Density depends on both the mass and the volume of an object. If you cut a bar of gold in half, you would have two bars with half the mass of the original bar. However, each bar would also have half the volume of the original bar. The density of gold does not change.
Answer:
Acetic acid 0,055M and acetate 0,095M.
Explanation:
It is possible to prepare a 0,15M buffer of acetic acid/acetate at pH 5,0 using Henderson-Hasselblach formula, thus:
pH = pka + log₁₀ [A⁻]/[HA] <em>-Where A⁻ is acetate ion and HA is acetic acid-</em>
Replacing:
5,0 = 4,76 + log₁₀ [A⁻]/[HA]
<em>1,7378 = [A⁻]/[HA] </em><em>(1)</em>
As concentration of buffer is 0,15M, it is possible to write:
<em>[A⁻] + [HA] = 0,15M </em><em>(2)</em>
Replacing (1) in (2):
1,7378[HA] + [HA] = 0,15M
2,7378[HA] = 0,15M
[HA] = 0,055M
Thus, [A⁻] = 0,095M
That means you need <em>acetic acid 0,055M</em> and <em>acetate 0,095M</em> to obtain the buffer you need.
i hope it helps!
Answer:
Explanation:
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Answer:
When the drill hits oil, some of the oil rises from the ground high into the air. This immediate release of oil is known as a "gusher." Once a reservoir has been located, pumps are used to extract the oil.
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
