Electric service has accumulated on an object is referred to as

horrorfan [7]

Answer:

I believe that it would be the third option down.

Explanation:

The thermal energy is in constant motion in the COOLER room, and when the desired temperature in the room is reached, it stops.

8 0

3 years ago

Which of the following best defines the term oxidation state?

eimsori [14]

C seems to be the best answer

4 0

3 years ago

Magnesium element has three isotopes: 24Mg, 25Mg, and 26Mg. The natural abundance of the isotopes on Earth is 78.99% 24Mg with a

Dafna11 [192]

Answer:

24.305 to the nearest thousandth.

Explanation:

That is 0.7899*23.985 + 0.10* 24.985+0.1101*25.9825

= 24.305.

3 0

4 years ago

If 26.4 g of hydrogen, H2, is produced from 100.0 g of methane, CH4, reacting with excess water as shown, what is the percentage

blondinia [14]

Answer: Percent yield represents the ratio between what is experimentally obtained and what is theoretically calculated, multiplied by 100%.

% yield

=

actual yield

theoretical yield

⋅

100

%

So, let's say you want to do an experiment in the lab. You want to measure how much water is produced when 12.0 g of glucose (

C

6

H

12

O

6

) is burned with enough oxygen.

C

6

H

12

O

6

+

6

O

2

→

6

C

O

2

+

6

H

2

O

Since you have a

1

:

6

mole ratio between glucose and water, you can determine how much water you would get by

12.0

g glucose

⋅

1 mole glucose

180.0 g

⋅

6 moles of water

1 mole glucose

⋅

18.0 g

1 mole water

=

7.20

g

This represents your theoretical yield. If the percent yield is 100%, the actual yield will be equal to the theoretical yield. However, after you do the experiment you discover that only 6.50 g of water were produced.

Since less than what was calculated was actually produced, it means that the reaction's percent yield must be smaller than 100%. This is confirmed by

% yield

=

6.50 g

7.20 g

⋅

100

%

=

90.3

%

You can backtrack from here and find out how much glucose reacted

65.0 g of water

⋅

1 mole

18.0 g

⋅

1 mole glucose

6 moles water

⋅

180.0 g

1 mole glucose

=

10.8

g

So not all the glucose reacted, which means that oxygen was not sufficient for the reaction - it acted as a limiting reagent.

Explanation:

6 0

3 years ago

C2H5OH(l)+CH3COOH(l) → CH3COOC2H5(l)+H2O(l) If the yield of ethyl ethanoate obtained when 20.00 g of ethanol is reacted with exc

vodomira [7]

Answer:

The percentage yield is 79.12%

Explanation:

The first thing we need to calculate here is the theoretical yield of ethyl ethanoate that can be obtained from 20g of ethanol.

Since the mole ratio is 1:1, then this is quite straightforward.

What we need to obtain at first here is the number of theoretical moles of ethanol reacting.

That would be mass of ethanol/molar mass of ethanol

Molar mass of ethanol is 46g/mol

Thus the number of moles is 20/46 = 0.4348 mole

Like it is indicated earlier, since the number of moles are equal from the balanced equation, it also means that 0.4348 mole of ethylethanoate is produced

Now, we need to know the mass of ethyl ethanoate produced

The mass can be calculated by multiplying the number of moles by the molar mass

The molar mass of ethyl ethanoate is 88g/mol

So the mass of it produced = 0.4348 * 88 = 38.2624 or let’s just say 38.26

Thus, the percentage yield will be;

Actual yield/Theoretical yield * 100%

From the question, our actual yield is 30.27g while our calculated theoretical yield is 38.26g

= 30.27/38.26 * 100% = 79.12%

5 0

3 years ago

Electric service has accumulated on an object is referred to as​

Electric service has accumulated on an object is referred to as