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Drupady [299]
3 years ago
6

ethanol is a common laboratory solvent and has a density of 0.789 g/mL. what is the mass, in grams, of 151 mL of ethanol

Chemistry
2 answers:
svlad2 [7]3 years ago
4 0
With the density given, you can use to convert volume (mL) to mass (g)_, and vice versa. you can use the density as a conversion factor.

so if the density is 0.780 g/mL, then 0.789 g= 1 mL

now let's use this conversion factor to convert the mL to grams.

151 mL ( \frac{0.789 grams}{1 mL} )= 119 grams


natta225 [31]3 years ago
3 0
<em>V = 151 mL = 151 cm³</em>
<em>d = 0,789 g/mL = 0,789 g/cm³</em>
--------------------------------------

d = m/V
m = d×V
m = 0,789×151
<u>m = 119,139g</u>

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The following reaction produces ethanoic acid (CHACOOH) from methanol (CH3OH) and carbon
tigry1 [53]

Answer:

\boxed{\text{300 g}}

Explanation:

We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

M_r:         32                          60

           CH₃OH + CO ⟶ CH₃COOH

m/g:        160

(a) Moles of CH₃OH

\text{Moles of CH$_{3}$OH} = \text{160 g CH$_{3}$OH }\times \dfrac{\text{1 mol CH$_{3}$OH }}{\text{32 g CH$_{3}$OH}}= \text{5.00 mol CH$_{3}$OH}

(b) Moles of CH₃COOH

\text{Moles of CH$_{3}$COOH} = \text{5.00 mol CH$_{3}$OH } \times \dfrac{\text{1 mol CH$_{3}$COOH}}{\text{1 mol CH$_{3}$OH }} = \text{5.00 mol CH$_{3}$COOH}

(c) Mass of CH₃COOH

\text{Mass of CH$_{3}$COOH} =\text{5.00 mol CH$_{3}$COOH} \times \dfrac{\text{60 g CH$_{3}$COOH}}{\text{1 mol CH$_{3}$COOH}} = \textbf{300 g CH$_{3}$COOH}\\\\\text{The maximum mass of ethanoic acid that can be produced is $\boxed{\textbf{300 g}}$}

3 0
3 years ago
Question 1
Nesterboy [21]
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3 0
2 years ago
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What is 256ml to liters
Fed [463]
That would be 0.26 liters

Hope it help!

~Mqddie
6 0
3 years ago
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the standard change in Gibbs free energy is Δ????°′=7.53 kJ/molΔG°′=7.53 kJ/mol . Calculate Δ????ΔG for this reaction at 298 K29
Vera_Pavlovna [14]

Answer:

ΔG = 16.218 KJ/mol

Explanation:

  • dihydroxyacetone phosphate ↔ glyceraldehyde-3-phosphate
  • ΔG = ΔG° - RT Ln Q

∴ ΔG° = 7.53 KJ/mol * ( 1000 J / KJ ) = 7530 J/mol

∴  R = 8.314 J/K.mol

∴ T = 298 K

∴ Q = [glyceraldehyde-3-phosphate] / [dihydroxyacetone phosphate]

⇒ Q = 0.00300 / 0.100 = 0.03

⇒ ΔG = 7530J/mol - (( 8.314 J/K.mol) * ( 298 K ) * Ln ( 0.03 ))

⇒ ΔG = 16217.7496 J/mol ( 16.218 KJ/mol )

5 0
3 years ago
Can someone please help I need help?!?!
anzhelika [568]

Answer:

a: 6 moles

b: 6.75 moles

c: 5 grams

Explanation:

a: mole ratio 2:3

b: mole ratio 2:3

c: mole ratio 2:2

7 0
3 years ago
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