Answer: HOPE THIS HELPED! :D
Water vapor is the most abundant and dominant GHG in the atmosphere. Its concentration depends on temperature and other meteorological conditions, and not directly upon human activities.
Explanation:
Answer is: Ksp for silver sulfide is 8.00·10⁻⁴⁸.
Reaction
of dissociation: Ag₂S(s) → 2Ag⁺(aq) + S²⁻(aq)<span>.
</span>s(Ag₂S) = s(S²⁻) = 1.26·10⁻¹⁶ M.
s(Ag⁺) = 2s(Ag₂S) = 2.52·10⁻¹⁶ M; equilibrium concentration of silver cations.
Ksp = s(Ag⁺)² · s(S²⁻).
Ksp = (2.52·10⁻¹⁶ M)² · 1.26·10⁻¹⁶ M.
Ksp = 6.35·10⁻³² M² · 1.26·10⁻¹⁶ M.
Ksp = 8.00·10⁻⁴⁸ M³.
Answer:
63.05% of MgCO3.3H2O by mass
Explanation:
<em>of MgCO3.3H2O in the mixture?</em>
The difference in masses after heating the mixture = Mass of water. With the mass of water we can find its moles and the moles and mass of MgCO3.3H2O to find the mass percent as follows:
<em>Mass water:</em>
3.883g - 2.927g = 0.956g water
<em>Moles water -18.01g/mol-</em>
0.956g water * (1mol/18.01g) = 0.05308 moles H2O.
<em>Moles MgCO3.3H2O:</em>
0.05308 moles H2O * (1mol MgCO3.3H2O / 3mol H2O) =
0.01769 moles MgCO3.3H2O
<em>Mass MgCO3.3H2O -Molar mass: 138.3597g/mol-</em>
0.01769 moles MgCO3.3H2O * (138.3597g/mol) = 2.448g MgCO3.3H2O
<em>Mass percent:</em>
2.448g MgCO3.3H2O / 3.883g Mixture * 100 =
<h3>63.05% of MgCO3.3H2O by mass</h3>
Given that 1 micrometer or micron (um) is equivalent by definition to 1 x 10^-6 m, this means that 1 square micron (um^2) is equivalent to (1 x 10^-6)^2 m^2, or 1 x 10^-12 m^2.
(2.60 um^2) * (1 x 10^-12 m^2 / 1 um^2) = 2.60 x 10^-12 m^2
Therefore the layer of graphene covers an area of 2.60 x 10^-12 m^2.
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Answer:
10.5 g
Explanation:
Step 1: Given data
- Molar concentration of the solution (C): 0.243 M
- Volume of solution (V): 0.580 L
Step 2: Calculate the moles of solute (n)
Molarity is equal to the moles of solute divided by the liters of solution.
M = n/V
n = M × V
n = 0.243 mol/L × 0.580 L = 0.141 mol
Step 3: Calculate the mass corresponding to 0.141 moles of KCl
The molar mass of KCl is 74.55 g/mol.
0.141 mol × 74.55 g/mol = 10.5 g
