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2 years ago

Why can’t you see the stars very well at night like you could 50 years ago

1 answer:

6 0

Answer:Artificial light from cities has created a permanent "skyglow" at night, obscuring our view of the stars. Here's their map of artificial sky brightness in North America, represented as a ratio of "natural" nighttime sky brightness. In the black areas, the natural night sky is still (mostly) visible.

Explanation:

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serious [3.7K]

A beta particle is an electron and it has a -1 charge and zero mass.

Beta decay by emitting an electron is called as β⁻ decay. When this happens, a neutron of the element converts into a proton by emitting an electron. Hence, the mass of daughter nucleus is same as parent atom but atomic number/number of protons is higher by 1 than atomic number of parent atom.

In a β⁻ decay, the symbol is used as ₋₁⁰β or ₋₁⁰e.
-1 is for charge
 0 is for the mass of the particle

4 0

2 years ago

Read 2 more answers

A sample of gas in a 14.6 L flexible container is at 25.0oC and 1.00atm. What is the volume of the sample when heated to 220.0oC

inn [45]

Answer: 24.1 L

Explanation:

To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where,

$V_1\text{ and }T_1$ are the initial volume and temperature of the gas.

$V_2\text{ and }T_2$ are the final volume and temperature of the gas.

We are given:

$V_1=14.6L\\T_1=25.0^oC=(25+273)K=298K\\V_2=?\\T_2=220.0^0C=(220+273)K=493K$

Putting values in above equation, we get:

$\frac{14.6}{298K}=\frac{V_2}{493}\\\\V_2=24.1L$

Thus the volume of the sample when heated to 220.0oC and the pressure is constant is 24.1 L

7 0

3 years ago

Using the Bohr model, determine the energy, in joules, necessary to ionize a ground-state hydrogen atom. Show your calculations.

lord [1]

Answer:

The energy required to ionize the ground-state hydrogen atom is 2.18 x 10^-18 J or 13.6 eV.

Explanation:

To find the energy required to ionize ground-state hydrogen atom first we calculate the wavelength of photon required for this operation.

It is given by Bohr's Theory as:

1/λ = Rh (1/n1² - 1/n2²)

where,

λ = wavelength of photon

n1 = initial state = 1 (ground-state of hydrogen)

n2 = final state = ∞ (since, electron goes far away from atom after ionization)

Rh = Rhydberg's Constant = 1.097 x 10^7 /m

Therefore,

1/λ = (1.097 x 10^7 /m)(1/1² - 1/∞²)

λ = 9.115 x 10^-8 m = 91.15 nm

Now, for energy (E) we know that:

E = hc/λ

where,

h = Plank's Constant = 6.625 x 10^-34 J.s

c = speed of light = 3 x 10^8 m/s

Therefore,

E = (6.625 x 10^-34 J.s)(3 x 10^8 m/s)/(9.115 x 10^-8 m)

E = 2.18 x 10^-18 J

E = (2.18 x 10^-18 J)(1 eV/1.6 x 10^-19 J)

E = 13.6 eV

5 0

3 years ago

Which rule for assigning oxidation numbers IS correct?

Vilka [71]

Answer:

B.) Oxygen is usually -2

Explanation:

Hydrogen is usually +1.

A pure group 1 element is not always +1.

A monoatomic ion can be a range of numbers. However, it must be a charge other than 0.

3 0

1 year ago

Breeder reactors are used to convert the nonfissionable nuclide U-238 to a fissionable product. Neutron capture by the U-238 is

m_a_m_a [10]

Answer:

Pu-239

Explanation:

Beta decay moves the element which undergoes the decay one place to the right in the periodic table since to conserve charge and being beta radiations an electron we convert a neutron into a proton and an electron. In neutron capture we increase the atomic mas by one unit. We that in mind, lets solve the question:

U-238 + ₁⁰ n ⇒ U-239 ⇒ Np -239 + ₋₁⁰β ⇒ Pu-239 + ₋₁⁰β

7 0

2 years ago