In the question, the number of atoms per unit cell is required for:
A) Polonium (Po)
In polonium, the structure is simple cubic, meaning there are 8 corner atoms, which add up to one atom per unit cell.
B) Manganese (Mn)
The structure of the Mn can be considered to be a body centered cubic (BCC) and the number of atoms for this is 8 corner atoms and 1 central atoms, making a total of 2 atoms per unit cell.
C) Silver (Ag)
Silver has a face centered cubic (FCC) unit cell structure, where there are 8 corner atoms and 6 atoms on the faces, so there are a total of 4 atoms per unit cell.
4 get of helium is 1 mole
Answer:
a) 7.0.
b) Nickel sulfate hepta hydrate.
c) 280.83 g/mol.
d) 44.9%.
Explanation:
<u><em>a) What is the formula of the hydrate?</em></u>
The mass of the hydrated sample (NiSO₄.xH₂O) = 5.0 g,
The mass of the anhydrous salt (NiSO₄) = 2.755 g,
The mass of water = 5.0 g - 2.755 g = 2.245 g.
∴ no. of moles of water = mass/molar mass = (2.245 g)/(18.0 g/mol) = 0.1247 mol.
∴ no. of moles of anhydrous salt (NiSO₄) = mass/molar mass = (2.755 g)/(154.75 g/mol) = 0.0178 mol.
∴ water of crystallization in the sample (x) = no. of moles of water/no. of moles of anhydrous salt (NiSO₄) = (0.1247 mol)/(0.0178 mol) = 7.0.
<u><em>b) What is the full chemical name for the hydrate?</em></u>
The name of the salt (NiSO₄.7H₂O) is Nickel sulfate hepta hydrate.
<u><em>c) What is the molar mass of the hydrate? </em></u>
(NiSO₄.7H₂O)
The molar mass = molar mass of NiSO₄ + 7(molar mass of H₂O) = (154.75 g/mol) + 7(18.0 g/mol) = 280.83 g/mol.
<em><u>d) What is the mass % of water in the hydrate?</u></em>
The mass % of water = (mass of water)/(mass of hydrated sample) x 100 = (2.245 g)/(5.0 g) x 100 = 44.9%.
A) energy is converted from one form to another
Answer:
The answer to your question is: 17.26% of carbon
Explanation:
Data
CxHy = 0.2121 g
BaCO₃ = 0.6006 g
Molecular mass BaCO₃ = 137 + 12 + 48 = 197 g
Reaction
CO₂ + Ba(OH)₂ ⇒ BaCO₃ + H₂O
Process
1.- Find the amount of carbon in BaCO₃
197 g of BaCO₃ --------------- 12 g of Carbon
0.6006 g ---------------- x
x = (0.6006 x 12) / 197
x = 0.0366 g of carbon
2.- Calculate the percentage of carbon in the organic compound
0.2121 g of organic compound --------------- 100%
0.0366g -------------- x
x = (0.0366 x 100) / 0.2121
x = 17.26%
