All categories

3 years ago

Omg posting takes away point WHAT.

Chemistry

1 answer:

Charra [1.4K]3 years ago

4 0

Answer:

yep

Explanation:

that why I have someone else post questions for me. But its a good incentive to help others to help you to help me to you to help others.

You might be interested in

Calculate the molarity of a 2.43 mol aqueous NaCl solution with a density of 2.00 g/mL.

anygoal [31]

Answer:

You need to add more points people will not answer if its this low

Explanation:

3 0

3 years ago

I need help with this someone help

allochka39001 [22]

Answer:

1. Element is Ca

2. 20 electrons

3. 2 valence electrons

4. 4 energy levels

6 0

2 years ago

Give five examples of structures with this formula (c6h12). at least one should contain a ring, and at least one should contain

hammer [34]

Answer: -

Following are five examples of structures with the chemical formula C₆H₁₂

Compound A is Hexene.

Compound B is 2-Hexene.

Compound C is 3-Hexene.

Compound D is Cyclohexane.

Compound E is Methylcyclopentane.

As we can see Hexene, 2- Hexene and 3-Hexene all have double bonds.

Cyclohexane and Methylcyclopentane contains a ring.

7 0

3 years ago

What is the difference between a 1s orbital and a 2s orbital is that

iVinArrow [24]

Answer:

The main difference is their energy level, 2s orbital is higher than 1s orbital.

4 0

3 years ago

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