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Nataliya [291]
4 years ago
11

Question 3 (1 point)

Chemistry
1 answer:
pentagon [3]4 years ago
5 0

Full question:

The IUPAC name for  CH3CH2C≡CCH3 is:

Answer:

2-pentyne

Explanation:

To name hydrocarbons, you first you have to identify the longest carbon chain.  There are 5 carbons in this chain, so we know the name is "pent".

You then have to identify the presence of any double or triple bonds. If double bonds, it is an alkene, if triple bonds, it is an alkyne. In this case there is a triple bond, so we know the hydrocarbon is pentyne.

You then number the chain to give the lowest number to the triple bond. It could either be 4 (countnig carbons from left to right) or 2 (from right to left). Therefore, the answer is 2-pentyne.

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What is Keq for the reaction N2 + 3H2 2NH3 if the equilibrium concentrations are [NH3] = 3 M, [N2] = 1 M, and [H2] = 2 M?
Licemer1 [7]

Answer : The correct option is, (D) k_{eq}=1.125

Explanation : Given,

Concentration of NH_3 = 3 M

Concentration of N_2 = 1 M

Concentration of H_2 = 2 M

The given balanced equilibrium reaction is,

N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)

The expression for equilibrium constant for this reaction will be,

K_{eq}=\frac{[NH_3]^2}{[N_2][H_2]^3}

Now put all the given values in this expression, we get the value of K_{eq}

K_{eq}=\frac{(3)^2}{(1)\times (2)^3}

K_{eq}=1.125

Therefore, the value of K_{eq} for the given reaction is, 1.125

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3 years ago
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CH2OCH2 is an organic compound with an oxygen atom bonded between the two carbon atoms, and it is not very soluble in water. Wha
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Answer:

Alcohol

Explanation:

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3 years ago
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What is an example of inert elements​
solmaris [256]
Neon, Helium, Krypton, Xenon, Argon
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if 3.0 grams of aluminum and 6.0 grams of bromine react to form AlBr3, how many grams of product would theoretically be produced
Gelneren [198K]
1) Chemical reaction: 2Al + 3Br₂ → 2AlBr₃.
m(Al) = 3,0 g.
m(Br₂) = 6,0 g.
n(Al) = m(Al) ÷ M(Al).
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n(Al) = 0,11 mol.
n(Br₂) = n(Br₂) ÷ m(Br₂).
n(Br₂) = 6 g ÷ 160 g/mol.
n(Br₂) = 0,0375 mol; limiting reagens.
n(Br₂) : n(AlBr₃) = 3 : 2.
n(AlBr₃) = 0,025 mol.
m(AlBr₃) = 0,025 mol · 266,7 g/mol.
m(AlBr₃) = 6,67 g.

2) m(Br₂) - all bromine reacts, so mass of bromine after reaction is zero grams (m(Br₂) = 0 g).
n(Al) = 0,11 mol - 0,025 mol = 0,085 mol.
m(Al) = 0,085 mol · 27 g/mol.
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n - amount of substance.
M - molar mass.

4 0
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