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cestrela7 [59]
3 years ago
5

If you need to measure the volume of liquid in a bottle of eyedrops, what unit would be the most practical?

Chemistry
2 answers:
julsineya [31]3 years ago
6 0

The correct answer is B. Milliliter

Explanation:

Volume, which refers to the space a liquid, solid, etc. occupies can be measured using different units such as milliliter that is one of the smallest units, liter that is the most common one and is equivalent to 1000 milliliters; dekaliter than is equivalent to 10 liter and kiloliter that is the equivalent to 1000 liters. This means, if you need to measure the volume of a liquid in a bottle of eye drops that is usually a small container, the best is to use the milliliter as a unit because this is the smallest unit from the options given and therefore the more suitable for measuring small volumes precisely. Indeed, most bottles of eye drops contain around 5 or 10 milliliters.

nexus9112 [7]3 years ago
4 0
The most practical would be milliliters
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Which excerpt from amy tan's "fish cheeks" is most clearly an example of conflict?
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Answer is: <span>d.and then they arrived - the minister's family and all my relatives in a clamor of doorbells and rumpled christmas packages. robert grunted hello, and i pretended he was not worthy of existence?
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yaeh

Explanation:

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7 0
3 years ago
Susan made observations about outside events she noticed throughout the day. The day began with rain. Susan saw puddles on the r
S_A_V [24]

Answer:

Rainfall - precipitation

disappeared puddles - evaporation

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Explanation:

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4 0
3 years ago
Calculate the volume of oxygen required to burn 12.00 l of ethane gas, c2h6, to produce carbon dioxide and water, if the volumes
kirza4 [7]
The  volume of   oxygen required  to  burn  12.00 L  ethane is calculated  as  follows

find the moles  of C2H6  used

At  STP  1 mole  is  always =  22.4 L, what about  12.00 L

= ( 12.00L  x 1 moles)  22.4 L = 0.536  moles

write the   reacting equation

2C2H6+  7O2 = 4CO2  + 6H2O
by  use  of mole  ratio  between  C2H6 :O2   which is 2:7  the  moles  of O2 

= 0.536  x7/2=  1.876  moles

again  at  STP  1mole =  22.4 L  what  about 1.876 moles

=    22.4 L x 1.876  moles/ 1 mole =  42.02 L


8 0
3 years ago
While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting e
sp2606 [1]

The question is incomplete, here is the complete question:

While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting ethylene with water vapor at elevated temperatures.

A chemical engineer studying this reaction fills a 1.5 L flask at 12°C with 1.8 atm of ethylene gas and 4.7 atm of water vapor. When the mixture has come to equilibrium she determines that it contains 1.16 atm of ethylene gas and 4.06 atm of water vapor.

The engineer then adds another 1.2 atm of ethylene, and allows the mixture to come to equilibrium again. Calculate the pressure of ethanol after equilibrium is reached the second time. Round your answer to 2 significant digits.

<u>Answer:</u> The partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm

<u>Explanation:</u>

We are given:

Initial partial pressure of ethylene gas = 1.8 atm

Initial partial pressure of water vapor = 4.7 atm

Equilibrium partial pressure of ethylene gas = 1.16 atm

Equilibrium partial pressure of water vapor = 4.06 atm

The chemical equation for the reaction of ethylene gas and water vapor follows:

                     CH_2CH_2(g)+H_2O(g)\rightleftharpoons CH_3CH_2OH(g)

<u>Initial:</u>                  1.8                4.7

<u>At eqllm:</u>           1.8-x             4.7-x

Evaluating the value of 'x'

\Rightarrow (1.8-x)=1.16\\\\x=0.64

The expression of K_p for above equation follows:

K_p=\frac{p_{CH_3CH_2OH}}{p_{CH_2CH_2}\times p_{H_2O}}

p_{CH_2CH_2}=1.16atm\\p_{H_2O}=4.06atm\\p_{CH_3CH_2OH}=0.64atm

Putting values in above expression, we get:

K_p=\frac{0.64}{1.16\times 4.06}\\\\K_p=0.136

When more ethylene is added, the equilibrium gets re-established.

Partial pressure of ethylene added = 1.2 atm

                     CH_2CH_2(g)+H_2O(g)\rightleftharpoons CH_3CH_2OH(g)

<u>Initial:</u>                2.36             4.06               0.64

<u>At eqllm:</u>           2.36-x        4.06-x             0.64+x

Putting value in the equilibrium constant expression, we get:

0.136=\frac{(0.64+x)}{(2.36-x)\times (4.06-x)}\\\\x=0.363,13.41

Neglecting the value of x = 13.41 because equilibrium partial pressure of ethylene and water vapor will become negative, which is not possible.

So, equilibrium partial pressure of ethanol = (0.64 + x) = (0.64 + 0.363) = 1.003 atm

Hence, the partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm

3 0
3 years ago
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