Answer:
23.71J is the work that the gas do.
Explanation:
The work that a gas do under isobaric conditions follows the formula:
W = P*ΔV
<em>Where W is work in atmL, P is the pressure and ΔV is final volume -Initial volume In Liters</em>
Replacing with the values of the problem:
W = P*ΔV
W = 0.600atm*(0.44000L - 0.0500L)
W = 0.234atmL
In Joules (1atmL = 101.325J):
0.234atmL × (101.325J / 1 atmL) =
<h3>23.71J is the work that the gas do.</h3>
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Answer:
Option D. pH= 1.3 strong acid
Explanation:
From the question given:
The hydrogen ion concentration [H+] = 0.05 M
pH = —Log [H+]
pH = —Log 0.05
pH = 1.3
Since the pH lies between 0 and 7, the solution is acidic. Since the pH value is low, the solution is a strong acid
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
H = V / LW = 0.01035 cm^3 / 73.15 cm / 30.48 cm
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.
Atomic radius gold = 174 pm
Diameter = 348 pm
46.4 nm x (1 m / 10^9 nm) x (10^12 pm / 1 m) x (1 Au atom / 248 pm) = 133 atoms of gold
<span>Energy is absorbed and then released to form an emission line.
When electrons absorb energy they increase there energy level. This is only temporary and the excited electron then relaxes back down to its original energy level releasing energy.
The energy is released in form of EM radiation of a specific frequency depending on the element and how many energy levels the electron relaxes.
This forms an emission line.</span><span />
.5336
using avogadro’s law
