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Viktor [21]
4 years ago
10

Please answer these as quick as possible. I think I know the answer, but I want to confirm. They’re simply 4 true or false quest

ions. ( q.5)

Chemistry
1 answer:
tino4ka555 [31]4 years ago
5 0

卂几丂山乇尺

Point "C" is correct and all the remaining are false.

Note, HomeAtomic and HeteroAtomic compounds are the types of compounds, but not homogenous and heterogenous.

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PLS I NEED HELP give me examples of an atom
Ugo [173]

Answer:

Neon (Ne)

Hydrogen (H)

Argon (Ar)

Iron (Fe)

Calcium (Ca)

Deuterium, an isotope of hydrogen that has one proton and one neutron.

Plutonium (Pu)

F-, a fluorine anion.

Explanation:

I got u

4 0
3 years ago
5.11 g of MgSO₄ is placed into 100.0 mL of water. The water's temperature increases by 6.70°C. Calculate ∆H, in kJ/mol, for the
cupoosta [38]

Answer: Thus ∆H, in kJ/mol, for the dissolution of MgSO₄ is -66.7 kJ

Explanation:

To calculate the entalpy, we use the equation:

q=mc\Delta T

where,

q = heat absorbed by water = ?

m = mass of water = {\text {volume of water}}\times {\text {density of water}}=100.0ml\times 1.00g/ml=100.0g

c = heat capacity of water = 4.186 J/g°C

\Delta T= change in temperature = 6.70^0C

q=100.0g\times 4.184J/g^0C\times 6.70^0C=2803.3J=2.8033kJ

Sign convention of heat:

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

The heat absorbed by water will be equal to heat released by MgSO_4

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass = 5.11 g

Molar mass  = 120 g/mol

Putting values in above equation, we get:

\text{Moles of }MgSO_4=\frac{5.11g}{120g/mol}=0.042mol

0.042 moles of MgSO_4 releases = 2.8033 kJ

1 mole of MgSO_4 releases = \frac{2.8033 kJ}{0.042}\times 1=66.7kJ

Thus ∆H, in kJ/mol, for the dissolution of MgSO₄ is -66.7 kJ

3 0
4 years ago
Assume that each atom is a sphere, and that the surface of each atom is in contact with its nearest neighbor. Determine the perc
tatiyna

Answer:

  • The percentage of unit cell volume that is occupied by atoms in a face- centered cubic lattice is 74.05%
  • The percentage of unit cell volume that is occupied by atoms in a body-centered cubic lattice is 68.03%  
  • The percentage of unit cell volume that is occupied by atoms in a diamond lattice is 34.01%

Explanation:

The percentage of unit cell volume = Volume of atoms/Volume of unit cell

Volume of sphere = \frac{4 }{3} \pi r^2

a) Percentage of unit cell volume occupied by atoms in face- centered cubic lattice:

let the side of each cube = a

Volume of unit cell = Volume of cube = a³

Radius of atoms = \frac{a\sqrt{2} }{4}

Volume of each atom = \frac{4 }{3} \pi (\frac{a\sqrt{2}}{4})^3 = \frac{\pi *a^3\sqrt{2}}{24}

Number of atoms/unit cell = 4

Total volume of the atoms = 4 X \frac{\pi *a^3\sqrt{2}}{24} = \frac{\pi *a^3\sqrt{2}}{6}

The percentage of unit cell volume = \frac{\frac{\pi *a^3\sqrt{2}}{6}}{a^3} =\frac{\pi *a^3\sqrt{2}}{6a^3} = \frac{\pi \sqrt{2}}{6} = 0.7405

= 0.7405 X 100% = 74.05%

b) Percentage of unit cell volume occupied by atoms in a body-centered cubic lattice

Radius of atoms = \frac{a\sqrt{3} }{4}

Volume of each atom =\frac{4 }{3} \pi (\frac{a\sqrt{3}}{4})^3 =\frac{\pi *a^3\sqrt{3}}{16}

Number of atoms/unit cell = 2

Total volume of the atoms = 2X \frac{\pi *a^3\sqrt{3}}{16} = \frac{\pi *a^3\sqrt{3}}{8}

The percentage of unit cell volume = \frac{\frac{\pi *a^3\sqrt{3}}{8}}{a^3} =\frac{\pi *a^3\sqrt{3}}{8a^3} = \frac{\pi \sqrt{3}}{8} = 0.6803

= 0.6803 X 100% = 68.03%

c) Percentage of unit cell volume occupied by atoms in a diamond lattice

Radius of atoms = \frac{a\sqrt{3} }{8}

Volume of each atom = \frac{4 }{3} \pi (\frac{a\sqrt{3}}{8})^3 = \frac{\pi *a^3\sqrt{3}}{128}

Number of atoms/unit cell = 8

Total volume of the atoms = 8X \frac{\pi *a^3\sqrt{3}}{128} = \frac{\pi *a^3\sqrt{3}}{16}

The percentage of unit cell volume = \frac{\frac{\pi *a^3\sqrt{3}}{16}}{a^3} =\frac{\pi *a^3\sqrt{3}}{16a^3} = \frac{\pi \sqrt{3}}{16} = 0.3401

= 0.3401  X 100% = 34.01%

4 0
3 years ago
What would be the correct name for element 119?
Kay [80]
Ununennium I believe
7 0
4 years ago
Why are solutions considered mixtures?
Alex_Xolod [135]
Solute and solvent form a solution, which means two different kinds of molecules/atoms/compounds are mixed together, therefore, it is a mixture.
6 0
3 years ago
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