Question 6 of 10

A chemist identifies compounds by identifying bright lines in their spectra. She does so by heating the compounds until they glo

padilas [110]

Answer:

a) λ = 189.43 10⁻⁹ m b) λ = 269.19 10⁻⁹ m

Explanation:

The diffraction network is described by the expression

d sin θ= m λ

Where m corresponds to the diffraction order

Let's use trigonometry to find the breast

tan θ = y / L

The diffraction spectrum is measured at very small angles, therefore

tan θ = sin θ / cos θ = sin θ

We replace

d y / L = m λ

Let's place in the first order m = 1

Let's look for the separation of the lines (d)

d = λ L / y

d = 501 10⁻⁹ 9.95 10⁻² / 15 10⁻²

d = 332.33 10⁻⁹ m

Now we can look for the wavelength of the other line

λ = d y / L

λ = 332.33 10⁻⁹ 8.55 10⁻²/15 10⁻²

λ = 189.43 10⁻⁹ m

Part B

The compound wavelength B

λ = 332.33 10⁻⁹ 12.15 10⁻² / 15 10⁻²

λ = 269.19 10⁻⁹ m

4 0

3 years ago

The British gold sovereign coin is an alloy of gold and copper having a total mass of 7.988 g, and is 22-karat gold 24 x (mass o

matrenka [14]

Answers:

(a) $0.0073kg$

(b) Volume gold: $3.79(10)^{-7}m^{3}$ , Volume cupper: $7.6(10)^{-8}m^{3}$

(c) $17633.554kg/m^{3}$

Explanation:

We are told the total mass $M$ of the coin, which is an alloy of gold and copper is:

$M=m_{gold}+m_{copper}=7.988g=0.007988kg$ (1)

Where $m_{gold}$ is the mass of gold and $m_{copper}$ is the mass of copper.

In addition we know it is a 22-karat gold and the relation between the number of karats $K$ and mass is:

$K=24\frac{m_{gold}}{M}$ (2)

Finding ${m_{gold}$ :

$m_{gold}=\frac{22}{24}M$ (3)

$m_{gold}=\frac{22}{24}(0.007988kg)$ (4)

$m_{gold}=0.0073kg$ (5) This is the mass of gold in the coin

<h2>(b) Volume of gold and cupper</h2><h2 />

The density $\rho$ of an object is given by:

$\rho=\frac{mass}{volume}$

If we want to find the volume, this expression changes to: $volume=\frac{mass}{\rho}$

For gold, its volume $V_{gold}$ will be a relation between its mass $m_{gold}$ (found in (5)) and its density $\rho_{gold}=19.30g/cm^{3}=19300kg/m^{3}$ :

$V_{gold}=\frac{m_{gold}}{\rho_{gold}}$ (6)

$V_{gold}=\frac{0.0073kg}{19300kg/m^{3}}$ (7)

$V_{gold}=3.79(10)^{-7}m^{3}$ (8) Volume of gold in the coin

For copper, its volume $V_{copper}$ will be a relation between its mass $m_{copper}$ and its density $\rho_{copper}=8.96g/cm^{3}=8960kg/m^{3}$ :

$V_{copper}=\frac{m_{copper}}{\rho_{copper}}$ (9)

The mass of copper can be found by isolating $m_{copper}$ from (1):

$M=m_{gold}+m_{copper}$

$m_{copper}=M-m_{gold}$ (10)

Knowing the mass of gold found in (5):

$m_{copper}=0.007988kg-0.0073kg=0.000688kg$ (11)

Now we can find the volume of copper:

$V_{copper}=\frac{0.000688kg}{8960kg/m^{3}}$ (12)

$V_{copper}=7.6(10)^{-8}m^{3}$ (13) Volume of copper in the coin

<h2>(c) Density of the sovereign coin</h2><h2 />

Remembering density is a relation between mass and volume, in the case of the coin the density $\rho_{coin$ will be a relation between its total mass $M$ and its total volume $V$ :

$\rho_{coin}=\frac{M}{V}$ (14)

Knowing the total volume of the coin is:

$V=V_{gold}+V_{copper}=3.79(10)^{-7}m^{3}+7.6(10)^{-8}m^{3}=4.53(10)^{-7}m^{3}$ (15)

$\rho_{coin}=\frac{0.007988kg}{4.53(10)^{-7}m^{3}}$ (16)

Finally:

$\rho_{coin}=17633.554kg/m^{3}}$ (17) This is the total density of the British sovereign coin

6 0

3 years ago

Which of the following describes the way heat is transferred in a geyser?

Iteru [2.4K]

Answer:

Heat is transferred by the hot air or water moving to a cooler area. The elements rotate in circular motions, giving the geyser pressure.

5 0

3 years ago

A book is at rest on a table. Identify the correct free-body diagram for this situation.

Law Incorporation [45]

Show a picture .. of what your doing

5 0

3 years ago

Read 2 more answers

When a net force of 17.0 newtons is applied to a dictionary placed on a frictionless table, it accelerates by 3.75 meters/second

4vir4ik [10]

Force = (mass) x (acceleration) (Newton's second law of motion)

Divide both sides of the equation by 'acceleration', and you have

Mass = (force) / (acceleration)

Mass = 17 newtons / 3.75 meters per second-sqrd = 4.533 kilograms (rounded)

8 0

3 years ago