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damaskus [11]
3 years ago
7

15 POINTS!!!!!

Chemistry
2 answers:
jekas [21]3 years ago
5 0

Answer:

no illustration, so its impossible to say- but on mooshiners they "hot that mash up usin a big ol wood fahr"

lilavasa [31]3 years ago
4 0
The source of energy is the substance.

Explanation : The gas comes from the substance, and the substance turns the gas into energy.
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Which equation is set up correctly to determine the volume of a 3.2 mole sample of oxygen gas at 50°C and 101.325 kPa?
77julia77 [94]
N = 3.2 moles, T = 50 + 273 = 323 K, P = 101.325 kPa,  R = 8.314 L.kPa/K.mol 

PV = nRT

V = nRT / P           substituting.

V = (3.2 mole)(8.314 L.kPa/K.mol )(323 K) / (<span>101.325 kPa)

That is the answer, but it is not among the options you provided. Check your options properly.</span>
7 0
3 years ago
Read 2 more answers
Nitrogen (N2) and hydrogen (H2) react to form ammonia (NH3). Consider a mixture of six nitrogen molecules and six hydrogen molec
Nezavi [6.7K]

Answer:

a)  No molecules of hydrogen

b) four molecules of ammonia

c) four left molecules of nitrogen.

Explanation:

The balanced reaction between nitrogen and hydrogen molecules to give ammonia molecules is:

N_{2}(g)+3H_{2}(g) -->2NH_{3}

Thus one molecule of nitrogen will react with three molecules of hydrogen to give two molecules of ammonia.

We have six molecules of each nitrogen and hydrogen in the closed container and they undergo complete reaction it means the limiting reagent is hydrogen. For six molecules of nitrogen, eighteen molecules of hydrogen will be required.

So six molecules of hydrogen will react with two molecules of nitrogen to give four molecules of ammonia.

The product mixture will have

a) No molecules of hydrogen

b) four molecules of ammonia

c) four left molecules of nitrogen.

7 0
3 years ago
Which element would most likely have chemical properties similar to that of fluorine (F)?
vredina [299]
The question is asking to state the element would most likely have chemical properties similar to that of fluorine and base on my research and further investigation, I would say that it would be Bromine. I hope you are satisfied with my answer and feel free to ask for more if you have questions and further clarifications 
8 0
4 years ago
Read 2 more answers
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
Plz can anyone help me I'm a bit stuck in this one ​
trasher [3.6K]

Answer:

just see if i am not wrong

learning balancing in chemistry it take time

hope i am correct

8 0
3 years ago
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