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Elan Coil [88]
3 years ago
8

Describe the processes of abrasion and plucking and how they result in the formation of a glacial horns.

Chemistry
2 answers:
kirill [66]3 years ago
6 0

Answer:

Plucking occurs when rocks and stones become frozen to the base or sides of the glacier and are plucked from the ground or rock face as the glacier moves. HOPE THIS HELPS!~

Explanation:

Alex3 years ago
6 0

Answer:

Abrasion occurs when the ice sheet scours and wears away at the underlying rock, gradually wearing it down over time. Plucking occurs when the chunks of rock become bonded to the ice sheet, so that when the ice sheet moves it carries the chunks of rock with it. Both of these processes wear away a rock body until a distinct formation develops. In this case, a glacier horn may develop.

Explanation:

straight from the teachers answer key your welcome

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Explanation:

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Mass x height x gravity is the formula to calculate:
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Answer:

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1. ___Mg +<br> O2 → __Mgo<br> What type of reaction is shown
Black_prince [1.1K]

Explanation:

In first one it should be 2Mg

and in second = 2Mgo

hope it helps....

8 0
3 years ago
The densities of gases a, b, and c at stp are 1.25 g/l, 2.86 g/l and 0.714 g/l, respectively. calculate the molar mass of each s
kicyunya [14]
The  molar  mass  of    a, b and  c at  STP is calculated  as  below

At  STP  T  is always=   273 Kelvin and ,P= 1.0 atm 

by  use of  ideal  gas  equation  that  is  PV =nRT
n(number   of moles) = mass/molar mass  therefore  replace   n  in  the  ideal   gas  equation

that  is Pv = (mass/molar mass)RT
multiply  both side  by molar  mass  and  then  divide  by  Pv  to  make  molar mass the  subject of the  formula

that is  molar  mass =  (mass x RT)/ PV

 density is always = mass/volume

therefore  by  replacing  mass/volume  in   the equation  by  density the equation
molar  mass=( density  xRT)/P  where R  =  0.082 L.atm/mol.K

the  molar mass  for  a
= (1.25 g/l  x0.082 L.atm/mol.k  x273k)/1.0atm = 28g/mol

the molar  mass of b
=(2.86g/l  x0.082L.atm/mol.k   x273  k) /1.0  atm  = 64  g/mol

the molar  mass of c

=0.714g/l  x0.082  L.atm/mol.K  x273 K) 1.0atm= 16 g/mol

therefore  the 
   gas  a  is  nitrogen N2   since 14 x2= 28 g/mol
   gas b =SO2  since  32 +(16x2)= 64g/mol
  gas c =   methaneCH4  since  12+(1x4) = 16 g/mol


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