Answer:
132.17 g
Explanation:
The reaction given , in the question is -
CH₄ (g ) + 4 S ( g ) ---> CS₂ ( g ) + 2H₂S ( g )
From the reaction , 4 mole of S is required for the production of 1 mole of CS₂ .
since ,
Moles of CS₂ = given mass of CS₂ / Molecular weight of CS₂
Since ,
the Molecular weight of CS₂ = 76
Given , mass of CS₂ = 72.57 g
Moles of CS₂ = 72.57 / 76 = 0.95 mol
Since ,
The yield is 92.0 % .
Moles of S required = 4 * 0.95 mol / 0.92 = 4.13 moles
Mass of S required = 4.13 * 32 = 132.17 g .
Explanation of the strategy: 1) calculate the mass of water and convert to number of moles, 2) convert the mass of anhydrous CuSO4 to moles, and, 3)calculate the mole ratio of water to CuSO4 anhydrous
1) Calculate the mass of water:
mass of water = mass of the hydrate CuSO4 - mass of the anhydrous compound
mass of water = 12.98 g - 9.70 g = 3.28g
2) Calculate the number of moles of water
number of moles = mass in grams / molar mass
molar mass of water = 18.01 g/mol
number of moles of water = 3.28 g / 18.01 g/mol = 0.182 mol
3) Calculate the number of moles of CuSO4 anhydrous
number of moles = mass in grams / molar mass
molar mass of Cu SO4 = 159.6 g/mol
number of moles of CuSO4 = 9.70g / 159.6 g/mol = 0.0608 moles
4) Calculate the ratio moles of water / moles of CuSO4
ratio = moles of water / moles of CuSO4 = 0.182 / 0.0608 = 2.99 ≈ 3
Therefore the molecular formula is CuSO4 . 3H2O
Name: copper(II) sulfate trihydrate.
Answer:
d = 43.5 g/cm³
Explanation:
Given data:
Mass of magnesium cube = 217.501 g
Volume of magnesium cube = 5.00 cm³
Density of magnesium cube = ?
Solution:
Formula:
d = m/v
d = density
m = mass
v = volume
by putting values,
d = 217.501 g/ 5.00 cm³
d = 43.5 g/cm³
The d subshell has 5 orbitals (with each being able to hold two electrons. As such, the d orbital can hold a maximum of 10 electrons.
<span>The "d" subshell can hold a maximum of _TEN_ electrons.</span>
