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3 years ago

The atomic number of an element is determined by the number of electrons in the nucleus of the atom.

Chemistry

2 answers:

iren [92.7K]3 years ago

7 0

A
The number of electrons/protons is the one that determines the atomic number of an element.

never [62]3 years ago

3 0

True beacuse the number of electrons is the atomic number

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What is 33 divided by 1862

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Nitrifying bacteria participate in the nitrogen cycle mainly by

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3 years ago

Ethyl butyrate, CH3CH2CH2CO2CH2CH3, is an artificial fruit flavor commonly used in the food industry for such flavors as orange

SIZIF [17.4K]

Answer:

A. 10.0 grams of ethyl butyrate would be synthesized.

B. 57.5% was the percent yield.

C. 7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

Explanation:

$CH_3CH_2CH_2CO_2H(l)+CH_2CH_3OH(l)+H^+\rightarrow CH_3CH_2CH_2CO_2CH_2CH_3(l)+H_2O(l)$

Moles of butanoic acid = $\frac{7.60 g}{88 g/mol}=0.08636 mol$

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

$\frac{1}{1}\times 0.08636 mol=0.08636 mol$ of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 100%

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

$100\%=\frac{\text{Experimental yield}}{10.0 g}\times 100$

Experimental yield = 10.0 g

10.0 grams of ethyl butyrate would be synthesized.

Theoretical yield of ethyl butyrate = 10.0 g

Experimental yield ethyl butyrate = 5.75 g

Percentage yield of the reaction = ?

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

$=\frac{5.75 g}{10.0 g}\times 100=57.5\%$

57.5% was the percent yield.

Moles of butanoic acid = $\frac{7.60 g}{88 g/mol}=0.08636 mol$

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

$\frac{1}{1}\times 0.08636 mol=0.08636 mol$ of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 78.0%

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

$78.0\%=\frac{\text{Experimental yield}}{10.0 g}\times 100$

Experimental yield = 7.80 g

7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

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3 years ago

Imagine you are making Kool Aid. Kool Aid comes in packets. The powder inside the packets is made up of very small particles. Th

lina2011 [118]

Answer:

Because the molecules have not been in water so they are not moving around each other

Explanation:

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A certain electrolyte solution contains 1 gram of salt for every 8 grams of sugar and every 200 grams of water. If the sugar to

12345 [234]

Answer:

The resulting solution contains approximately 666 g of water.

Explanation:

In the initial solution we have:

1g salt : 8g sugar : 200g water

This means that the ratios are:

$\frac{salt}{sugar} = \frac{1}{8} \\\\\frac{sugar}{water} = \frac{8}{200} =\frac{1}{25}$

In the final solution we have:

5g salt: xg sugar: yg water

The new ratios are:

$\frac{salt}{sugar} = \frac{3}{8} \\\\\frac{sugar}{water} = \frac{1}{50}$

Now we can calculate the amount of sugar in the final solution:

$\frac{salt}{sugar} = \frac{5}{x} =\frac{3}{8} \\\\X = 13.333 g$

Finally, we calculate the amount of water:

$\frac{sugar}{water} = \frac{13.333}{y} = \frac{1}{50} \\y = 666.667 g$

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3 years ago