All categories

3 years ago

The frequency light of 3.52*10-7

Chemistry

1 answer:

zimovet [89]3 years ago

7 0

Answer:

The speed of light, c, equals the wavelength, λ (pronounced lambda), times the frequency, ν, (pronounced noo).

c=λν

c is a constant. It is usually given as 3.00×108 m/s or 3.00×1010 cm/s rounded to three significant figures.❤

You might be interested in

Mario wants to split a beam of white light into different colors. Which tool will work best?

cestrela7 [59]

The correct answer --> A

7 0

3 years ago

2. What is the name of the compound AIPO?

Andru [333]

Answer:

Aluminum phosphate

Explanation:

7 0

2 years ago

A sample of gas occupies 7.80 liters at 425°C? What will be the volume of the gas at 35°C if the pressure does not change?

balandron [24]

From ideal gas equation PV = nRT, V/T = nR/P ==> V/T = constant. Therefore V1/T1 = V2/T2 ==> 7.8/698 = V2/308. V = 3.44L {TEMPERATURE IN KELVIN = 273 + 425 AND 35 = 698 AND 308}

4 0

3 years ago

If you start a chemical reaction with 5 atoms of hydrogen, how many hydrogen atoms will you have when it is finished? Group of a

mote1985 [20]

Answer:

5 atoms

Explanation:

According to the law of conservation of mass, "matter is neither created nor destroyed in the cause of a chemical reaction".

We finish with what we start with in a chemical reaction. Although new species might form, the number of atoms on both sides of the expression will still be maintained.

All chemical reactions obey this law of conservation.

8 0

3 years ago

Consider the following reaction:

adell [148]

Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

2H₂O₂ ⟶ 2H₂O + O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

d[H₂O]/dt = 2d[O₂]/dt = 2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = 6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂

(a) Initial moles of H₂O₂

$\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }$

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

$\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }$

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

$\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}$

8 0

3 years ago