Answer:
The smallest ballon is the ballon X
Explanation:
It is possible to answer this question by using Graham's law:
Where v is the speed of effusion and MW is molar weight of each compound.
This equation is showing that speed is inversely related to the square root of its molar mass. As carbon dioxide has a bigger MW than carbon monoxide, the speed of effusion of carbon dioxide is lower doing its ballon biggest than carbon monoxide ballon, thus: <em>The smallest ballon is the ballon X</em>
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I hope it helps!
A=dv/ dt
a=(v2-v1)/(t2-t1)
a=(60-50)/(10-0)
a=10/10
a=1
it's acceleration is 1 km/s^(2) or 1000 m/s^(2)
Hope I can help u
False. A mixture represents elements or molecules which are not chemically combined.
Answer:
By writing an experiment report
Explanation:
Explanation:
Molar mass
The mass present in one mole of a specific species .
The molar mass of a compound , can easily be calculated as the sum of the all the individual atom multiplied by the number of total atoms .
(a) )hydrogen fluoride, HF
Since ,
Molar mass of of the atoms are -
Hydrogen , H = 1 g/mol
Fluorine, F = 19 g/mol.
Molar mass of HF = ( 1 g/mol ) + ( 19 g/mol ) = 20 g/mol.
(b) ammonia, NH₃
Molar mass of of the atoms are -
nitrogen, N = 14 g/mol
and Hydrogen , H = 1 g/mol
Molar mass of NH₃ = 14 + ( 3 x 1 ) g/mol = 17 g/mol.
(c) nitric acid , HNO₃
Molar mass of of the atoms are -
oxygen , O = 16 g/mol.
Hydrogen , H = 1 g/mol
nitrogen, N = 14 g/mol
Molar mass of HNO₃ = (1 g/mol ) + ( 14 g /mol) + ( 3 x 16 ) = 63 g/mol.
(d)Silver sulfate, Ag₂SO₄
Molar mass of of the atoms are -
silver, Ag = 108 g/mol
sulfur, S = 32 g/mol.
oxygen , O = 16 g/mol.
molar mass of Ag₂SO₄ = ( 2 x 108 g/mol ) + ( 32 g /mol )+ (4 x16 g/mol )
=312 g/mol.
(e )boric acid , B(OH)₃
Molar mass of of the atoms are -
boron , B = 11 g/mol
oxygen , O = 16 g/mol.
Hydrogen , H = 1 g/mol
Molar mass of B(OH)₃ = ( 11 ) + ( 3x16 ) + ( 3 x 1 ) = 62 g/mol.