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timofeeve [1]
2 years ago
7

Allotropes are different molecular forms of an element, such as dioxygen (O₂) and ozone (O₃). (b) Calculate the ratio of densiti

es, d₀₃/d₀₂, and explain the significance of this number.
Chemistry
1 answer:
kipiarov [429]2 years ago
4 0

The ratio of densities, d₀₃/d₀ is 3:2

We know that at STP all gas have the same volume that is $22.4 \mathrm{~L}$.

Dioxygen $1 \mathrm{mLe} \mathrm{O}_{2}=32 \mathrm{gm} / \mathrm{moL} .$

Ozone 1 mole  $\mathrm{O}_{3}=48 \mathrm{gm} / \mathrm{mol}$.

$\begin{aligned} \therefore \text { density of } \mathrm{O}_{2} \text { at STP } &=\frac{\text { mass of } \mathrm{O}_{2}}{\text { volume }} \\ \text { or, } d_{\mathrm{O}_{2}} &=\frac{32 \mathrm{gm} / \mathrm{mol}_{\mathrm{o}}}{22.4 \mathrm{~L}} \end{aligned}$

$\therefore$The ratio of densities,

$\frac{d_{3}}{d o_{2}}=\frac{\frac{48 \mathrm{gm} / \mathrm{mol}}{22.4 \mathrm{~L}}}{\frac{32 \mathrm{gm} / \mathrm{mol}}{22.4 \mathrm{~L}}}$

$\frac{d 0_{3}}{d 0_{2}}=\frac{3}{2}$

do3: do2 = 3:2

Oxygen is found naturally as a molecule. two oxygen atoms strongly bind together with a covalent double bond to shape dioxygen or O2. Oxygen is normally observed as a molecule. it is referred to as dioxygen.

After hydrogen and helium, oxygen is the third most plentiful element in the universe and the most established element in the world. two atoms of the factors combine to generate dioxygen, an odourless and colourless diatomic gasoline with the formulation O2, at ordinary temperature and stress.

Learn more about dioxygen brainly.com/question/19905677

#SPJ4

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jasenka [17]
B the answer is letter b
8 0
3 years ago
Read 2 more answers
C3H8 (g) + O2(g) → CO2 (g) + H2O(g)
lorasvet [3.4K]

Answer:

11.55 mol ≈ 11.6 mol CO2

Explanation:

                                  C3H8 (g) + 5O2(g) → 3CO2 (g) + 4H2O(g)

from reaction            1 mol                           3 mol

given                          3.85 mol                    x mol

Solving proportion

x =(3.85*3)/1 = 11.55 mol ≈11.6 mol

8 0
3 years ago
Based on the activity series provided, which reactants will form products? F > Cl > Br > I CuI2 Br2 Right arrow. Cl2 Al
AleksAgata [21]

The reactants based on the reactivity capable of forming the products are \rm CuI_2\;and\;Br_2. Thus, option A is correct.

The reactivity series has been the arrangement of the elements based on their reactivity. The highly reactive element has been capable of replacing the less reactive element in the compound, and form the product.

In the following reactions, the new product has been formed by:

  • \rm CuI_2\;and\;Br_2

From the reactivity series, Br has been more reactive than I, and thereby replaces iodine in the compound for the formation of product.

  • \rm Cl_2\;and\;AlF_3

The compound has been formed with F. Cl has been less reactive than F, and will not be able to form the product.

  • \rm Br_2\;and\;NaCl

The compound has been formed with Cl. Br has been less reactive than Cl, thus will not displaces Cl for the formation of new product.

  • \rm CuF_2\;and\;I_2

The compound has been formed with F. I has been less reactive than F, and will not be able to form the product.

The reactants that will be able to form the products have been \rm CuI_2\;and\;Br_2. Thus, option A is correct.

For more information about the reactivity series, refer to the link:

brainly.com/question/10443908

5 0
3 years ago
What is the approximate pH of a .06 M solution of CH3COOH given that Ka= 1.78*10-5
Step2247 [10]
CH₃COOH ⇔ CH₃COO⁻ + H⁺

[CH₃COO⁻] = [H⁺] = x

K=\frac{|CH_{3}COO^{-}|*|H^{+}|}{|CH_{3}COOH|}=\frac{x^{2}}{|CH_{3}COOH|}\\\\
1,78*10^{-5}=\frac{x^{2}}{0,06} \ \ |*0,06\\\\
0,1068*10^{-5}=x^{2}\\\\
x_{1}\approx0,001 \ \land \ \ x_{2}=\approx-0,001\\\\
pH=-log|H^{+}|=-log0,001=3
4 0
3 years ago
Un globo lleno de helio tenia un volumen de 8.5 L en el suelo a 20°C y a una presión de 750 torr. Cuando se le soltó, el globo s
Ray Of Light [21]

Answer:

El volumen del gas era 12.95 L

Explanation:

Se relaciona la presión y el volumen mediante la ley de Boyle, que dice:

“El volumen ocupado por una determinada masa gaseosa a temperatura constante, es inversamente proporcional a la presión”

La ley de Boyle se expresa matemáticamente como:  P*V=k

Por otro lado, la Ley de Charles consiste en la relación que existe entre el volumen y la temperatura absoluta de una cierta cantidad de gas ideal, el cual se mantiene a una presión constante. Esta ley dice que cuando la cantidad de gas y de presión se mantienen constantes, el cociente que existe entre el volumen y la temperatura siempre tendrán el mismo valor:  

\frac{V}{T}=k

Por último, la Ley de Gay Lussac dice que la temperatura absoluta y la presión son directamente proporcionales. Es decir, cuando se mantiene todo lo demás constante, mientras suba la temperatura de un gas subirá también su presión. Y mientras la temperatura del gas baje, lo mismo ocurrirá con la presión:

\frac{P}{T}=k

Combinado las mencionadas tres leyes se obtiene:

\frac{P*V}{T} =k

Cuando se desean estudiar dos diferentes estados, uno inicial y una final de un gas, se puede aplicar:

\frac{P1*V1}{T1} =\frac{P2*V2}{T2}

Recordando que la temperatura debe usarse en grados Kelvin, conoces los siguientes datos:

  • P1: 750 torr
  • V1: 8.5 L
  • T1: 20°C= 293°K (siendo 0°C=273°K)
  • P2: 425 torr
  • V2: ?
  • T2: -20°C= 253 °K

Reemplazando:

\frac{750 torr*8.5 L}{293K} =\frac{425 torr*V2}{253 K}

Resolviendo:

V2=\frac{750 torr*8.5 L}{293K} *\frac{253 K}{425 torr}

V2= 12.95 L

<u><em>El volumen del gas era 12.95 L</em></u>

<u><em></em></u>

5 0
3 years ago
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