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Alexxx [7]
3 years ago
14

Which mineral leaves a green-black powder when rubbed against an unglazed porcelain plate

Chemistry
1 answer:
solong [7]3 years ago
4 0

<u>Answer:</u>

Pyrite leaves behind a green-black streak when it is rubbed against an unglazed porcelain plate as a part of the streak test process.

<u>Explanation:</u>

Different minerals produce different coloured streaks when rubbed against a white ceramic or porcelain streak plate. This streak test is done to identify the mineral and distinguish the same from other minerals that look similar in colour and texture.

It must be ensured that the test is done on clean and fresh specimens of the mineral and that there must be no contaminants. Pyrite specimens are usually brass-yellow colour but it leaves a green-black streak when the streak test is done.

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The unequal sharing of electrons between the oxygen and hydrogen atoms within a water molecule makes water a ____________ molecu
Ivanshal [37]

Answer:

Polar

Explanation:

8 0
2 years ago
Which of the following elements has the strongest ionization energy: Mg or Cl? Define ionization
Mila [183]

Answer:

Cl

Explanation:

The element Cl will have the strongest ionization energy from the given choices. Most non-metals have higher ionization energy compared to metals.

Ionization energy is the energy required to remove the most loosely held electron from the gaseous phase of an atom.

  • As you go from left to right on the periodic table, it increases progressive
  • From top to bottom, the ionization energy reduces significantly.
  • The attractive force between the protons in the nucleus and the electrons plays a very important role.
  • In metals, they have very large atomic radius, the attractive force on the outer electrons is very weak.
  • This is not the case in non-metals
5 0
3 years ago
For the vaporization reaction Br2(l) → Br2(
oksian1 [2.3K]
    The  temperature  at   which  the  process  be   spontaneous  is  calculated  as  follows

delta  G  =  delta H  -T delta S

let  delta G  be =0

therefore  delta H- T  delta s =0

therefore  T=  delta  H/  delta  S
convert  31   Kj  to  J  =  31  x1000=  31000 j/mol

T=31000j/mol /93 j/mol.k =333.33K


3 0
3 years ago
The edge length of the unit cell of KCl (NaCl-like structure, FCC) is 6.28 Å. Assuming anion-cation contact along the cell edge,
Mila [183]

Answer:

1.33 Å

Explanation:

Given that the edge length , a of the KCl which forms the FCC lattice = 6.28 Å

Also,

For the FCC lattice in which the anion-cation contact along the cell edge , the ratio of the radius of the cation to that of anion is 0.731.

Thus,

\frac {r^+}{r^-}=0.731 .................1

Also, the sum of the radius of the cation and the anion in FCC is equal to half of the edge length.

Thus,

r^++r^-=\frac {a}{2}  ...................2

Given that:

Cl^-\ (r^-) = 1.82\ \dot{A}

To find,

K^+\ (r^+) = ? \dot{A}

Using 1 and 2 , we get:

1.731\ r^+=0.731\times \frac {6.28}{2}

<u>Size of the potassium ion = 1.33 Å</u>

4 0
3 years ago
29 Points!!
atroni [7]
The last one would be false
3 0
3 years ago
Read 2 more answers
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