What is the approximate wavelength of radio waves

If an electron travels at a velocity of 1.000 x 107 m/s and has a mass of 9.109 x 10-28 g, what is its wavelength?

alina1380 [7]

Good luck i don’t even know

3 0

3 years ago

Use the periodic table to choose the element that matches each description.

nika2105 [10]

Using the periodic table to choose the element that matches each description include the following below:

Halogen: iodine

Group IIA: magnesium

Nonreactive: argon

Alkali metal: potassium.

<h3>What is a Periodic table?</h3>

This contains elements which are arranged according to the order of their atomic number in a tabular form. There are 18 groups which are the vertical columns present while there are 8 periods which are the horizontal rows present in the periodic table.

Example of an alkali metal is potassium while the non reactive ones include argon, neon etc. Examples of halogens include chlorine, iodine etc. are the ones which have seven electrons in their outer electron shells thereby just requiring one electron to achieve to obtain a stable octet configuration.

These are therefore the elements which match the descriptions provided in this case and is the most appropriate choice.

Read more about Periodic table here brainly.com/question/15987580

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8 0

2 years ago

Determine the value of the equilibrium constant, Kgoal, for the reactionN2(g)+H2O(g)⇌NO(g)+12N2H4(g), Kgoal=?by making use of th

Black_prince [1.1K]

Answer:

$K_{goal}=1.793*10^{-33}$

Explanation:

N2(g)+O2(g)⇌2NO(g), $K_1 = 4.10*10^{-31}$

N2(g)+2H2(g)⇌N2H4(g), $K_2 = 7.40*10^{-26}$

2H2O(g)⇌2H2(g)+O2(g), $K_3 = 1.06*10^{-10}$

If we add above reaction we will get:

2N2(g)+2H2O(g)⇌2NO(g)+N2H4(g) Eq (1)

Equilibrium constant for Eq (1) is $K_1*K_3*K_3$

Divide Eq (1) by 2, it will become:

N2(g)+H2O(g)⇌NO(g)+1/2N2H4(g) Eq (2)

Equilibrium constant for Eq (2) is $(K_1*K_3*K_3)^{1/2}$

$Equilibrium constant =K_{goal}= (K_1*K_2*K_3)^{1/2}\\K_{goal}= (4.10*10^{-31} *7.40*10^{-26}*1.06*10^{-10})^{1/2}\\K_{goal}=1.793*10^{-33}$

7 0

3 years ago

The basic principle in balancing a chemical equation is to ______. View Available Hint(s) The basic principle in balancing a che

OlgaM077 [116]

Answer:

have the same number of atoms of each element in the reactants and in the products

Explanation:

<em>The basic principle in balancing a chemical equation would simply be to have the same number of atoms of each element in the reactants and in the products.</em>

<u>A balanced chemical equation is one that has the same number of atoms of each element on the reactant and the product's side of the equation.</u> For example, consider the equation below:

$H_2 + O_2 --> H_2O$

On the reactant's side, there are 2 atoms of H and O while there are 2 atoms of H and 1 atom of O on the product's side. This is an imbalanced equation. In order for it to be balanced, the number of atoms of H and O on the reactant side must be equal to the number of H and O on the product side as below.

$2H_2 + O_2 --> 2H_2O$

5 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.

azamat

Answer:

(a) $m_{gold}=7.322g$

(b)

$V_{gold}=0.379cm^3$

$V_{copper}=0.122cm^3$

(c) $\rho _{coin}=15.94g/cm^3$

Explanation:

Hello,

(a) In this case, with the given formula we easily compute the mass of gold contained in the sovereign as shown below:

$m_{gold}=\frac{m_{tota}*karats}{24}=\frac{7.988g*22}{24}=7.322g$

(b) Now, by knowing the density of gold and copper, 19.32 and 8.94 g/cm³ respectively, we compute each volume, by also knowing that the rest of the coin contains copper:

$V_{gold}=\frac{m_{gold}}{\rho_{gold}} =\frac{7.322g}{19.32g/cm^3}=0.379cm^3$

$m_{copper}=7.988g-7.322g=1.09g\\V_{copper}=\frac{m_{copper}}{\rho_{copper}}=\frac{1.09g}{8.94g/cm^3} \\\\V_{copper}=0.122cm^3$

(c) Finally, the volume is computed by dividing the total mass over the total volume containing both gold and copper:

$\rho _{coin}=\frac{m_{total}}{V_{gold}+V_{copper}}=\frac{7.988 g}{0.379cm^3+0.122cm^3}\\ \\\rho _{coin}=15.94g/cm^3$

Best regards.

3 0

3 years ago

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