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inna [77]
3 years ago
11

Imagine that you're doing an experiment that requires you to smell the odor of a chemical solution in a beaker. Which procedure

should always be followed? A. Wave the fumes toward your nose with your hand. B. You should never smell chemicals. C. Hold the beaker directly beneath your nose. D. Pour some of the solution on a paper towel and then sniff the towel.
Chemistry
2 answers:
mixas84 [53]3 years ago
7 0

The correct answer really is B.

If you are directed to break that rule then you better be in a high level chemistry class. When I taught things like that I insisted that students just wait until the chemical permeated the fume cabinet and even then I was always very nervous.

Sometimes you have to know when to ignore a bad direction. If you are working with chlorine, for example, you should be especially careful. That stuff was used in WWI as part of a chemical warfare technique. Many men suffered grotesque deaths by breathing it in, particularly if they were in trenches. Chlorine is heavier than air. It sinks to the lowest level.

Travka [436]3 years ago
3 0
Well, I would say A. Wafting is recommended, but you shouldn’t smell an unknown chemical though.
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Heating curve shows temperature verses energy gain. Which parts of the curve represent a gain in potential energy?
Brilliant_brown [7]

Answer:

Those two horizontal lines.

Explanation:

Hello there!

In this case, when focusing on these heating curves, it is important to say they tend to have two constant-temperature sections and three variable-temperature sections. Thus, from lower to higher temperature, the first constant-temperature section corresponds to melting and the second one vaporization, whereas the three variable-temperature sections correspond to the heating of the solid until melting, the liquid until vaporization and the gas until the critical point.

In such a way, we infer that the boxes referred to constant temperature are referred to a gain in potential energy, that is, the two horizontal lines.

Regards!

6 0
2 years ago
Which projectiles will be visibly affected by air resistance when they fall?
Anika [276]
All of the above will be affected by air resistance, but the most obvious will be the balloon or leaf.
Hope it helps somewhat!
8 0
3 years ago
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How would you prepare 1.00 L of a 0.50 M solution of each of the following?
Lunna [17]

Answer:

Take approx 41.7 mL of 12-M HCl in a 1.00-L flask and fill the rest of the volume with distilled water.

Explanation:

Hello,

In this case, for the dilution process from concentrated  12-M hydrochloric acid to 1.00 L of the diluted 0.50M hydrochloric acid, the volume of concentrated HCl you must take is computed by considering that the moles remain constant for all dilution processes as shown below:

n_1=n_2

Which can also be written in terms of concentrations and volumes:

M_1V_1=M_2V_2

Thus, solving for the initial volume or aliquot that must be taken from the 12-M HCl, we obtain:

V_1=\frac{M_2V_2}{M_1} \\\\V_1=\frac{1.00L*0.5M}{12M}\\ \\V_1=0.0417L=41.7mL

It means that you must take approx 41.7 mL of 12-M HCl in a 1.00-L flask and fill the rest of the volume with distilled water for such preparation.

Best regards.

6 0
3 years ago
which quantity can be calculated for a solid compound, given only the formula of the compound and the Periodic Table of elements
Kryger [21]
You can calculate the gram formula mass of this compound by adding the atomic mass of every element in this compound according to the periodic table of elements.
7 0
3 years ago
Calculate the amount of heat released when 27.0 g H2O is cooled from a liquid at 314 K to a solid at 263 K. The melting point of
BlackZzzverrR [31]

Answer : The amount of heat released, 45.89 KJ

Solution :

Process involved in the calculation of heat released :

(1):H_2O(l)(314K)\rightarrow H_2O(l)(273K)\\\\(2):H_2O(l)(273K)\rightarrow H_2O(s)(273K)\\\\(3):H_2O(s)(273K)\rightarrow H_2O(s)(263K)

Now we have to calculate the amount of heat released.

Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+\Delta H_{fusion}+[m\times c_{p,s}\times (T_{final}-T_{initial})]

where,

Q = amount of heat released = ?

m = mass of water = 27 g

c_{p,l} = specific heat of liquid water = 4.184 J/gk

c_{p,s} = specific heat of solid water = 2.093 J/gk

\Delta H_{fusion} = enthalpy change for fusion = 40.7 KJ/mole = 40700 J/mole

conversion : 0^oC=273k

Now put all the given values in the above expression, we get

Q=[27g\times 4.184J/gK\times (314-273)k]+40700J+[27g\times 2.093J/gK\times (273-263)k]

Q=45896.798J=45.89KJ     (1 KJ = 1000 J)

Therefore, the amount of heat released, 45.89 KJ

7 0
3 years ago
Read 2 more answers
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