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Mars2501 [29]
3 years ago
10

How many moles of neon occupy a volume of 14.3 l at stp? how many moles of neon occupy a volume of 14.3 l at stp? 1.57 moles 0.6

38 moles 36.7 moles 32.0 moles 6.45 moles?
Chemistry
1 answer:
andriy [413]3 years ago
5 0
The  number   of  neon  moles  that  occupy  a volume of 14.3 l  at STP is calculated as follows

At STP 1  mole = 22.4 liters

what about  14.3 liters

by cross  multiplication
= (1 mole x 14.3 l)/22.4 l =0.638  moles  of  neon


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A gas occupies a volume at 34.2 mL at a temperature of 15.0 C and a pressure of 800.0 torr. What will be the volume of this gas
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The answer is 34.1 mL.
Solution:
Assuming ideal behavior of gases, we can use the universal gas law equation
     P1V1/T1 = P2V2/T2
The terms with subscripts of one represent the given initial values while for terms with subscripts of two represent the standard states which is the final condition.
At STP, P2 is 760.0torr and T2 is 0°C or 273.15K. Substituting the values to the ideal gas expression, we can now calculate for the volume V2 of the gas at STP:
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Read 2 more answers
What is the name for this molecule? a skeletal model of an 8-carbon zig zag chain. there is a double bond between the first and
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The name of the compound by using the <u>IUPAC nomenclature of organic compounds</u> is 1 -octene. The correct option is the last option - 1-octene.

<h3>Nomenclature of Organic compounds</h3>

From the question, we are to determine the name of the given molecule.

To name the compound, we will follow the IUPAC rules.

Some of IUPAC rules are

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  • For Alkenes (organic compounds with double bond), number the chain of carbons that includes the C=C so that the C=C has the lower position number. Change “ane” to “ene” and assign a position number to the first carbon of the C=C.

The given compound has 8 carbons and a double bond. The root name of the compound is octane.

By <u>IUPAC rules</u>, the compound is an <u>Octene</u>.

Since the double bond is between carbon-1 and carbon-2. The compound becomes 1-octene.

Hence, the name of the compound by using the <u>IUPAC nomenclature of organic compounds</u> is 1 -octene. The correct option is the last option - 1-octene.

Learn more on Nomenclature of Organic compounds here: brainly.com/question/26754333

The diagram for the compound is attached below.

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What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb
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0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

(\text{CH}_3)_3\text{N} in this question acts as a weak base. As seen in the equation in the question, (\text{CH}_3)_3\text{N} produces \text{OH}^{-} rather than \text{H}^{+} when it dissolves in water. The concentration of \text{OH}^{-} will likely be more useful than that of \text{H}^{+} for the calculations here.

Finding the value of [\text{OH}^{-}] from pH:

Assume that \text{pK}_w = 14,

\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}.

[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}.

Solve for [(\text{CH}_3)_3\text{N}]_\text{initial}:

\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}

Note that water isn't part of this expression.

The value of Kb is quite small. The change in (\text{CH}_3)_3\text{N} is nearly negligible once it dissolves. In other words,

[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}.

Also, for each mole of \text{OH}^{-} produced, one mole of (\text{CH}_3)_3\text{NH}^{+} was also produced. The solution started with a small amount of either species. As a result,

[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}.

\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3},

[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b},

[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}.

8 0
3 years ago
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