I need help like rn how many moles are in 44 L of NO2 gas

The pressure on a 200 milliliter sample of CO2 (g) at constant temperature is increased from

balu736 [363]

The answer for the following problem is mentioned below.

Therefore the final volume of the gas is 100 ml.

Explanation:

Given:

Initial pressure ( $P_{1}$ ) = 600 mm of Hg

Final pressure ( $P_{2}$ ) = 1200 mm of Hg

Initial volume ( $V_{1}$ ) = 200 ml

To find:

Final volume ( $V_{2}$ )

We know;

According to the ideal gas equation,

P × V = n × R × T

Where;

P represents the pressure of the gas

V represents the volume of the gas

n represents the no of moles of the gas

R represents the universal gas constant

T represents the temperature of the gas

So,

From the above mentioned equation,

P × V = constant

$\frac{P_{1} }{P_{2} }$ = $\frac{V_{1} }{V_{2} }$

Where,

( $P_{1}$ ) represents the initial pressure of the gas

( $P_{2}$ ) represents the final pressure of the gas

( $V_{1}$ ) represents the initial volume of the gas

( $V_{2}$ ) represents the final volume of the gas

So;

$\frac{600}{1200}$ = $\frac{V_{2} }{200}$

$V_{2}$ = 100 ml

Therefore the final volume of the gas is 100 ml.

5 0

3 years ago

Relate dark matter to the development of the universe after the Big Bang. In 3-5 sentences, speculate on how the development of

Alenkinab [10]

Answer:

Dark matter makes up 85% of the mass of the universe. Dark matter is not directly observable because it doesn't interact with any electromagnetic wave. In the development of the universe, without dark matter, the universe will not function, move or rotate as it does now (this speculation led to the quest to find the anomaly of mass and energy in the known universe, eventually leading to the idealization of dark matter) and will not have enough gravitational force to hold it together. After the big bang, the presence of dark matter and energy ensured that the newly formed universe didn't just float away, rather, it provided enough gravitational force to hold the universe while still allowing it to expand sufficiently.

The development of the universe would have been different without the universe in the sense that the young universe won't have enough mass to hold it together, and the universe would have simply floated apart. The behavior of the universe would have been different from what we observe now, and some physical laws that applies now will not apply to the universe.

6 0

3 years ago

Rank the following elements in order of decreasing atomic radius (1 is biggest, 5 is smallest)

AlexFokin [52]

There are various kind of elements that are present in periodic table. Some elements are harmful, some are radioactive, some are noble gases. The atomic radius in decreasing order is Bi>Sb>As>N>O.

<h3>What is periodic table?</h3>

Periodic table is a table in which we find elements with properties like metals, non metals, metalloids and radioactive element arranges in increasing atomic number.

Along the period, the size of elements decreases. Down the group the size of elements increases. The atomic radius in decreasing order is Bi>Sb>As>N>O.

Therefore, atomic radius in decreasing order is Bi>Sb>As>N>O.

Learn more about periodic table, here:

brainly.com/question/11155928

#SPJ1

3 0

1 year ago

Zinc, sulphur, oxygen What would this compound be called?

Nataliya [291]

Zinc sulphate

Zinc sulphateZnSO₄

This is the answer.

6 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