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Kryger [21]
3 years ago
5

CH=C-CH=CH– CH = CH, what is the name of this molecule?​

Chemistry
1 answer:
Kobotan [32]3 years ago
3 0

Answer:

pent-3-ene-1-yne

Explanation:

1 2 3 4 5

CH ≡ C - CH = CH - CH3

IUPAC name : Pent-3-ene-1-yne

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A pea plant has two dominant alleles for plant height – tall. We say that this plant is?:
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NaOH (aq) + HCl(aq) H2O + NaCl (aq)
yuradex [85]

Answer:

The answer to your question is 0.005

Explanation:

Data

Volume of NaOH = 25 ml

[NaOH] = 0.2 M

moles of NaOH = ?

To solve this problem is not necessary to have the chemical reaction. Just use the formula of Molarity and solve it for moles.

Formula

Molarity = moles / volume

-Solve for moles

moles = Molarity x volume

-Convert volume to liters

          1000 ml ---------------- 1 l

              25 ml ---------------- x

               x = (25 x 1) / 1000

               x = 0.025 l

-Substitution

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-Result

moles = 0.005

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3 years ago
Which of these is a property of bases?
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3 years ago
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The reaction is proceeding at a rate of 0.0080 Ms-1 in 50.0 mL of solution in a system with unknown concentrations of A and B. W
Leokris [45]

Answer:

0.0010 mol·L⁻¹s⁻¹  

Explanation:

Assume the rate law is  

rate = k[A][B]²

If you are comparing two rates,

\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \dfrac{k_{2}\text{[A]}_2[\text{B]}_{2}^{2}}{k_{1}\text{[A]}_1[\text{B]}_{1}^{2}}= \left (\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}}\right ) \left (\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}\right )^{2}

You are cutting each concentration in half, so

\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}} = \dfrac{1}{2}\text{ and }\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}= \dfrac{1}{2}

Then,

\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \left (\dfrac{1}{2}\right ) \left (\dfrac{1}{2}\right )^{2} = \dfrac{1}{2}\times\dfrac{1}{4} = \dfrac{1}{8}\\\\\text{rate}_{2} = \dfrac{1}{8}\times \text{rate}_{1}= \dfrac{1}{8}\times \text{0.0080 mol$\cdot$L$^{-1}$s$^{-1}$} = \textbf{0.0010 mol$\cdot$L$^{-1}$s$^{-1}$}\\\\\text{The new rate is $\large \boxed{\textbf{0.0010 mol$\cdot$L$^\mathbf{{-1}}$s$^{\mathbf{-1}}$}}$}

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Answer:

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