Which of the following is a radioisotope used to date rock formations older than 50,000 years old?

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]

Answer: The percent yield of the 1-bromobutane is 48.65 %

Explanation:

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

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

For NaBr:

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

Difference between pure and applied research

nata0808 [166]

Answer:

pure is water made by us

Explanation:

lap water are pured also but somtimes they are dirty

6 0

2 years ago

Read 2 more answers

What is the endpoint of a titration?Select one:When there is no acid and all base.When the amount of acid and base are equal.Whe

user100 [1]

Answer:

The correct option is: When the amount of acid and base are equal

Explanation:

Titration is an analytic method that is used to determine the concentration of an unknown solution, called titrand.

In this method, standard solution of known concentration, called titrant, is taken in the burette and added drop-wise to the titrand solution in the flask, until the endpoint is reached.

In case of an acid-base titration, a pH indicator is used, which changes the color of the solution when the endpoint is reached.

The endpoint indicates the equivalence point of an acid-base titration, where the concentration of the acid and base is equal.

Therefore, the correct option is: When the amount of acid and base are equal

3 0

3 years ago

The compound trimethylamine, (CH3)3N, is a weak base when dissolved in water. Write the Kb expression for the weak base equilibr

Zina [86]

Answer:

Kb = [OH⁻] . [C₃H₉NH⁺] / [ C₃H₉N ]

Explanation:

The equation for the reaction of trimethylamine when it is dissolved in water is:

C₃H₉N + H₂O ⇄ C₃H₉NH⁺ + OH⁻ Kb

1 mol of trimethylamine catches a proton from the water in order to produce trimethylamonium.

It is a base, because it give OH⁻ to the medium

Expression for Kb (Molar concentration)

Kb = [OH⁻] . [C₃H₉NH⁺] / [ C₃H₉N ]

7 0

3 years ago

How many grams of sulfur are in 3.54 g of H2S?

ira [324]

Answer:

3.329 g

Explanation:

First you need to determine the molar mass of H2S which is 34.1 g/mol.

With that we know that to find the moles of H2S we just divide the mass of sample with the molar mass.

3.54 g / 34.1 g/mol = 0.103812317 mol of H2S

This means that there is also 0.103812317 mol of sulfur since there is 1 mole of sulfur per 1 mole of H2S.

The molar mass of sulfur is 32.065 g/mol and to find the mass of sulfur you need to multiply the molar mass with the moles of the compound.

0.103812317 mol * 32.065 g/mol = 3.329 g of sulfur

Let me know if you get something else or if something is unclear in the comments so that we can figure it out.

5 0

3 years ago