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11 months ago

What is the mass of 0.20 mole of C₂H6O (ethanol)?

Chemistry

1 answer:

lesya [120]11 months ago

6 0

Answer:

9.2-g of ethanol

Explanation:

The question is asking for us to solve for the mass of ethanol given the moles of it.

We can do this by going from mol -> grams using the molar mass of ethanol (46.068g).

Start off with the moles given (0.20mol) and multiply it by the molar mass of ethanol (46.068g).

This would give the mass of 9.214-g, but since our question has only 2 significant figures, round off your final answer to the same number of sig. figs.

So your final answer would be 9.2-g of ethanol.

***Molar mass is calculated by adding the masses of each element (found in periodic table) by how many times they are in the molecules.

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2 years ago

Which statement describes an eletron

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3 years ago

Determine the volume, in liters, occupied by 0.015 molecules of oxygen at STp?

Arlecino [84]

5.58 X $10^{-25}$ Litres is the volume, in liters, occupied by 0.015 molecules of oxygen at STP.

Explanation:

Data given:

molecules of oxygen = 0.015

number of moles of oxygen =?

temperature at STP = 273 K

Pressure at STP = 1 atm

volume = ?

R (gas constant) = 0.08201 L atm/mole K

to convert molecules to moles,

number of moles = $\frac{molecules}{Avagadro's number}$

number of moles = 2.49 x $10^{-26}$

Applying the ideal gas law since the oxygen is at STP,

PV = nRT

rearranging the equation:

V = $\frac{nRT}{P}$

putting the values in the rearranged equation:

V = $\frac{2.49 X10^{-26} X 0.08201 X 273}{1}$

V = 5.58 X $10^{-25}$ Litres.

5 0

3 years ago

25 g of a compound is added to 500 mL of water if the freezing point of the resulting solution is

vlabodo [156]

Answer:

a) 119 g/mol

Explanation:

-We apply the formula for freezing point depression to obtain the molality of the solution:

$\bigtriangleup T_f=K_fm, \ \ K_f=1.36\textdegree C/m\\\\\therefore m=\frac{\bigtriangleup T_f}{K_f}\\\\=\frac{0.57\textdegree C}{1.36\textdegree C}\\\\=0.4191\ mol/Kg\\\\$

#We use the molality above to calculate the molar mass:

$m=\frac{0.4191\ mol}{1\ Kg}=\frac{25\ g}{0.5\ Kg}\\\\\therefore 1 \ mol=\frac{25\ g}{0.5\ Kg}\times\frac{1\ Kg}{ 0.4191}\\\\=119.3033\approx 119\ g/mol$

Hence, the molar mass of the compound is 119 g/mol

5 0

3 years ago

A beaker with 155 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjug

OverLord2011 [107]

Answer:

ΔpH = 0.296

Explanation:

The equilibrium of acetic acid (CH₃COOH) in water is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺

Henderson-Hasselbalch formula to find pH in a buffer is:

pH = pKa + log₁₀ [CH₃COO⁻] / [CH₃COOH]

Replacing with known values:

5.000 = 4.740 + log₁₀ [CH₃COO⁻] / [CH₃COOH]

0.260 = log₁₀ [CH₃COO⁻] / [CH₃COOH]

1.820 = [CH₃COO⁻] / [CH₃COOH] (1)

As total molarity of buffer is 0.100M:

[CH₃COO⁻] + [CH₃COOH] = 0.100M (2)

Replacing (2) in (1):

1.820 = 0.100M - [CH₃COOH] / [CH₃COOH]

1.820[CH₃COOH] = 0.100M - [CH₃COOH]

2.820[CH₃COOH] = 0.100M

[CH₃COOH] = 0.100M / 2.820

[CH₃COOH] = 0.035M

Thus: [CH₃COO⁻] = 0.100M - 0.035M = 0.065M

5.40 mL of a 0.490 M HCl are:

0.0054L × (0.490mol / L) = 2.646x10⁻³ moles HCl.

Moles of CH₃COO⁻ are: 0.155L × (0.065mol / L) = 0.0101 moles

HCl reacts with CH₃COO⁻ thus:

HCl + CH₃COO⁻ → CH₃COOH

After reaction, moles of CH₃COO⁻ are:

0.0101 moles - 2.646x10⁻³ moles = 7.429x10⁻³ moles of CH₃COO⁻

Moles of CH₃COOH before reaction are: 0.155L × (0.035mol / L) = 5.425x10⁻³ moles of CH₃COOH. As reaction produce 2.646x10⁻³ moles of CH₃COOH, final moles are:

5.425x10⁻³ moles + 2.646x10⁻³ moles = 8.071x10⁻³ moles of CH₃COOH. Replacing these values in Henderson-Hasselbalch formula:

pH = 4.740 + log₁₀ [7.429x10⁻³ moles] / [8.071x10⁻³ moles]

pH = 4.704

As initial pH was 5.000, change in pH is:

ΔpH = 5.000 - 4.740 = 0.296

4 0

3 years ago