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Harlamova29_29 [7]

2 years ago

12

Q3.What did you observe about the wavelengths and frequencies of the different colors of light?

1 answer:

Natalija [7]2 years ago

6 0

Answer:

Explanation:

Q3.What did you observe about the wavelengths and frequencies of the different colors of light?

Q4. Do the frequencies of colors of light increase from red to violet?

Q5. What did you observe about the corresponding energies from red to violet?

wavelength times frequency =c=the speed of light

the higher the frequency, the shorter the wavelength

the different colors of visible light are our good friend ROY G BIV

RED ORANGE YELLOW GREEN BLUE INDIGO VIOLET

As we go from red to violet, the wavelength decreases and the energy increases because the wave is vibrating at a higher number of vibrations per secomd. This is what "frequency' means, how many vibrations are there per second, how FREQUENT are thee.

the higher the frequency the more energy because the wave is vibrating faster.

the equation that relates this is Energy = h times frequency, where h is

Planck's constant = 6.262 x 10^-27 ERG-SEC

the frequency of vibration and therefor the energy of visible electromagnetic radiation (light) increase as we go across

ROY G BIV

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timama [110]

<span>I am sure these answers are correct cause I've checked all of them. Here they are :

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8 0

2 years ago

The radioactive isotope of lead, Pb-209, decays at a rate proportional to the amount present at time t and has a half-life of 3.

Marina86 [1]

Answer:

Therefore it will take 7.66 hours for 80% of the lead decay.

Explanation:

The differential equation for decay is

$\frac{dA}{dt}= kA$

$\Rightarrow \frac{dA}{A}=kdt$

Integrating both sides

ln A= kt+c₁

$\Rightarrow A= e^{kt+c_1}$

$\Rightarrow A=Ce^{kt}$ [ $e^{c_1}=C$ ]

The initial condition is A(0)= A₀,

$\therefore A_0=Ce^{0.k}$

$\Rightarrow C=A_0$

$\therefore A=A_0e^{kt}$ .........(1)

Given that the $Pb_{209}$ has half life of 3.3 hours.

For half life $A=\frac12 A_0$ putting this in equation (1)

$\frac12A_0=A_0e^{k\times3.3}$

$\Rightarrow ln(\frac12)= 3.3k$ [taking ln both sides, $ln \ e^a=a$ ]

$\Rightarrow k=\frac{ln \frac12}{3.3}$

⇒k= - 0.21

Now A₀= 1 gram, 80%=0.8

and A= (1-0.8)A₀ = (0.2×1) gram = 0.2 gram

Now putting the value of k,A and A₀in the equation (1)

$\therefore 0.2=1e^{(-0.21)\times t}$

$\Rightarrow e^{-0.21t}=0.2$

$\Rightarrow -0.21t= ln(0. 2)$

$\Rightarrow t= \frac{ln (0.2)}{-0.21}$

⇒ t≈7.66

Therefore it will take 7.66 hours for 80% of the lead decay.

3 0

2 years ago

How can electronegativity be used to explain the reaction we see in terms of the numbers and types of bonded being formed in lit

kolezko [41]

Explanation:

Lithium is an electropositive element that readily loses electrons.

Oxygen is electronegative and it will readily accept electrons.

Due to this significant electronegativity differences between the two species they form electrovalent or ionic bonds between them.

2atoms of Li lose two electrons:

Li → Li²⁺ + e⁻

Lithium isoelectronic with helium

For oxygen;

O + 2e⁻ → O²⁻

Oxygen is isoelectronic with Neon

Two ions of the lithium combines with the oxygen to form the bond;

4Li + O₂ → 2Li₂O

The electrostatic attraction between the two ions forms the ionic bond

5 0

2 years ago

Part a use these data to calculate the heat of hydrogenation of buta-1,3-diene to butane. c4h6(g)+2h2(g)→c4h10(g)

Reptile [31]

<u>Answer:</u> The heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_{(product)}]-\sum [n\times \Delta H_{(reactant)}]$

For the given chemical reaction:

$C_4H_6(g)+2H_2(g)\rightarrow C_4H_{10}(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(1\times \Delta H_{(C_4H_{10})})]-[(1\times \Delta H_{(C_4H_6)})+(2\times \Delta H_{(H_2)})]$

We are given:

$\Delta H_{(C_4H_{10})}=-2877.6kJ/mol\\\Delta H_{(C_4H_6)}=-2540.2kJ/mol\\\Delta H_{(H_2)}=-285.8kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(1\times (-2877.6))]-[(1\times (-2540.2))+(2\times (-285.8))]\\\\\Delta H_{rxn}=234.2J$

Hence, the heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

4 0

2 years ago

We can change a gas to liquid by the temperature and the pressure. NextReset

USPshnik [31]

We can change a gas to a liquid by INCREASING the temperature and DECREASING the pressure

7 0

2 years ago

Read 2 more answers