Calculate the molar mass for Mn2O3, an oxide of manganese.

A student isolated 7.2 g of 1-bromobutane reacting equimolar amounts of 1-butanol (10 ml) and NaBr (11.1 g) in the presence of s

Alla [95]

<u>Answer:</u> The percent yield of the 1-bromobutane is 48.65 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For NaBr:</u>

Given mass of NaBr = 11.1 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

$\text{Moles of NaBr}=\frac{11.1g}{103g/mol}=0.108mol$

The chemical equation for the reaction of 1-butanol and NaBr is:

$\text{1-butanol + NaBr}\rightarrow \text{1-bromobutane}$

By Stoichiometry of the reaction

1 mole of NaBr produces 1 mole of 1-bromobutane

So, 0.108 moles of NaBr will produce = $\frac{1}{1}\times 0.108=0.108$ moles of 1-bromobutane

Now, calculating the mass of 1-bromobutane from equation 1, we get:

Molar mass of 1-bromobutane = 137 g/mol

Moles of 1-bromobutane = 0.108 moles

Putting values in equation 1, we get:

$0.108mol=\frac{\text{Mass of 1-bromobutane}}{137g/mol}\\\\\text{Mass of 1-bromobutane}=(0.108mol\times 137g/mol)=14.80g$

To calculate the percentage yield of 1-bromobutane, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of 1-bromobutane = 7.2 g

Theoretical yield of 1-bromobutane = 14.80 g

Putting values in above equation, we get:

$\%\text{ yield of 1-bromobutane}=\frac{7.2g}{14.80g}\times 100\\\\\% \text{yield of 1-bromobutane}=48.65\%$

Hence, the percent yield of the 1-bromobutane is 48.65 %

5 0

3 years ago

The ksp of pbbr2 is 6.60× 10–6. what is the molar solubility of pbbr2 in 0.500 m kbr solution

Alex_Xolod [135]

The solubility of PbBr₂(s) with the presence of 0.500 M of KBr is 2.64 x 10⁻⁵ M.

4 0

3 years ago

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Direct electron transfer from a singlet reduced species to a triplet oxidizing species is quantum-mechanically forbidden

posledela

Direct electron transfer from a a singlet reduced species to a triplet oxidizing species is quantum-mechanically forbidden.

<h3><u>Transfer from singlet to triplet:</u></h3>

Either an excited singlet state or an excited triplet state will occur when an electron in a molecule with a singlet ground state is stimulated (through radiation absorption) to a higher energy level.
All electron spins in a molecule electronic state known as a singlet are coupled.
In other words, the ground state electron and the stimulated electron's spin are still coupled (a pair of electrons in the same energy level must have opposite spins, per the Pauli exclusion principle).
The excited electron and ground state electron are parallel in a triplet state because they are no longer coupled (same spin).
It is less likely that a triplet state would arise when the molecule absorbs radiation since excitation to a triplet state necessitates an additional "forbidden" spin transfer.

To view more questions on quantum mechanism, refer to:

brainly.com/question/13639384

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8 0

2 years ago

Which compound are ionic and which are covalent? (N2) (CCl4) (SiO2) (AlCl3) (CaCl2) (LiBr)

muminat

Covalent compounds: N2, CCl4, SiO2 and AlCl3.

Ionic compounds: CaCl2 and LiBr.

Hope this helps!

3 0

4 years ago

When solutions of silver nitrate and sodium chloride are mixed, silver chloride precipitates out of the solution according to th

kipiarov [429]

The concentration of the sodium chloride would be 0.082 M

<h3>Stoichiometric calculations</h3>

From the equation of the reaction, the ratio of AgCl produced to NaCl required is 1:1.

Mole of 46.6 g AgCl produced = 46.6/143.32 = 0.325 moles

Equivalent mole of NaCl = 0.325 moles.

Molarity of 0.325 moles, 3.95 L NaCl = mole/volume = 0.325/3.95 = 0.082 M

More on stoichiometric calculations can be found here: brainly.com/question/27287858

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3 0

2 years ago

Read 2 more answers