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monitta
3 years ago
11

What is the percent by volume of ethanol (C2H60, or ethyl alcohol) in the final solution when 90 mL of ethanol is diluted to a v

olume of 550 mL with water?

Chemistry
1 answer:
r-ruslan [8.4K]3 years ago
8 0

Answer:

16.0%.

Explanation:

Volume percent of a substance is the ratio of the substance volume to the solution volume multiplied by 100.

V % of ethanol = (volume of ethanol / volume of the solution) x 100.

volume of ethanol = 90.0 mL, volume of the solution = 550.0 mL.

∴ V % of ethanol = (90.0 mL / 550.0 mL) x 100 = 16.36% ≅ 16.0%.

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Calculate the unit cell edge length for an 78 wt% Ag- 22 wt% Pd alloy. All of the palladium is in solid solution, and the crysta
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Answer:

The unit cell edge length for the alloy is 0.405 nm

Explanation:

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concentration of Ag, C_{Ag} = 78 wt%

concentration of Pd, C_P_d = 22 wt%

density of Ag = 10.49 g/cm³

density of Pd = 12.02 g/cm³

atomic weight of Ag, A_A_g = 107.87 g/mol

atomic weight of iron, A_P_d = 106.4 g/mol

Step 1: determine the average density of the alloy

\rho _{Avg.} = \frac{100}{\frac{C_A_g}{\rho _A_g} + \frac{C__{Pd}}{\rho _{Pd}} }

\rho _{Avg.} = \frac{100}{\frac{78}{10.49} + \frac{22}{12.02} } = 10.79 \ g/cm^3

Step 2: determine the average atomic weight of the alloy

A _{Avg.} = \frac{100}{\frac{C_v}{A _A_g} + \frac{C__{Pd}}{A _{Pd}} }

A _{Avg.} = \frac{100}{\frac{78}{107.87} + \frac{22}{106.4} } = 107.54 \ g/mole

Step 3: determine unit cell volume

V_c=\frac{nA_{avg.}}{\rho _{avg.} N_a}

for a FCC crystal structure, there are 4 atoms per unit cell; n = 4

V_c=\frac{4*107.54}{ 10.79*6.022*10^{23}} = 6.620*10^{-23} \ cm^3/cell

Step 4: determine the unit cell edge length

Vc = a³

a = V_c{^\frac{1}{3} }\\\\a = (6.620*10^{-23}}){^\frac{1}{3}}\\\\a = 4.05 *10^{-8} \ cm= 0.405 nm

Therefore, the unit cell edge length for the alloy is 0.405 nm

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3 years ago
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