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Vsevolod [243]
3 years ago
11

Which is the IUPAC name for NO?

Chemistry
2 answers:
kykrilka [37]3 years ago
8 0

Answer:

nitrogen monoxide

Explanation:

for those taking quiz on edge

Vanyuwa [196]3 years ago
3 0

Answer:

Nitrogen (ii) oxide

Explanation:

To know the IUPAC name for NO, we shall determine the oxidation number of N in NO.

NOTE: The oxidation number of oxygen (O) is always – 2.

Thus the oxidation number of N in NO can be obtained as follow:

N + O = 0 (ground state)

N + (– 2) = 0

N – 2 = 0

Collect like terms

N = 0 + 2

N = +2

Thus, the oxidation number of Nitrogen (N) in NO is +2.

Therefore, the IUPAC name for NO is Nitrogen (ii) oxide

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The characteristics that distinguish one substance from another are called properties. A physical property is a characteristic of matter that is not associated with a change in its chemical composition. Familiar examples of physical properties include density, color, hardness, melting and boiling points, and electrical conductivity. Some physical properties, such as density and color, may be observed without changing the physical state of the matter. Other physical properties, such as the melting temperature of iron or the freezing temperature of water, can only be observed as matter undergoes a physical change

Figure :Copper and nitric acid undergo a chemical change to form copper nitrate and brown, gaseous nitrogen dioxide. During the combustion of a match, cellulose in the match and oxygen from the air undergo a chemical change to form carbon dioxide and water vapor. Cooking red meat causes a number of chemical changes, including the oxidation of iron in myoglobin that results in the familiar red-to-brown color change.

chemical change always produces one or more types of matter that differ from the matter present before the change. The formation of rust is a chemical change because rust is a different kind of matter than the iron, oxygen, and water present before the rust formed. The explosion of nitroglycerin is a chemical change because the gases produced are very different kinds of matter from the original substance. Other examples of chemical changes include reactions that are performed in a lab (such as copper reacting with nitric acid), all forms of combustion (burning), and food being cooked, digested, or rotting. Note that the number of atoms for a given element is calculated by multiplying the coefficient of any formula containing that element by the element’s subscript in the formula. If an element appears in more than one formula on a given side of the equation, the number of atoms represented in each must be computed and then added together. For example, both product species in the example reaction, CO2 and H2O, contain the element oxygen, and so the number of oxygen atoms on the product side of the equation is

(1 CO2 molecule × 2 O atoms per CO2 molecule) + (2 H2O molecules × 1 O atom per H2O(molecule) = 4 O atoms

The equation for the reaction between methane and oxygen to yield carbon dioxide and water is confirmed to be balanced per this approach, as shown here: CH 4 + 2O2⟶CO2 + 2H2O

Element Reactants Products Balanced?

C 1×1 = 1 1×1 = 1 1 = 1, yes

H 4×1 = 4 2×2 = 4 4 = 4, yes

O 2×2 = 4 (1×2) + (2×1) = 4 4 = 4, yes

A balanced chemical equation often may be derived from a qualitative description of some chemical reaction by a fairly simple approach known as balancing by inspection. Consider as an example the decomposition of water to yield molecular hydrogen and oxygen. This process is represented qualitatively by an unbalanced chemical equation: H2O⟶H2 + O2 (unbalanced)

Comparing the number of H and O atoms on either side of this equation confirms its imbalance:

Element Reactants Products Balanced?

H 1×2 = 2 1×2 = 2 2 = 2, yes

O 1×1 = 1 1×2 = 2 1 ≠ 2, no

H2O to H2 O2 would yield balance in the number of atoms, but doing so also changes the reactant’s identity (it’s now hydrogen peroxide and not water). The O atom balance may be achieved by changing the coefficient for H2O to 2.

2H2O⟶H2 + O2 (unbalanced)

Element Reactants Products Balanced?

H 2×2 = 4 1×2 = 2 4 ≠ 2, no

O 2×1 = 2 1×2 = 2 2 = 2, yes

The H atom balance was upset by this change, but it is easily reestablished by changing the coefficient for the H2 product to 2.

2H2O⟶2H2 + O2 (balanced)

Element Reactants Products Balanced?

H 2×2 = 4 2×2 = 4 4 = 4, yes

O 2×1 = 2 1×2 = 2 2 = 2, yes

These coefficients yield equal numbers of both H and O atoms on the reactant and product sides, and the balanced equation is, therefore:

2H2O⟶2H2 + O2

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7 0
2 years ago
How many liters of space are in 4.00x10^22 of O2 occupy at STP
irinina [24]

Answer:

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Explanation:

  • 1 mol ≡ 6.022 E23 molecules

∴ molecules O2 = 4.00 E22 molecules

⇒ moles O2 = (4.00 E22 molecules O2)×(mol O2/6.022 E23 molecules)

⇒ moles O2 = 0.0664 moles

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∴ T = 25°C ≅ 298 K

∴ P = 1 atm

assuming ideal gas:

∴ V = RTn/P

⇒ V O2 = ((0.082 atm.L/K.mol)(298 K)(0.0664 mol))/( 1 atm)

⇒ V O2 = 1.623 L

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Deffense [45]

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So the answer is True

Explanation:

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