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Marrrta [24]
2 years ago
7

2,2,2,3,3,5,7,8 mean median mode

Chemistry
1 answer:
Nesterboy [21]2 years ago
6 0

Answer:it is 3 just look at the middle numbers

Explanation:

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A change in matter that produces one or more new substances is a ???
Dvinal [7]
Chemical reaction, generally speaking. It's a vague definition, but it's easy to remember.

8 0
3 years ago
Read 2 more answers
How many grams of sulfur must be burned to give 100.0 g of So2
andriy [413]

Answer:

50 g of S are needed

Explanation:

To star this, we begin from the reaction:

S(s) + O₂ (g) →  SO₂ (g)

If we burn 1 mol of sulfur with 1 mol of oxygen, we can produce 1 mol of sulfur dioxide. In conclussion, ratio is 1:1.

According to stoichiometry, we can determine the moles of sulfur dioxide produced.

100 g. 1mol / 64.06g = 1.56 moles

This 1.56 moles were orginated by the same amount of S, according to stoichiometry.

Let's convert the moles to mass

1.56 mol . 32.06g / mol = 50 g

4 0
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)

A

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.

B

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.

C

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.

8 0
3 years ago
Help me with this please!!! Smart ones help me 18 points
Helen [10]
Hey! The answer would be 1/2.
4 0
3 years ago
Read 2 more answers
What is the correct answer please
Anestetic [448]

Answer:

I think it's the second answer --If you increase the acidity..

I hope answer I can answer your question!

7 0
3 years ago
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