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Lelechka [254]
3 years ago
5

At a pressure of 782.3 mm Hg and 34.4 °C, a certain gas has a volume of 362.4 mL. What will

Chemistry
1 answer:
german3 years ago
7 0

Answer: v2=331.289mL

Explanation:

Formula for ideal gas law is p1v1/T1=p2v2/T2

P1=782.3mmHg

P2=769mmHg at STP

V1=362.4mL

V2=?

T1=273+34.4=307.4k

T2=273k at STP

Then apply the formula and make v2 the subject of formula

V2= 782.3×362.4×273/760×307.4

V2=77397006.96/233624

V2=331.289mL

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What is the answers and pls show work if possible!!
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D = m / V


It even gives you the density of gold in the problem. Major hint. Once you know the volume (using V = m / D) then you can calculate the height (thickness) from the equation...


V = L x W x H

Volume = Length x Width x Height


start by converting 200.0 mg into grams

1000 mg = 1 g

200. mg x (1 g / 10^3 mg) = 0.200 g


V = m / D

V = 0.200 g / (19.32 g/cm^3)

V = 0.01035 cm^3


Convert 2.4 ft and 1 ft to cm

2.4 ft x (12 in / 1 ft) x (2.54 cm / 1 in) = 73.15 cm

1 ft = 30.48 cm


Compute the height (thickness)

V = LxWxH

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H = 4.64 x 10^-6 cm


Convert to nanometers

4.64 x 10^-6 cm x (1 m / 100 cm) x (10^9 nm / 1 m) = 46.4 nm


Knowing the atomic radius of gold, I might have asked my students for the minimum number of gold atoms in this thickness of gold. This would assume that the gold atoms are all in a row. This would give the minimum number of gold atoms.


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nlexa [21]

Answer:

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