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AysviL [449]
3 years ago
6

How many liters of 3.5 M solution can be made using 23 moles of LiBr *must show work to get credit*

Chemistry
1 answer:
MrMuchimi3 years ago
5 0

<u>Answer:</u> 6.57 L of solution can be made.

<u>Explanation:</u>

Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:

\text{Molarity of solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (L)}} .....(1)

Given values:

Molarity of LiBr = 3.5 M

Moles of LiBr = 23 moles

Putting values in equation 1, we get:

3.5mol/L=\frac{23mol}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{23mol}{3.5mol/L}=6.57L

Hence, 6.57 L of solution can be made.

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Answer:

C₃H₈(g) + 6 H₂O(g) ⇒ + 10 H₂(g) + 3 CO₂(g)

Explanation:

Propane can be turned into hydrogen by the two-step reforming process.

In the first step, propane and water react to form carbon monoxide and hydrogen. The balanced chemical equation is:

C₃H₈(g) + 3 H₂O(g) ⇒ 3 CO(g) + 7 H₂(g)

In the second step, carbon monoxide and water react to form hydrogen and carbon dioxide. The balanced chemical equation is:

CO(g) + H₂O(g) ⇒ H₂(g) + CO₂(g)

In order to get the net chemical equation for the overall process, we have to multiply the second step by 3 and add it to the first step. Then, we cancel what is repeated.

C₃H₈(g) + 3 H₂O(g) + 3 CO(g) + 3 H₂O(g) ⇒ 3 CO(g) + 7 H₂(g) + 3 H₂(g) + 3 CO₂(g)

C₃H₈(g) + 6 H₂O(g) ⇒ + 10 H₂(g) + 3 CO₂(g)

4 0
3 years ago
How is n1 in the Rydberg equation related to the quantum number n in the Bohr model?How is n1 in the Rydberg equation related to
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If you assume this reaction is driven to completion because of the large excess of one ion, what is the concentration of [Fe(SCN
viktelen [127]

Answer : The concentration of [Fe(SCN)]^{2+} is, 4.32\times 10^{-4}M

Explanation :

When we assume this reaction is driven to completion because of the large excess of one ion then we are assuming limiting reagent is SCN^- and Fe^{3+} is excess reagent.

First we have to calculate the moles of KSCN.

\text{Moles of }KSCN=\text{Concentration of }KSCN\times \text{Volume of solution}

\text{Moles of }KSCN=0.00180M\times 0.006L=1.08\times 10^{-5}mol

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Now we have to calculate the concentration of [Fe(SCN)]^{2+}

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