If a 2.46 lb ball of aluminum is coated with a 614 micrometer layer of gold, what is the density (in g/mL) of the gold-coated aluminum ball? The densities of aluminum and gold are 2.70 g/mL and 19.32 g/mL, respectively. Type your numerical answer to 2 decimal places (ex N.NN) without units.
The answer was 3.35.
I m a bit confused with this problem because the mass is a gold-coated aluminum ball, so wouldn t the densities be different in two parts of the mass? The density of the gold coat would be that of the gold and the density of the aluminum part would be that of the aluminum. However, the question is asking for one density so I m assuming my professor meant a gold/aluminum mixed ball. I m confused as to how
Need help with this chemistry dimensional analysis problem!?
2019-02-25 4:53 pm
回答 (1)
2019-02-25 6:08 pm
1lb = 453.59 g
Mass of the Al ball = (2.46 lb) × (453.59 g/lb) = 1116 g
Volume of the ball, V = (1116 g) / (2.7 g/mL) = 413 mL = 413.3 cm³
V = (4/3) π r³
Radius of the ball, r = ∛[3 V / (4 π)] = ∛[3 × 413.3 / (4 × 3.14)] cm = 4.622 cm
Volume of the gold-coated aluminum ball = (4/3) × 3.14 × (4.622 + 614 × 10⁻⁴)³ cm³ = 430.1 cm³
Volume of the gold layer = (430.1 - 413.3) cm³ = 16.8 mL
Mass of the gold layer = (19.32 g/mL) × (16.8 g/mL) = 325 g
Density of the gold-coated aluminum ball in g/mL :
= (1116 + 325) / 430.1
= 3.35
Mass of the Al ball = (2.46 lb) × (453.59 g/lb) = 1116 g
Volume of the ball, V = (1116 g) / (2.7 g/mL) = 413 mL = 413.3 cm³
V = (4/3) π r³
Radius of the ball, r = ∛[3 V / (4 π)] = ∛[3 × 413.3 / (4 × 3.14)] cm = 4.622 cm
Volume of the gold-coated aluminum ball = (4/3) × 3.14 × (4.622 + 614 × 10⁻⁴)³ cm³ = 430.1 cm³
Volume of the gold layer = (430.1 - 413.3) cm³ = 16.8 mL
Mass of the gold layer = (19.32 g/mL) × (16.8 g/mL) = 325 g
Density of the gold-coated aluminum ball in g/mL :
= (1116 + 325) / 430.1
= 3.35
收錄日期: 2021-05-01 22:29:21
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