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3 years ago

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Put the following steps of specimen preparation and staining in order: I. Application of staining dyes II. Heat fixation III. Sm

ear preparation A. II, III, I B. III, II, I C. I, III, II D. III, I, II

1 answer:

MrRissso [65]3 years ago

3 0

Answer:

Explanation:

II

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Mandarinka [93]

VOLUME= 5cm*10cm*2cm =100cm^3
but density of iron=7.874g/cm^3
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mass of that block = 787.4g

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If 65.5 moles of an ideal gas is at 9.15 atm at 50.30 °C, what is the volume of the gas?

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To calculate for the volume, we need a relation to relate the number of moles (n), pressure (P), and temperature (T) with volume (V). For simplification, we assume the gas is an ideal gas. So, we use PV=nRT.

PV = nRT where R is the universal gas constant
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3 years ago

Match the following with their correct molecular weight. 2-butanone Propyl acetate 4-methyl-2-pentanone Butyl acetate Methanol E

Hatshy [7]

Answer:

2-butanone = 72.11 g/mol (option F)

Propyl acetate = 102.13 g/mol (option C)

4-methyl-2-pentanone = 100.16 g/mol (option D)

Butyl acetate = 116.16 g/mol (option B)

Methanol = 32.04 g/mol (option E)

Ethanol = 46.07 g/mol (option A)

Explanation:

Step 1: Data given

Molar mass of C = 12.01 g/mol

Molar mass of H = 1.01 g/mol

Molar mass of O = 16.00 g/mol

Step 2:

2-butanone = C4H8O

⇒ 4*12.01 + 8*1.01 + 16.00 = 72.11 g/mol (option F)

Propyl acetate = C5H10O2

⇒ 5*12.01 + 10*1.01 + 2*16.00 = 102.13 g/mol (option C)

4-methyl-2-pentanone = C6H12O

⇒ 6*12.01 + 12*1.01 + 16.00 = 100.16 g/mol (option D)

Butyl acetate = C6H12O2

⇒ 6*12.01 + 12*1.01 + 2*16.00 = 116.16 g/mol (option B)

Methanol = CH3OH = CH4O

⇒ 12.01 + 4*1.01 + 16.00 = 32.04 g/mol (option E)

Ethanol = C2H5OH = C2H6O

⇒ 2*12.01 + 6*1.01 + 16.00 = 46.07 g/mol (option A)

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Answer:

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Explanation:

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