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Mazyrski [523]
3 years ago
10

Which best explains why ice floats

Chemistry
2 answers:
Misha Larkins [42]3 years ago
7 0
It lowers density thats why its able too float :)
Vadim26 [7]3 years ago
6 0
Ice floats because of the small air particles that are inside of the ice. The fewer air particles and the ice won't float. The more and the ice will float.
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For each of the following pairs of complexes, identify which one you would predict to have the larger Δo value, and explain why.
mash [69]

Answer:

a) [Fe(H2O)6]3+

b) [Fe(CN)6]3−

c) [Ru(CN)6]3-

Explanation:

. [Mn(H2O)6]2+ or [Fe(H2O)6]3+

The both complexes are d5 complexes with the same ligand , water. Water is a weak ligand and note that Mn^2+ often have a crystal field stabilization energy of zero hence

[Fe(H2O)6]3+ will possess a greater ∆o value.

The splitting of d orbitals according to the crystal field theory depends on the;

i)geometry of the complex

ii) nature of the metal ion,

iii)charge on the metal ion,

iv) ligands that surround the metal ion.

When the geometry and the ligands are held constant, the order of crystal field splitting is as follows;

Pt4+ > Ir3+ > Rh3+ > Co3+ > Cr3+ > Fe3+ > Fe2+ > Co2+ > Ni2+ > Mn2+

[Fe(H2O)6]3+ or [Fe(CN)6]3−

[Fe(CN)6]3− will have a greater ∆o because the cyanide ion is a strong field ligand compared to water. A strong field ligand causes a greater splitting of the octahedral crystal field compared to a weak field ligand.

. [Fe(CN)6]3− or [Ru(CN)6]3-

[Ru(CN)6]3- will exhibit a greater crystal field splitting. Crystal field splitting increases with the second and third row transition elements when compared to the crystal field splitting of the first row transition elements. Note that, there is an increase of approximately 30%–50% in Δo on going from a first-row transition metal to a second-row metal and another 30%–50% increase on going from a second-row to a third-row metal when they have the same geometry and oxidation state.

4 0
3 years ago
Idk how to do this, if you know the answer plz explain
KatRina [158]
The temperature is constant which makes it the independent variable
7 0
3 years ago
Read 2 more answers
Determine the temperature of 1.42 mol of a gas contained in a 3.00-L container at a pressure of 123 kPa
Rashid [163]

Answer:

The temperature is 30,92K

Explanation:

We use the formula PV=nRT. We convert the unit of pressure in kPa into atm.

101,325kPa----1atm

121kPa-------x=(121,3kPax 1 atm)/101,325kPa=1, 2 atm

PV=nRT---->T= (PV)/(RT)

T=(1,2 atm x 3L)/(1,42 mol x 0,082 l atm/K mol )= 30, 91721058 K

6 0
3 years ago
The half-life of 131I is 80 days. How much of a 500.0 mg sample remains after 3 half lives?
I am Lyosha [343]

Answer:

62.5 mg

Explanation:

Just multiply the original amount by  1/2   three times:

500 mg  x  1/2  x  1/2  x  1/2 = 62.5 mg

6 0
2 years ago
Please select the word from the list that best fits the definition<br><br> growth from seed to plant
Misha Larkins [42]
Is there a picture for this question ?
8 0
3 years ago
Read 2 more answers
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