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Gnesinka [82]
3 years ago
8

What is the mass of a cube of aluminum that is 3.0 cm on each edge? The density of aluminum is 2.7 g/cm3.

Chemistry
1 answer:
n200080 [17]3 years ago
8 0
Do you know the formula? I can’t find it.
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(a) Calculate the mass of CaCl2·6H2O needed to prepare 0.125 m CaCl2(aq) by using 500. g of water. (b) What mass of NiSO4·6H2O m
Alinara [238K]

Answer:

(a) 13.7 g.

(b) 28.91 g.

Explanation:

  • molality (m) is the no. of moles of solute dissolved in 1.0 kg of solvent.

∴ m = (no. of moles of solute)/(mass of water (kg))

<em>∴ m = (mass/molar mass of solute)/(mass of water (kg)).</em>

<em />

<u><em>(a) Calculate the mass of CaCl₂·6H₂O needed to prepare 0.125 m CaCl₂(aq) by using 500. g of water.</em></u>

∵ m = (mass/molar mass of CaCl₂·6H₂O)/(mass of water (kg)).

m = 0.125 m, molar mass of CaCl₂·6H₂O = 219.0757 g/mol, mass of water = 500.0 g = 0.5 kg.

∴ 0.125 m = (mass of CaCl₂·6H₂O / 219.0757 g/mol)/(0.5 kg).

∴ mass of CaCl₂·6H₂O = (0.125 m)(219.0757 g/mol)(0.5 kg) = 13.7 g.

<u><em>(b) What mass of NiSO₄·6H₂O must be dissolved in 500. g of water to produce 0.22 m NiSO₄(aq)?</em></u>

∵ m = (mass/molar mass of NiSO₄·6H₂O)/(mass of water (kg)).

m = 0.22 m, molar mass of NiSO₄·6H₂O = 262.84 g/mol, mass of water = 500.0 g = 0.5 kg.

∴ 0.125 m = (mass of NiSO₄·6H₂O / 262.84 g/mol)/(0.5 kg).

∴ mass of NiSO₄·6H₂O = (0.22 m)(262.84 g/mol)(0.5 kg) = 28.91 g.

6 0
3 years ago
The _______ are found on the right side of the arrow in a chemical reaction. A. reactants B. products C. subscripts D. coefficie
julia-pushkina [17]

Answer:

B. products are found on the write side of the arrow in a chemical reaction.

8 0
3 years ago
Read 2 more answers
What are some limitations of this experiment? How could it be improved?
solong [7]

Answer: Limitation: They may be more expensive and time consuming than lab experiments. Limitation: There is no control over extraneous variables that might bias the results. This makes it difficult for another researcher to replicate the study in exactly the same way.

Explanation:

Hope this helps!

3 0
2 years ago
Read 2 more answers
Consider this reaction:
nadya68 [22]
11.7 g hope this helps and have a great day
4 0
2 years ago
A plot of binding energy per nucleon Eb A versus the mass number (A) shows that nuclei with a small mass number have a small bin
Illusion [34]

Answer:

6He =   4.90 MeV/Nucleon

8Li =     5.18  MeV / Nucleon

62Ni =   8.82 MeV / Nucleon

115 In =  8.54 MeV / Nucleon

Explanation:

Our strategy here is to remember that when the mass of a given nuclei is calculated from the sum of the mass of its  protons and neutrons, this mass is greater than  the actual  value . This is the mass defect.

Now this mass defect we can convert to energy  by utilizing Einstein´s equation, E = mc².This is  the binding energy.

For 6He with actual mass 6.0189 u ( He has Z = 2, that is 2 protons )

mass protons =   2 x 1.0078 u  =  2.0516 u

mass neutrons = 4 x 1.0087 u  =  4.0348 u

predicted mass = (2.0516 + 4.0348) u = 6.0504 u

mass defect = (6.0504 - 6.0189) u = 0.0315 u

Now we need to convert this mass expressed in atomic mass units to kilograms ( 1 u = 1.66054 x 10⁻²⁷ Kg )

0.0315 u x 1.66054 x 10⁻²⁷ Kg =5.231 x 10⁻²⁹ Kg

E =5.231x 10⁻²⁸ Kg x (3 x 10⁸ m/s )² = 4.707 x 10⁻¹²J

Finally we will convert this energy in Joules to eV

E = 4.707 x 10⁻¹²  J x 6.242 x 10¹⁸ eV/J = 2.94 x 10⁷ eV = 29.4 MeV

E per nucleon for 6He = 29.4 MeV / 6 =4.90 MeV / Nucleon

Now the calculations for the rest of the nuclei are performed in similar manner with the following results:

8Li = 5.18 MeV / Nucleon

62Ni = 8.82 MeV / Nucleon

115 In =  8.54 MeV / Nucleon

8 0
3 years ago
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