The correct IUPAC name of the following structure CH3-CH2-CH=CH2 is But - 1 - ene
<h3>IUPAC rules for alkene nomenclature</h3>
- The parent alkene is chosen to have the longest carbon chain containing the carbon-carbon double bond.
- The alkane's suffix "ane" is changed to "ene."
- An alkene is referred to as a diene or triene depending on whether a double bond appears twice or three times in the parent chain.
- the location of side chains or double bonds denoted by the numbers 1, 2, 3, etc.
- The carbon atom in the double bond, which is written directly before the suffix "ene," receives the lowest number because it is the longest chain and is numbered from that end.
- If both sides of a double bond receive the same number as the chain is being numbered, the carbon chain is numbered so that the substituent receives the lowest number.
Hence the structure CH3-CH2-CH=CH2 is named on the basis of the above rule as but - 1 - ene.
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Answer:
[H2] = [I2] = 0.64M; [HI] = 4.72M
Explanation:
Based on the reaction:
H2 + I2 ⇄ 2HI
The K is defined as:
k = 54.3 = [HI]² / [H2] [I2]
<em>Where [] is molar concentration of each reactant at equilibrium</em>
As the initial concentration of HI is 6mol/dm^3 = 6M the equilibrium concentration of each reactant is:
[H2] = X
[I2] = X
[HI] = 6 - 2X
<em>Where X is reaction coordinate</em>
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Replacing:
54.3 = [6-2X]² / [X] [X]
54.3X² = 4X² - 24X + 36
0 = -50.3X² - 24X + 36
Solving for X:
X = -1.12. False solution, produce negative concentrations
X = 0.64M. Right solution
Replacing:
[H2] = 0.64M
[I2] = 0.64M
[HI] = 6 - 2*0.64M = 4.72M
Equilibrium concentrations are:
<h3>[H2] = [I2] = 0.64M; [HI] = 4.72M</h3>
Answer:In alpha decay, shown in Fig. 3-3, the nucleus emits a 4He nucleus, an alpha particle. Alpha decay occurs most often in massive nuclei that have too large a proton to neutron ratio. An alpha particle, with its two protons and two neutrons, is a very stable configuration of particles. Alpha radiation reduces the ratio of protons to neutrons in the parent nucleus, bringing it to a more stable configuration. Many nuclei more massive than lead decay by this method.
Consider the example of 210Po decaying by the emission of an alpha particle. The reaction can be written 210Po Æ 206Pb + 4He. This polonium nucleus has 84 protons and 126 neutrons. The ratio of protons to neutrons is Z/N = 84/126, or 0.667. A 206Pb nucleus has 82 protons and 124 neutrons, which gives a ratio of 82/124, or 0.661. This small change in the Z/N ratio is enough to put the nucleus into a more stable state, and as shown in Fig. 3-4, brings the "daughter" nucleus (decay product) into the region of stable nuclei in the Chart of the Nuclides.
In alpha decay, the atomic number changes, so the original (or parent) atoms and the decay-product (or daughter) atoms are different elements and therefore have different chemical properties.
Upper end of the Chart of the Nuclides
In the alpha decay of a nucleus, the change in binding energy appears as the kinetic energy of the alpha particle and the daughter nucleus. Because this energy must be shared between these two particles, and because the alpha particle and daughter nucleus must have equal and opposite momenta, the emitted alpha particle and recoiling nucleus will each have a well-defined energy after the decay. Because of its smaller mass, most of the kinetic energy goes to the alpha particle.
Answer:
Explanation:
you would use d=m over v which means density = mas divide by volume
Answer:
Where is your question? I do not see it.
Explanation: