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Ghella [55]
3 years ago
14

Which has more volume a kilogram of feathers or a kilogram of bricks please answer quickly thank you

Chemistry
2 answers:
NARA [144]3 years ago
7 0
1 kilogram of feathers has more volume than 1 kilogram of brick. feathers have less density than brick therefore they take up more space in a container.
density & buoyancy
when you place an object in a liquid such as water and it floats, it is said to have buoyancy.
the more density an object has, the less buoyancy it has, and it will sink.
swim bladder
a fish controls its depth by adding or subtracting gas from its swim bladder. by squeezing the bladder and releasing gas the fish becomes more dense and therefore it sinks deaper. if the fish adds gas to the bladder it becomes less dence, increasing its buoyancy and it will rise up in the water.
elena55 [62]3 years ago
4 0
I'll answer the first question: the answer is....a kilogram of bricks.
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How many moles of oxygen atoms are present in 5 H20
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Explanation:

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Anton thinks that glucose is a reactant of photosynthesis, but his classmate Jamila disagrees.
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Which of the following sets represents a pair of isotopes? 14C and 14N 206Pb and 208Pb O2 and O3 32S and 32S2-
ludmilkaskok [199]
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3 0
2 years ago
How many cubic centimeters of an ore containing only 0.22% gold (by mass) must be processed to obtain $100.00 worth of gold? The
bezimeni [28]

Answer:

The cubic centimetres of the ore containing 0.22% gold (by mass) that must be processed to obtain the $100.00 worth of gold is approximately 216 cm³

Explanation:

The percentage by mass of gold in the ore = 0.22%

The density of the ore = 8.0 g/cm³

The price of the gold = $818 per troy ounce

14.6 troy oz = 1.0 pound

1 lb = 454 g

Given that one troy ounce = $818

$100 worth of gold = 1/818 ×100 troy ounce = 100/818 troy ounce

1 troy oz = 1.0/14.6 lb

100/818 troy oz =  100/818 × 1.0/14.6 lb = 250/29857 lb ≈ 0.0084 lb

1 lb = 454 g

250/29857 lb = 454 × 250/29857 g ≈ 3.8015 g

$100 = 3.8015 g worth of gold

The mass, M, of the ore containing 3.8015 g of gold is given as follows;

0.22% of M = 3.8015 g

0.22/100 × M = 3.8015 g

M = 3.8015 g × 100/0.22 = 1727.933 g

The volume, V, of the ore containing 3.8015 g of gold is given as follows;

Density of ore = Mass of ore/(Volume of ore)

Volume of ore = Mass of ore /(Density of ore)

The density of the ore = 8.0 g/cm³

Volume of ore = 1727.933 g /(8.0 g/cm³) = 215.99 cm³ ≈ 216 cm³

Therefore, the cubic centimetres of the ore containing 0.22% gold (by mass) that must be processed to obtain the $100.00 worth of gold ≈ 216 cm³.

5 0
3 years ago
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