A student is given a sample of a blue copper sulfate hydrate. He weighs the sample in a dry covered porcelain crucible and got a
1 answer:
Answer:
There are present 5,5668 moles of water per mole of CuSO₄.
Explanation:
The mass of CuSO₄ anhydrous is:
23,403g - 22,652g = 0,751g.
mass of crucible+lid+CuSO₄ - mass of crucible+lid
As molar mass of CuSO₄ is 159,609g/mol. The moles are:
0,751g × = 4,7052x10⁻³ moles CuSO₄
Now, the mass of water present in the initial sample is:
23,875g - 0,751g - 22,652g = 0,472g.
mass of crucible+lid+CuSO₄hydrate - CuSO₄ - mass of crucible+lid
As molar mass of H₂O is 18,02g/mol. The moles are:
0,472g × = 2,6193x10⁻² moles H₂O
The ratio of moles H₂O:CuSO₄ is:
2,6193x10⁻² moles H₂O / 4,7052x10⁻³ moles CuSO₄ = 5,5668
That means that you have <em>5,5668 moles of water per mole of CuSO₄.</em>
I hope it helps!
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Answer:
The wavelength of the emitted photon will be approximately 655 nm, which corresponds to the visible spectrum.
Explanation:
In order to answer this question, we need to recall Bohr's formula for the energy of each of the orbitals in the hydrogen atom:
, where:
[tex]m_{e}[tex] = electron mass
e = electron charge
[tex]\epsilon_{0}[tex] = vacuum permittivity
[tex]\hbar[tex] = Planck's constant over 2pi
n = quantum number
[tex]E_{1}[tex] = hydrogen's ground state = -13.6 eV
Therefore, the energy of the emitted photon is given by the difference of the energy in the 3d orbital minus the energy in the 2nd orbital:
[tex]E_{3} - E_{2} = -13.6 eV(\frac{1}{3^{2}} - \frac{1}{2^{2}})=1.89 eV[tex]
Now, knowing the energy of the photon, we can calculate its wavelength using the equation:
[tex]E = \frac{hc}{\lambda}[tex], where:
E = Photon's energy
h = Planck's constant
c = speed of light in vacuum
[tex]\lambda[tex] = wavelength
Solving for [tex]\lambda[tex] and substituting the required values:
[tex]\lambda = \frac{hc}{E} = \frac{1.239 eV\mu m}{1.89 eV}=0.655\mu m = 655 nm[tex], which correspond to the visible spectrum (The visible spectrum includes wavelengths between 400 nm and 750 nm).
Answer:
92.04%
Explanation:
Given:
Mass of CO₂ obtained = 53.0 grams
Mass of calcium carbonate heated = 1.31 grams
Now,
the molar mass of the calcium carbonate = 100.08 grams
The number of moles heated in the problem = Mass / Molar mass
= (1.31 grams) / (100.08 grams/moles)
= 0.013088 moles
now,
1 mol of calcium carbonate yields 1 mol of CO₂
thus,
0.013088 moles of calcium carbonate will yield = 0.013088 mol of CO₂
now,
Theoretical mass of 0.013088 moles of CO₂ will be
= Number of moles × Molar mass of CO₂
= 0.013088 × 44 = 0.5758 grams
Thus, the percent yield for this reaction =
or
the percent yield for this reaction =
or
the percent yield for this reaction = 92.04%