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andrew11 [14]
3 years ago
13

A student inflates a balloon for a party inside her house, where the temperature is 23°C. She then takes the balloon outside, wh

ere the temperature is 12°C. What most likely will happen to the balloon?
Chemistry
2 answers:
salantis [7]3 years ago
7 0

The balloon will lose volume since Charles law says if temperature decreases then volume decreases and going from 23 degrees to 12 degrees is a temperature decrease.


Most likely the balloon will decrease in volume.

Anton [14]3 years ago
6 0

The answer is that it will shrink

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Starting with acetylene and bromoethane, show how you would use reagents from the table to synthesize 3-hexanone.
Jobisdone [24]

solution:

A = 192 x (1/2) ^ (15/5) = 192 x (1/2) ^3 = 192 x 1/8 = 24 mg

 Starting by hitting acetylene with NaNH2 to deprotonate, this C-- will attack the C connected to the Br Sn2 style to lengthen the chain by two carbons.  

Do this same thing again with the other CH of the acetylene and another bromoethaneto get a six carbon chain, namely, 3-hexyne.  

Now, reduce the alkyne to an alkene via H2/Pd/C, and that gives 3-hexene.  


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3 years ago
The breaking buffer that we use this week contains 10mM Tris, pH 8.0, 150mM NaCl. The elution buffer is breaking buffer that als
olasank [31]

Answer:

A breakdown of the breaking buffer was first listed with its respective component and their corresponding value; then a table was made for the stock concentrations in which the volume that is being added was determined by using the formula M_1*V_1 = M_2*V_2. It was the addition of these volumes altogether that make up the 0.25 L (i.e 250 mL)  with water

Explanation:

Given data includes:

Tris= 10mM

pH = 8.0

NaCl = 150 mM

Imidazole = 300 mM

In order to make 0.25 L solution buffer ; i.e (250 mL); we have the following component.

Stock Concentration             Volume to be             Final Concentration

                                               added            

1 M Tris                                     2.5 mL                         10 mM

5 M NaCl                                  7.5 mL                        150 mM

1 M Imidazole                           75 mL                         300 mM    

M_1*V_1 = M_2*V_2. is the formula that is used to determine the corresponding volume that is added for each stock concentration

The stock concentration of Tris ( 1 M ) is as follows:

M_1*V_1 = M_2*V_2.

1*V_1= 0.01 M *250mL\\V_1 = 2.5mL

The stock concentration of NaCl (5 M ) is as follows:

M_1*V_1 = M_2*V_2.

1*V_1= 0.15 M *250mL\\V_1 = 7.5mL

The stock concentration of Imidazole (1 M ) is as follows:

M_1*V_1 = M_2*V_2.

1*V_1= 0.03 M *250mL\\V_1 = 75mL

Hence, it is the addition of all the volumes altogether that make up 0.25L (i.e 250 mL) with water.

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Answer:

A

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