Here's my best guess
the volume of the unit cell is (385*10^-12)^3=5.7066*10^-29 m^3
multiply by density to get mass
mass = (7 g/cm^3)*(100^3 cm^3 / 1^3 m^3) * 5.7066*10^-29 m^3= 3.99466*10^-22 g
covert to moles
3.99466*10^-22 g * 1 mol / 239.82 g = 1.6657 *10^-24 mol
convert to number of units
1.6657 *10^-24 mol * 6.23*10^23 units/mol = 1.04
385 pm = 3.85*10^(-8) cm
The volume of the unit cell is the cube of that, which is 5.71*10^(-23) cm^3. Since the ratio of mass to volume (i.e. the density) must be the same no matter what amount of TlCl you have, you can say:
7 = x/(5.71*10^(-23)), where x is the mass of the unit cell. Solving for x, you get 4*10^(-22) g.
The mass of a molecule of TlCl is 240 amu, which in grams is 4*10^(-22) g. The mass of the unit cell and the mass of a molecule of TlCl is the same. Therefore there is one formula unit of TlCl per unit cell.
Answer:
26.2g = Mass of water in the calorimeter
Explanation:
The heat absorbed for the water is equal to the heat released for the metal. Based on the equation:
Q = m*C*ΔT
<em>Where Q is heat, m is the mass of the sample, C is specific heat of the material and ΔT is change in temperature</em>
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Replacing we can write:

13.9g * 0.449J/g°C * (54.2°C-15.6°C) = m(H₂O) * 4.184J/g°C * (15.6°C-13.4°C)
240.9J = m(H₂O) * 9.2J/g
<h3>26.2g = Mass of water in the calorimeter</h3>
Do you think black schools had budgets to purchase update textbooks
Number of oxygen atoms present in 7.15×10²³ molecules of mannose are 1.92≈2 a m u .
<h3>What is mannose sugar?</h3>
D-mannose is a simple sugar found in many fruits. It is related to glucose. In some cells it occurs naturally in the body.
Mannose is a six-carbon sugar found in a variety of fruits and vegetables. This sugar is not found free in foods. It is a part of polysaccharide chains attached to a variety of proteins.
To calculate number of oxygen atoms present in 7.15×10²³ molecules of mannose-
Since,
1 molecule of oxygen=2.303×10²³ no. of atoms
Here, total number of molecules of oxygen is 6.
Therefore, 6 molecule of oxygen =6×2.303×10²³÷7.15×10²³
=1.92≈2 a m u .
Hence, the total number of oxygen atoms in 7.15×10²³ molecules of mannose is 1.92 ≈2 a m u.
Learn more about mannose sugar, here:
brainly.com/question/7008130
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Answer:
Explanation:
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