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Chemistry

1 answer:

Alexxx [7]3 years ago

6 0

At room temperature oxygen, $O_2$ is present in the gas phase.

A gas is defined as a substance which is characterized by the rapidly movement of molecules which are widely separated. The shape and volume of gas is not definite that is they take on the shape of the container in which it is placed. The gas molecules have minimum inter-nuclear attractions and thus have high kinetic energy.

Thus, the sentences that describes oxygen at room temperature are:

It has no definite volume and takes the shape of its container.
Its particles move fast enough to overcome the attraction between them.
It has more energy than it would be at a cooler temperature as the kinetic energy is directly proportional to the temperature that is kinetic energy increases on increasing the temperature.

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How much heat is required to raise the temperature of 1,500 g of water from 25°C to 52°C? The specific heat of water is 4.184 J/

seropon [69]

Mass=1500g=m
∆T=52-25=27°C
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Apply thermodynamics

$\\ \rm\rightarrowtail Q=mc\delta T$

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1 year ago

Read 2 more answers

The average density of a carbon-fiber-epoxy composite is 1.615 g/cm3. the density of the epoxy resin is 1.21 g/cm3 and that of t

ra1l [238]

The composite material is composed of carbon fiber and epoxy resins. Now, density is an intensive unit. So, to approach this problem, let's assume there is 1 gram of composite material. Thus, mass carbon + mass epoxy = 1 g.

Volume of composite material = 1 g / 1.615 g/cm³ = 0.619 cm³
Volume of carbon fibers = x g / 1.74 g/cm³
Volume of epoxy resin = (1 - x) g / 1.21 g/cm³

a.) V of composite = V of carbon fibers + V of epoxy resin
0.619 = x/1.74 + (1-x)/1.21
Solve for x,
x = 0.824 g carbon fibers
1-x = 0.176 g epoxy resins

Vol % of carbon fibers = [(0.824/1.74) ÷ 0.619]*100 = 76.5%

b.) Weight % of epoxy = 0.176 g epoxy/1 g composite * 100 = 17.6%
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4 0

2 years ago

How many moles are present in 45.0 g of lithium oxide (Li2O)?

BigorU [14]

Molar mass Li2O = 7.0 x 2 + 16 => 30 g/mol

1 mole Li2O ---------- 30 g
( moles Li2O ) -------- 45.0 g

moles Li2O = 45.0 x 1 / 30

moles Li2O = 45.0 / 30

= 1.50 moles of Li2O

hope this helps!

4 0

2 years ago