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Anton [14]
1 year ago
5

Calculate the grams of so2 gas present at stp in a 5. 9 l container. (atomic weights: s=32, o=16).

Chemistry
1 answer:
Solnce55 [7]1 year ago
8 0

The 5.9 L container at STP contains 16.83 g of SO2.

To begin, we must determine how many moles of SO2 are contained in the 5.9 L container.

At standard temperature and pressure (STP), one mole of SO2 equates to 22.4 liters.

5.9 L = 5.9 / 22.45.9 L = 0.26 moles of SO2 as a result.

The result is that 0.263 moles of SO2 are present in the container.

Following that, we shall determine the mass of 0.263 moles of SO2.

The following are ways to do this:

The mass of SO2 is 32 + (16 2) = 32 + 32= 64 g/mol,

where SO2 Mole = 0.263 Molar.

Molar + Molar Mass Equals Mass

SO2 mass is 16.83 g, or 0.263 of a kilogram.

Therefore, 16.83 g of SO2 are present in the 5.9 L container at STP.

to know more about mass, visit to:-

brainly.com/question/15959704

#SPJ4

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The electron will have anti-clockwise notation.

We know from the Pauli exclusion principle, no two electrons in an atom can have all the four quantum numbers i.e. principal quantum number (n), azimuthal quantum number (l), magnetic quantum number (m) and spin quantum number (s) same. The importance of the principle also restrict the possible number of electrons may be present in a particular orbital.

Let assume for an 1s orbital the possible values of four quantum numbers are n = 1, l = 0, m = 0 and s = \frac{+}{-}\frac{1}{2}.

The exclusion principle at once tells us that there may be only two unique sets of these quantum numbers:

1, 0, 0, +\frac{1}{2} and 1, 0, 0, -\frac{1}{2}.

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6 0
3 years ago
What are the traditional states of matter, and what type of property is a matter's state?
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6 0
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Read 2 more answers
An important reaction in the formation of photochemical smog is the photodissociation of NO2: NO2 + hv >>> NO(g) + O(g)
vaieri [72.5K]

Explanation:

NO_2 + hv\rightarrow NO(g) + O(g)

The maximum wavelength of light that can cause this reaction is 420 nm.

a) The wavelength given lies in the range of visible light range that is from 400 nano meters to 700 nano meters.

The light with wavelength of 420 nm is found in the range of visible light.

b)The maximum strength of a bond :

E=\frac{h\times c}{\lambda}

where,

E = energy of photon = Energy required to break single molecule of nitrogen dioxide

h = Planck's constant = 6.626\times 10^{-34}Js

c = speed of light = 3\times 10^8m/s

\lambda = wavelength = 420 nm=420\times 10^{-9}m

E=\frac{6.626\times 10^{-34}Js\times 3\times 10^8 m/s}{420\times 10^{-9}m}

E=4.733\times 10^{-19} J

Energy required to break 1 mole of nitrogen dioxide molecules:

= E\times N_A=4.733\times 10^{-19} J\times 6.022\times 10^{23} mol^{-1}

=285,012.66 J/mol=285.13 kJ/mol

(1 J = 0.001 kJ )

285.13 is the maximum strength of a bond, in kJ/mol, that can be broken by absorption of a photon of 420-nm light.

c) the photodissociation reaction showing Lewis-dot structures is given in an image attached.

6 0
3 years ago
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