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3 years ago

How many molecules are in 28 grams of nitrogen gas in balloons?

Chemistry

1 answer:

melomori [17]3 years ago

6 0

Answer:

$6.022\cdot 10^{23}$ molecules of nitrogen gas

Explanation:

In order to convert the mass of a given compound into the number of molecules, we need to:

identify the number of moles of a given compound dividing its mass by the molar mass, $n = \frac{m}{M}$ ,
multiply the number of moles by the Avogadro's constant which tells us that 1 mole contains $N_A = 6.022\cdot 10^{23}$ molecules.

In this problem:

$m_{N_2} = 28~g$

$M_{N_2} = 28.01~g/mol$

The number of particles is then found by:

$N_{N_2} = n_{N_2}\cdot N_A = \frac{m_{N_2}}{M_{N_2}} N_A = \frac{28 g}{28.01 g/mol}\cdot 6.022\cdot 10^{23} mol^{-1} = 6.022\cdot 10^{23}$

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Why do atoms combine in an ionic bond

victus00 [196]

Answer:

Ionic bonds are formed through the exchange of valence electrons between atoms, typically a metal and a nonmetal. The loss or gain of valence electrons allows ions to obey the octet rule and become more stable. Therefore, ions combine in ways that neutralize their charges.

Explanation:

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3 years ago

The hypothetical elements shown here (figures a–d) do not include hydrogen or helium. Which element would you expect to bond cov

Marizza181 [45]

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3 years ago

To calculate the freezing point depression ___ solution's freezing point and the freezing point of it's pure solvent.

murzikaleks [220]

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3 years ago

What are the names of the following three compounds (see attached)

balu736 [363]

Answer:

1. 4-ethyl-1-heptene

2. 6-ethyl-2-octene

3. 1-butyne

Explanation:

The compounds are named according to IUPAC rules.

Compound 1:

Identify the longest carbon chain. This chain is called the parent chain.
Identify all of the substituents (groups appending from the parent chain).
The parent chain is numbered so that the multiple bonds have the lowest numbers (double has the priority over alkyl substituents).
The longest chain contains 7 carbon atoms, so taken the name hept.
The double bond between C1 and C2, so take no. 1 and add the suffix ene to hept "1-heptene".
The ethyl group is the alkyl substituent on position 4.
So the name is 4-ethyl-1-heptene.

Compound 2:

Identify the longest carbon chain. This chain is called the parent chain.
Identify all of the substituents (groups appending from the parent chain).
The parent chain is numbered so that the multiple bonds have the lowest numbers (double has the priority over alkyl substituents).
The longest chain contains 8 carbon atoms, so taken the name oct.
The double bond between C2 and C3, so take no. 2 and add the suffix ene to oct "2-octene".
The ethyl group is the alkyl substituent on position 6.

So the name is 6-ethyl-2-octene.

Compound 3:

Identify the longest carbon chain. This chain is called the parent chain.
Identify all of the substituents (groups appending from the parent chain), there is no substituents.
The parent chain is numbered so that the multiple bonds have the lowest numbers (Triple bond here take the lowest number).
The longest chain contains 4 carbon atoms, so taken the name but.
The triple bond between C1 and C2, so take no. 1 and add the suffix yne to but "1-butyne".

So the name is 1-butyne.

7 0

3 years ago

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

Stolb23 [73]

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