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2 years ago

Find the energy of a photon of light if the wavelength of the light wave is 6.88 x 10–7m

1 answer:

6 0

Use the formula E=hv, h=plancks constant and v=frequency
use the formula c=v*lambda to find v
the answer will be 2.88*10^-23J

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Adding hydrogen atoms to an unsaturated fatty acid will make it __________.

xxMikexx [17]

The answer is (a.) more solid

Adding hydrogen atoms to an unsaturated fatty acid will make it more solid. This process is called hydrogenation. In this process, hydrogen is added to an unsaturated fatty acid to make it more saturated and thereby more solid at a room temperature.

3 0

3 years ago

Read 2 more answers

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

What is the molar mass, in grams, of a mole of an element equal to?

prohojiy [21]

Technically molar mass cannot be in grams, it is in grams per mole. and it refers to a specific number of molecules of a substance, therefore substances have different molar masses because the elements have different weights. for example having 10 water molecules would be a lot heavier than having 10 air molecules

6 0

2 years ago

Lastly, Snape thinks we should try one more calculation. What is the retention factor if the distance traveled by the solvent fr

Neporo4naja [7]

Answer:

0.1 is the retention factor.

Explanation:

Distance covered by solvent , $d_s= 2.0 cm$