Asked by randy
Show that for 1 kg of pure gold the volume of water displaced is 51.8 cm^3
I know that the density of pure gold is 19.32 gm/cm^3 so 1 kg/19.32 gm/cm^3= 0.05175 ?
I have come along way with this problem and feel like I am close, but now I need help.
How do I get kg and gm to cancel?
How do I get this to be 51.8cm^3?
I know that the density of pure gold is 19.32 gm/cm^3 so 1 kg/19.32 gm/cm^3= 0.05175 ?
I have come along way with this problem and feel like I am close, but now I need help.
How do I get kg and gm to cancel?
How do I get this to be 51.8cm^3?
Answers
Answered by
Damon
1 kg = 1,000 grams
so
1,000 grams (1 cm^3/19.32 grams) = 51.8 cm^3
always multiply by a ratio that is the same top and bottom but in different units so that the units cancel
for example if I were given this the other way as 51.8 cm^3 of gold, then
51.8 cm^3 (19.32 grams/1cm^3) = 1000 grams because the cm^3 cancels top and bottom and we are left with grams.
so
1,000 grams (1 cm^3/19.32 grams) = 51.8 cm^3
always multiply by a ratio that is the same top and bottom but in different units so that the units cancel
for example if I were given this the other way as 51.8 cm^3 of gold, then
51.8 cm^3 (19.32 grams/1cm^3) = 1000 grams because the cm^3 cancels top and bottom and we are left with grams.
Answered by
Anonymous
chicken
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