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2 years ago

Which two types of information are found in an elements box in the periodic table

Physics

2 answers:

Pani-rosa [81]2 years ago

4 0

Answer:

Each box represents an element and contains its atomic number, symbol, average atomic mass, and (sometimes) name.

Explanation:

nekit [7.7K]2 years ago

4 0

Answer:

An element's period and group

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From the lab iodine clock reaction: explain the reason this result by comparing what is involved in changing Fe2+ into Fe 3+ to

uranmaximum [27]

What is involved in C02 is carbon dioxide

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2 years ago

Violet light of wavelength 400 nm ejects electrons with a maximum kinetic energy of 0.860 eV from sodium metal. What is the bind

Travka [436]

Answer:

Binding Energy = 2.24 eV

Explanation:

First, we need to find the energy of the photon of light:

E = hc/λ

where,

E = Energy of Photon = ?

h = Plank's Constant = 6.626 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λ = wavelength of light = 400 nm = 4 x 10⁻⁷ m

Therefore,

E = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(4 x 10⁻⁷ m)

E = (4.97 x 10⁻¹⁹ J)(1 eV/1.6 x 10⁻¹⁹ J)

E = 3.1 eV

Now, from Einstein's Photoelectric Equation:

E = Binding Energy + Kinetic Energy

Binding Energy = E - Kinetic Energy

Binding Energy = 3.1 eV - 0.86 eV

<u>Binding Energy = 2.24 eV</u>

4 0

3 years ago

A student wants to get real and inverted

sveta [45]

Answer:

concave mirror

Explanation:

5 0

2 years ago

A sealed tank containing seawater to a height of 10.5 mm also contains air above the water at a gauge pressure of 2.95 atmatm. W

weqwewe [10]

Answer:

The water is flowing at the rate of 28.04 m/s.

Explanation:

Given;

Height of sea water, z₁ = 10.5 m

gauge pressure, $P_{gauge \ pressure}$ = 2.95 atm

Atmospheric pressure, $P_{atm}$ = 101325 Pa

To determine the speed of the water, apply Bernoulli's equation;

$P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2$

where;

P₁ = $P_{gauge \ pressure} + P_{atm \ pressure}$

P₂ = $P_{atm}$

v₁ = 0

z₂ = 0

Substitute in these values and the Bernoulli's equation will reduce to;

$P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 + \frac{1}{2}\rho (0)^2 = P_2 + \rho g(0) + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 = P_2 + \frac{1}{2}\rho v_2^2\\\\P_{gauge} + P_{atm} + \rho gz_1 = P_{atm} + \frac{1}{2}\rho v_2^2\\\\P_{gauge} + \rho gz_1 = \frac{1}{2}\rho v_2^2\\\\v_2^2 = \frac{2(P_{gauge} + \rho gz_1)}{\rho} \\\\v_2 = \sqrt{ \frac{2(P_{gauge} + \rho gz_1)}{\rho} }$