C?

Somebody N.
Featured 9 hours ago · Trigonometry

${26.54}^{\circ}$

#### Explanation:

$A = \left[\begin{matrix}3 \\ 8 \\ - 1\end{matrix}\right]$

$B = \left[\begin{matrix}4 \\ - 3 \\ - 1\end{matrix}\right]$

$C = A - B = \left[\begin{matrix}3 \\ 8 \\ - 1\end{matrix}\right] - \left[\begin{matrix}4 \\ - 3 \\ - 1\end{matrix}\right] = \left[\begin{matrix}3 - 4 \\ 8 - \left(- 3\right) \\ - 1 - \left(- 1\right)\end{matrix}\right] = \left[\begin{matrix}- 1 \\ 11 \\ 0\end{matrix}\right]$

We can find the angle between vectors using the Dot Product

The dot product states that for vectors a and b:

$a \cdot b = | | a | | \cdot | | b | | \cdot \cos \left(\theta\right)$

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.

$\left(a + b\right) \left(c + d\right) = a c + a d + b c + b d$

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

$\left(a + b + c\right) \cdot \left(d + e + f\right) = a d + b e + c f$

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

Let $a = \left[\begin{matrix}x \\ y \\ z\end{matrix}\right]$

Magnitude of $a = | | a | |$

$| | a | | = \sqrt{{x}^{2} + {y}^{2} + {z}^{2}}$

Now to our example:

First find the product of:

$A \cdot C$

$\left[\begin{matrix}3 \\ 8 \\ - 1\end{matrix}\right] \cdot \left[\begin{matrix}- 1 \\ 11 \\ 0\end{matrix}\right] = \left[\begin{matrix}3 \times - 1 \\ 8 \times 11 \\ - 1 \times 0\end{matrix}\right]$

$= \left[\begin{matrix}- 3 \\ 88 \\ 0\end{matrix}\right] = - 3 + 88 + 0 = 85$

Now find the magnitudes of A and C:

$| | A | | = \sqrt{{\left(3\right)}^{2} + {\left(8\right)}^{2} + {\left(- 1\right)}^{2}} = \sqrt{74}$

$| | C | | = \sqrt{{\left(- 1\right)}^{2} + {\left(11\right)}^{2} + {\left(0\right)}^{2}} = \sqrt{122}$

So we have for:

$a \cdot b = | | a | | \cdot | | b | | \cdot \cos \left(\theta\right)$

$85 = \sqrt{74} \cdot \sqrt{122} \cdot \cos \left(\theta\right)$

$\cos \left(\theta\right) = \frac{85}{\sqrt{74} \cdot \sqrt{122}}$

$\theta = \arccos \left(\cos \left(\theta\right)\right) = \arccos \left(\frac{85}{\sqrt{74} \cdot \sqrt{122}}\right) = {26.54}^{\circ}$
( 2 .d.p.)

So the angle between vectors A and C is ${26.54}^{\circ}$

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.

1

## Distance between (4,5) and (4,-3)? *

Tom M.
Featured 9 hours ago · Algebra

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.

$\therefore \text{Distance} = 5 - \left(- 3\right)$
$= 8$

1

## I heard that two electrons can exist in the same spot in time and space as long as they are spinning in opposite directions. What happens when they spin in the same direction? Is this more dangerous than a black hole?

Mark C.
Featured an hour ago · Astronomy

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

#### Explanation:

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.

2

## How do you graph f(x)=(2x^2+1)/(x^3-x) using holes, vertical and horizontal asymptotes, x and y intercepts?

Alexander L.
Featured 8 hours ago · Algebra

We can graph this equation using a sign chart.

#### Explanation:

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 $y = 0$.

Vertical Asymptotes
In order to find vertical asymptotes, we muct factor the denominator. Factoring out an $x$ from ${x}^{3} - x$, we will get $x \left({x}^{2} - 1\right)$. Equating both answers to 0 will allow us to find our asymptotes.

$x = 0$

${x}^{2} - 1 = 0$
${x}^{2} = 1$
$\sqrt{{x}^{2}} = \sqrt{1}$
$x = \pm 1$
$x \ne - 1 , 0 , 1$

$x$ can not equal any of these values as they all result in a 0 in the denominator. This will make the function undefined aka asymptotes.

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 $f \left(x\right)$ and solve.

$0 = \frac{2 {x}^{2} + 1}{{x}^{3} - x}$

There are no x-intercepts as the equation above would result in imaginary numbers as answers ($\frac{i \sqrt{2}}{\sqrt{2}} , - \frac{i \sqrt{2}}{\sqrt{2}}$).

Y-Intercepts
To find y-intercepts substitute 0 for $x$ and solve.

$y = \frac{2 {\left(0\right)}^{2} + 1}{{\left(0\right)}^{3} - \left(0\right)}$

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 $- \propto$ and $\propto$.

We will then select a value for $x$ in between the critical numbers.

We will test the $x$ values for positivity by substituting them for $x$ in the formula and write either the positive or negative symbol in the spaces we created. I'll do $x = 10$ as an example below.
$y = \frac{2 {\left(10\right)}^{2} + 1}{{\left(10\right)}^{3} - \left(10\right)}$
$y = \frac{67}{330}$

We don't necessarily care about the value, just whether it is positive or negative. Since it is positive, we will mark a $+$ between $1$ and $\propto$. We will do this for all the values we have chosen.

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.

1

## Evaluate analytically: lim_(h->0) ((x+h)^(1/3) - x^(1/3)) / h Do you multiply it by the conjugate? I'm not getting the right answer when I do

Jim H
Featured 10 hours ago · Calculus

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

#### Explanation:

The conjugate of ${x}^{\frac{1}{2}} - {y}^{\frac{1}{2}}$ is ${x}^{\frac{1}{2}} + {y}^{\frac{1}{2}}$ (and vice versa) because

$\left(a - b\right) \left(a + b\right) = {a}^{2} - {b}^{2}$.

So if either of $a$ or $b$ involves a square root, then its square does not involve a square root..

The analog for cube roots is based on the fact that

$\left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right) = {a}^{3} - {b}^{3}$.

So the conjugate of ${x}^{\frac{1}{3}} - {y}^{\frac{1}{3}}$ $\text{ }$ is $\text{ }$ ${x}^{\frac{2}{3}} + {\left(x y\right)}^{\frac{1}{3}} + {y}^{\frac{2}{3}}$ (and vice versa)

(((x+h)^(1/3) - x^(1/3))) / h * (((x+h)^(2/3)+(x(x+h))^(1/3) + x^(2/3)))/ (((x+h)^(2/3)+(x(x+h))^(1/3) + x^(2/3))) = ((x+h)-x)/( h((x+h)^(2/3)+(x(x+h))^(1/3) + x^(2/3))

$= \frac{1}{{\left(x + h\right)}^{\frac{2}{3}} + {\left(x \left(x + h\right)\right)}^{\frac{1}{3}} + {x}^{\frac{2}{3}}}$

And, now we can evaluate the limit

${\lim}_{h \rightarrow 0} \frac{1}{{\left(x + h\right)}^{\frac{2}{3}} + {\left(x \left(x + h\right)\right)}^{\frac{1}{3}} + {x}^{\frac{2}{3}}} = \frac{1}{3 {x}^{\frac{2}{3}}}$