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2 years ago

Suggest two materials produced from hydrocarbons by the petrochemical industry.

Chemistry

1 answer:

Masteriza [31]2 years ago

5 0

Answer:

The major hydrocarbon products produced from petroleum by refining are: liquefied petroleum gas, gasoline, diesel fuel, kerosene, fuel oil, lubricating oil, and paraffin wax.

Explanation:

hope i helped :)

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Consider the image of a fossil below.

Alchen [17]

Answer:That would be to find the age of the fossil. Its called radiometric dating.

Explanation:

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1 year ago

Evaluate<br> 0.411 + 9.9 + 0.8<br> and round the answer appropriately,

Lemur [1.5K]

Answer:

the answer is 11.111 and when you round it to the nearest one it’s 11, but to the nearest tenth it’s 10

Explanation:

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2 years ago

I need help with number two

Komok [63]

The temperature is held constant at (b) and (d). At these points, the substance is changing states. B is changing from solid to liquid and D is changing from liquid to gas

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2 years ago

How many milliliters of 2.55 M NaOH is needed to make 125 mL of 0.75 M NaOH?

photoshop1234 [79]

As i am reading the problem, i notice they give two concentrations (M), a volume and asking for the other volume; this should be a hint that we need to use the dilution formula---> M1V1 = M2V2

M1= 2.55 M
V1= ?
M2= 0.75 M
V2= 125 mL

now, we plug in the values into the formula

(2.55 x V1) = (0.75 x 125)

<span>V1= 36.8 mL</span>

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3 years ago

If you hit the surface of Iron with a photon of energy and find that the ejected electron has a wavelength of .75 nm, what is th

lubasha [3.4K]

Answer:

The wavelength of the incoming photon is 172.8 nm

Explanation:

The wavelength of the incoming photon can be calculated with the photoelectric equation:

$KE = h\frac{c}{\lambda_{p}} - \phi$ (1)

Where:

KE: is the kinetic energy of the electron

h: is Planck's constant = 6.62x10⁻³⁴ J.s

c: is the speed of light = 3.00x10⁸ m/s

$\lambda_{p}$ : is the wavelength of the photon =?

Φ: is the work function of the surface (Iron) = 4.5 eV

The kinetic energy of the electron is given by:

$KE = \frac{p^{2}}{2m} = \frac{(\frac{h}{\lambda_{e}})^{2}}{2m}$ (2)

Where:

p: is the linear momentum = h/λ

m: is the electron's mass = 9.1x10⁻³¹ kg

$\lambda_{e}$ : is the wavelength of the electron = 0.75 nm = 0.75x10⁻⁹ m

Hence, the wavelength of the photon is:

$\frac{(\frac{h}{\lambda_{e}})^{2}}{2m} = h\frac{c}{\lambda_{p}} - \phi$

$\lambda_{p} = \frac{hc}{\frac{h^{2}}{2m\lambda_{e}^{2}} + \phi} = \frac{6.62 \cdot 10^{-34} J.s*3.00\cdot 10^{8} m/s}{\frac{(6.62 \cdot 10^{-34} J.s)^{2}}{2*9.1 \cdot 10^{-31} kg*(0.75 \cdot 10^{-9} m)^{2}} + 4.5 eV*\frac{1.602 \cdot 10^{-19} J}{1 eV}} = 1.728 \cdot 10^{-7} m = 172.8 nm$

Therefore, the wavelength of the incoming photon is 172.8 nm.

I hope it helps you!

3 0

2 years ago