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Lynna [10]
2 years ago
7

Calculate to three significant digits the density of dinitrogen monoxide gas at exactly and exactly . You can assume dinitrogen

monoxide gas behaves as an ideal gas under these conditions.
Chemistry
1 answer:
Ulleksa [173]2 years ago
5 0

Answer:

see explanation below

Explanation:

You miss the part of the temperature and pressure. According to what I found this is held under 30 °C (or 303 K) and 1 atm.

The problem states that we can treat this gas as an ideal gas, therefore, we can use the equation of an ideal gas which is:

PV = nRT (1)

Now, the density (d) is calculated as:

d = m/V (2)

We can rewrite (2) in function of mass of volume so:

m = d*V (3)

Now, the moles (n) of (1) can be calculated like this:

n = m /MM (4)

If we replace it in (1) and then, (3) into this we have the following:

PV = mRT/MM ----> replacing (3):

PV = dVRT/MM ----> V cancels out so finallly:

P = dRT/MM

d = P * MM / RT (5)

The molar mass of N2O is 44 g/mol So, replacing all the data we have:

d = 1 * 44 / 0.082 * 303

d = 1.77 g/L

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Answer:

A jump occurs when a core electron is removed.

Explanation:

A jump in ionization energy occurs when a core electron is removed. A large jump in the ionization energy easily be seen from the electronic configuration of an element.

For Beryllium, the electronic configuration of is 1s2 2s2.

There are two valence electrons in the outermost shell hence the ionization energy data for beryllium will show a sudden jump or increase in going from the second to the third ionization energy owing to the removal of a core electron

The electronic configuration for Nitrogen is 1s2 2s2 2p3. Five valence electrons are found in the outermost shell so the ionization energy data for nitrogen will show a sudden jump or increase in going from the fifth to sixth ionization energy because of the removal of a core electron

The electronic configuration of oxygen is 1s2 2s2 2p4. There are six valence electrons hence ionization energy for oxygen atom will show a sudden jump or increase in going from the sixth to the seventh ionization energy because of the removal of a core electron

The electronic configuration of Lithium is 1s2 2s1

There is one valence electron in its outermost shell so its ionization energy data will show a sudden jump or increase in going from the first to the second ionization energy because of the removal of a core electron.

8 0
3 years ago
The quantity 44 liters expressed in cubic meters is ____.
Lerok [7]

D. 0.044 m3

hope this helped

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Which statement about a chemical reaction is true?
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3 years ago
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For the reaction H₂O(g) + Cl₂O(g) → 2HClO(g), you know ΔS°rxn and S° of HClO(g) and of H₂O(g). Write an expression that can be u
expeople1 [14]

The expression for  S°(Cl2O)(g) is given by ,

S°(Cl2O)(g) = 2×S°(HClO)(g) - S°(H2O)(g) - ΔS°rxn

Given ,

The balanced chemical equation is ,

H2O (g) + Cl2O (g) → 2HClO (g)

The general standard entropy of the reaction is given by ,

ΔS°rxn = sum of the S°(Product ) - sum of the S°(reactants )

ΔS°rxn = ΣmS°products  -  ΣnS°reactants

where , m and n are the stoichiometry coefficients of each product and reactants .

Thus , the standard entropy of the given reaction is given by ,

ΔS°rxn = 2×S° (HClO)(g)  - [ S°(H2O)(g) + S°(Cl2O)(g) ]

ΔS°rxn = 2×S° (HClO)(g) - S°(H2O)(g) - S°(Cl2O)(g)

Thus the S°(Cl2O) is given by ,

S°(Cl2O)(g) = 2×S°(HClO)(g) - S°(H2O)(g) - ΔS°rxn

Hence the expression for the standard entropy of Cl2O (g) is given by ,

S°(Cl2O)(g) = 2×S°(HClO)(g) - S°(H2O)(g) - ΔS°rxn

<h3>What is a balance chemical reaction ?</h3>

A balanced chemical reaction is a type of reaction which include the reactants and products in same amount i.e. the no of mole on both side of the reaction is remains same or balanced .

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