All categories

2 years ago

How many moles are in 68.5 liters of oxygen gas at STP?

Chemistry

2 answers:

mr Goodwill [35]2 years ago

8 0

Answer:

3.0580 moles are in 68.5 liters of oxygen gas at STP

Explanation:

Let the number of moles of oxygen gas be n

Volume occupied by n moles of oxygen gas at STP = 68.5 L

At STP, 1 mol of gas occupies 22.4 L

Then 68.4 L of volume will be occupied by:

$\frac{1}{22.4} \times 68.5 L=3.0580 mol$

3.0580 moles are in 68.5 liters of oxygen gas at STP

givi [52]2 years ago

3 0

Since we know that one mole of any gas at STP is equal to 22.4 L we can multiply 135L by the following conversion: 1 mole/22.4L. When you set up the problem it looks like this…: (135L)x 1 mole/22.4L =6.03 moles of oxygen gas The liters cancel out and you are left with moles as your units.

So your answer is then 3.058

You might be interested in

In an ideal situation where no heat energy is produced, what is the relationship between the chemical energy provided by the bat

kow [346]

Answer:

See explanation

Explanation:

The principle of conservation of energy states that energy can neither be created nor destroyed but can be converted from one form to another. Hence, chemical energy in a battery can be converted to electrical energy.

Usually, the conversion of energy from one form to another is not 100% efficient according to the second law of thermodynamics. Some energy is wasted in the process, sometimes as heat.

Hence, in an ideal situation where no heat energy is produced; all the chemical energy is converted to electrical energy (100% energy conversion). There will be no energy loss if no heat is produced.

6 0

2 years ago

How can you tell if glucose is present?

rosijanka [135]

Answer:

Plants use photosynthesis to make glucose. Glucose is also know as sugar. You can tell it is present if the plant receives sunlight as well as water.

Explanation:

MARK BRAINLIEST

4 0

2 years ago

What are valency and radicals

PolarNik [594]

Valency- it means the combing capacity if an element.
<span> radical- it is an atom, molecule, or ion that has unpaired valence electrons or an open electron shell.
</span>

6 0

2 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

What type of reaction is most likely to occur when barium reacts with fluorine? write the chemical equation for the reaction?

scZoUnD [109]

Answer is: this is synthesis reaction.<span>
</span>Chemical reaction: Ba + F₂ → BaF₂.
Synthesis reaction<span> is a type of reaction in which multiple reactants combine to form a single product.
</span>In barium fluoride, barium has oxidation number +2 and fluorine has oxidation number -1, so compound has neutral charge.

6 0

2 years ago