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BartSMP [9]
3 years ago
11

A chemist has 2.0 mol of methanol (CH₃OH). The molar mass of methanol is 32.0 g/mol. What is the mass, in grams, of the sample?

Chemistry
2 answers:
CaHeK987 [17]3 years ago
8 0
I believe the correct answer from the choices listed above is the last option. The mass of the methanol sample would be 64 grams. It was calculated by multiplying the number of moles to the molar mass of methanol. Hope this answers the question. Have a nice day.
lina2011 [118]3 years ago
7 0

Answer:

64

Explanation:

Mole of a compound = mass of a compound / molar mass of a compound

Mass of a compound = mole of a compound x molar mass of a compound

Given,

Mole of methanol = 2 mol

Molar mass of methanol = 32 g/mol

Mass of methanol = Mole of methanol x Molar mass of methanol  

                               = 2 mol x 32 g/mol = 64 g

Therefore, the mass, in grams, of the sample = 64

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Explanation:

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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:

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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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