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3 years ago

Which type of reaction creates more energy nuclear or chemical

Chemistry

2 answers:

vaieri [72.5K]3 years ago

7 0

Answer:

It depends on the atomic number of the element, fusion produces much more energy for lighter elements and fission produces more energy for heavier elements.

Explanation:

lubasha [3.4K]3 years ago

5 0

Answer:

Nuclear

Explanation:

Nuclear produces more energy than chemical.

You might be interested in

How do you determine how many molecules are in a chemical equation

V125BC [204]

Answer:

Determine the mass of substance and it's molar mass. Divide the given mass by it's molar mass to get moles and multiply time

6.002*1023 molecules 1mole

5 0

2 years ago

The closeness of a measurement to its true value is a measure of its _____.

svlad2 [7]

The closeness of a measurement to its true value is a measure of its accuracy. This term is the degree of which a certain measurement conforms to the correct value or the standard value. It is not the same with the term precision. Precision, on the other hand, is a measure used to characterize the closeness of the data measured.

7 0

3 years ago

Read 2 more answers

You should know the location of safety equipment

s2008m [1.1K]

I would say you should use or test it once a week to ensure it is working properly in an active laboratory since it is a workplace with significant chemical hazards so it would give peace of mind to know on a quite regular basis that it can be relied on in case of an emergency like an eye flush for example.

7 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

2 years ago

Draw all four products obtained when 2-ethyl-3-methyl-1,3-cyclohexadiene is treated with HBr at room temperature and show the me

LenKa [72]

Answer:

See explanation below

Explanation:

In this case we have reaction of addition. In this case a diene reacting with an acid as HBr. This reaction is known as Hydrohalogenation, and, as we have a diene, this kind of reaction can be done as 1,4 addition. Which means that the reaction will be undergoing with an adition in the carbon 1, and carbon 4.

At room temperature we can expect that this reaction can be done in thermodynamic conditions, Now, as the problem states that is forming 4 products, we can expect products of a 1,2 addition too. This product can be formed if the reaction is taking place in the most stable carbocation, and then, by resonance, we can expect the 1,4 product too.

Now, the HBr can be attacked by the double bond of the first position, giving two possible products or by the double bond of the third position giving the other two products. These products are all possible, obviously the most stable will be the major of all of them, but the other three are perfectly possible. One product is formed without doing much, and the other by resonance. Same happens with the other double bond.

In the picture below, you have the mechanism for all the 4 products.

Hope this helps

5 0

2 years ago