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3 years ago

The 3 components of air

Chemistry

2 answers:

jok3333 [9.3K]3 years ago

8 0

Nitrogen (around 78%), Oxygen (around 21%), and Argon (around 1%).

Hope this helps :)

Lady bird [3.3K]3 years ago

7 0

The Main 3 components are Nitrogen, Oxygen, and Argon.

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HELPPP ME WITH SICENCE PLZZZZZZ

Goryan [66]

Answer:

molecules

Explanation:

molecules because the summer like atoms and Sciences sciency to be honest

7 0

3 years ago

Read 2 more answers

what mass of carbon dioxide will be produced when 12.9 g of butane reacts with an excess of oxygen in the following reaction?

Nadya [2.5K]

39.1 gCO2
I think if that’s an option

3 0

3 years ago

Какая масса хлора и какое количество вещества необходимо для реакции с 12г магния при получении хлорида магния?

Naddika [18.5K]

Mg + Cl₂ = MgCl₂M(Mg) = 24г/моль m 1 моль Mg = 24 г.По условию задачи дано 12г. Mg Количество вещества n(Mg) =12÷24=0,5 мольРассуждаем: по уравнению реакции с 1 моль магния реагирует 1 моль хлора, следовательно с 0,5 моль будет реагировать 0,5 моль хлора.1 моль хлора при н.у. занимает объем 22,4л. , тогда 0,5 моль хлора займет:0,5х22,4л.= 11,2л. Ответ: Для взаимодействии 12 г. магния потребуется 11,2 л. хлора.

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4 years ago

If the visible light spectrum is from 400 to 700 nm, would light with an energy of 2.79 x 10^-19 J be visible with the naked eye

sladkih [1.3K]

Answer:

713 nm. It is not visible with the naked eye.

Explanation:

Step 1: Given data

Energy of light (E): 2.79 × 10⁻¹⁹ J

Planck's constant (h): 6.63 × 10⁻³⁴ J.s

Speed of light (c): 3.00 × 10⁸ m/s

Wavelength (λ): ?

Step 2: Calculate the wavelength of the light

We will use the Planck-Einstein equation.

E = h × c / λ

λ = h × c / E

λ = 6.63 × 10⁻³⁴ J.s × 3.00 × 10⁸ m/s / 2.79 × 10⁻¹⁹ J

λ = 7.13 × 10⁻⁷ m

Step 3: Convert "λ" to nm

We will use the relationship 1 m = 10⁹ nm.

7.13 × 10⁻⁷ m × (10⁹ nm/1 m) = 713 nm

This light is not in the 400-700 nm interval so it is not visible with the naked eye.

5 0

3 years ago

Electron transfer is the term for this process. The resulting anion and cation are attracted by Coulombic forces and an ionic co

Nitella [24]

<h2>The compound formed is NaCl and $CaCl_{2}$ </h2>

Explanation:

PART 1:

Atomic number of sodium is 11.
Sodium(Na, Z = 11, Group 1A) will lose 1 electron to become $Na^{1+}$ which is isoelectronic to Neon (Ne ,Z = 10).
The symbol for the sodium ion is $Na^{+}$ .
Chlorine atom ( $Cl$ , Z = 17, Group 7A or 17) gains 1 elecron to be isoelectronic to neon.
The symbol for the compound formed is NaCl.
PART 2:
Atomic number of calcium is 20
Calcium(Ca, Z = 20 ,Group 2A) will lose two electrons to become $Ca^{2+}$ which is isoelectronic to argon.
The symbol for the ion is $Ca^{2+}.$
Two chlorine atoms (Cl, Z = 17, Group 7A or 17) each gains one electron to be isoelectronic to argon(Z = 18)
The symbol for the compound formed is $CaCl_{2}$ .

8 0

4 years ago