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Yuri [45]
9 months ago
12

What is the number of molecules in 500m cube of Oxygen at room temperature

Chemistry
1 answer:
d1i1m1o1n [39]9 months ago
5 0

4.95 is the number of molecules in 500m cube of Oxygen at room temperature and pressure

we have got data

pressure=1atm

temperature=298 k

volume=500 m^3

gas constant, R=8.31

now, by applying ideal gas equation

PV=nRT

1×500=n×8.31×298

n=2476.38/500

n=4.95

It is the lowest part of the material and has chemical elemental properties. Atoms do not independently exist; rather, they combine to form ions and molecules, which then combine to form the material that we can see, feel, and touch.

Molecules are made up of one or more atoms connected by covalent (chemical) connections. Atoms can be visualized as circles with a nucleus in the center (made up of protons and neutrons) and one or more concentric circles around it that indicate the "shells" or "levels" in which the electrons surrounding the atom's nucleus are located, as well as markings that identify the electron. each level

To know more about  molecules visit : brainly.com/question/11405437

#SPJ9

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Given the following balanced equation, if the rate of O2 loss is 3.64 × 10-3 M/s, what is the rate of formation of SO3? 2 SO2(g)
Fynjy0 [20]

Answer:

Rate of formation of SO₃ [\frac{d[SO_{3}] }{dt}] = 7.28 x 10⁻³ M/s

Explanation:

According to equation   2 SO₂(g) + O₂(g) → 2 SO₃(g)

Rate of disappearance of reactants = rate of appearance of products

                     ⇒ -\frac{1}{2} \frac{d[SO_{2} ]}{dt} = -\frac{d[O_{2} ]}{dt}=\frac{1}{2} \frac{d[SO_{3} ]}{dt}  -----------------------------(1)

    Given that the rate of disappearance of oxygen = -\frac{d[O_{2} ]}{dt} = 3.64 x 10⁻³ M/s

             So the rate of formation of SO₃ [\frac{d[SO_{3}] }{dt}] = ?

from equation (1) we can write

                                   \frac{d[SO_{3}] }{dt} = 2 [-\frac{d[O_{2}] }{dt} ]

                                ⇒ \frac{d[SO_{3}] }{dt} = 2 x 3.64 x 10⁻³ M/s

                                ⇒ [\frac{d[SO_{3}] }{dt}] = 7.28 x 10⁻³ M/s

∴ So the rate of formation of SO₃ [\frac{d[SO_{3}] }{dt}] = 7.28 x 10⁻³ M/s

7 0
3 years ago
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