All categories

1 year ago

If you have an altitude of 55 mi. (miles), which layer of the atmosphere are you in?

Chemistry

1 answer:

Kruka [31]1 year ago

8 0

$Answer:$ $You$ $would$ $be$ $in$ $the$ $D-region$

$The$ $reason$ $I$ $understand$ $this$ $is$ $because$ $the$ $D-region$ $stands$ $55$ $miles$ $in$ $altitude$ , $despite$ $it$ $will$ $likely$ $vanishes$ $at$ $night$ .

You might be interested in

P is the pressure in atmospheres (atm), V is the volume in liters (L), n is the number of moles, R is the gas constant (0.0821 L

yKpoI14uk [10]

Answer:

The mass of the neon gas m = 1.214 kg

Explanation:

Pressure = 3 atm = 304 k pa

Volume = 0.57 L = 0.00057 $m^{3}$

Temperature = 75 °c = 348 K

Universal gas constant = 0.0821 $\frac{L . atm}{mol K}$

We have to change the unit of this constant. it may be written as

Universal gas constant = 8.314 $\frac{KJ}{mol K}$

Gas constant for neon = $\frac{8.314}{20}$ = 0.41 $\frac{KJ}{kg K}$

From ideal gas equation,

P V = m R T ------- (1)

We have all the variables except m. so we have to solve this equation for mass (m).

⇒ 304 × $10^{3}$ × 0.00057 = m × 0.41 × 348

⇒ 173.28 = 142.68 × m

⇒ m = 1.214 kg

This is the mass of the neon gas.

4 0

2 years ago

What is one reason that scientists who study weather need extremely advanced technology?

Anton [14]

Answer: C) To monitor weather event restricted several days in advance.

Explanation:

5 0

2 years ago

Read 2 more answers

pyrite is a mineral composed of 46.5 mass % iron and 53.5 mass % sulfur. determine the empirical formula for calcocite.

padilas [110]

The empirical formula for pyrite is FeS2.

HOW TO CALCULATE EMPIRICAL FORMULA:

The empirical formula represents the simplest whole number ratio of constituents element of a compound. The empirical formula of pyrite can be calculated as follows:

46.5 mass % Fe = 46.5g of Fe

53.5 mass % S = 53.5g of S

Next, we divide each element's mass value by its molar mass

Fe = 46.5g ÷ 56g/mol = 0.83mol

S = 53.5g ÷ 32g/mol = 1.67mol

Next, we divide each mole value by the smallest (0.83mol)

Fe = 0.83mol ÷ 0.83 = 1

S = 1.67mol ÷ 0.83 = 2.014

Approximately, the ratio of Fe to S is 1:2. Therefore, the empirical formula of pyrite is FeS2.

Learn more at: brainly.com/question/14044066?referrer=searchResults

3 0

1 year ago

Is this correct? also for the mass i have to round it to the closest whole number

11Alexandr11 [23.1K]

A lithium atom has 3 protons and 3 electrons. It can lose one of its electrons, making it an ion. It now has more positive protons than electrons so it has an overall positive charge. Overall charge is +1

8 0

2 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