It looks like you overlooked the provided image necessary to address this inquiry. Nevertheless, I found it and have the solution. Referring to the topographical map of a segment of Charleston, SC, the feature located at the spot marked with an X represents the peak of a hill. The correct answer is option D.
Response:
To reach the answer, 465.6 mg of MgI₂ is required.
Detailed Explanation:
We need to establish the moles of ion I⁻ in the resulting solution.
C = n/V -> n = C x V = 0.2577 (L) x 0.1 (mol/L) = 0.02577 mol.
In the initial solution, there was 0.087 M KI, which we can similarly convert into moles, yielding 0.02242 mol.
This indicates we require an additional amount of 0.02577 - 0.02242 = 0.00335 mol of I⁻. Since each molecule of MgI₂ produces two I⁻ ions, we divide 0.00335 by 2 to determine the moles of MgI₂, giving us 0.001675 mol.
Consequently, the quantity of MgI₂ to be added is:
Weight of MgI₂ = 0.001675 mol x 278 g/mol = 0.4656 g = 465.6 mg
Answer:
10 °C
Explanation:
Greetings,
As per atmospheric conditions, unsaturated air cools consistently at a rate of 10 °C for every 1000 meters ascended, whereas saturated air cools at a slower pace beyond the lifting condensation level due to latent heat release during the wet adiabatic process.
Best regards.
Answer:
Explanation:
AgNO3 + NaCl --> AgCl + NaNO3
Moles
of AgNO3
= molarity * volume
= 1 * 0.01
= 0.01 mol
for NaCl
= 0.01 * 1
= 0.01 mol.
According to stoichiometry, one mole of silver nitrate corresponds to one mole of NaCl reacted. Hence,
Moles of AgCl generated = 0.01 × 1
= 0.01 mol AgCl produced.
Heat gained by the solution as precipitation occurs:
Solution mass = density × volume
= 1 × 20
= 20 g.
Using q = m * Cp * (T2 - T1)
= 20 * 4.18 * (32.6 - 25.0)
= 635 J
The absorbed heat of 635 J indicates the reaction released -635 J
Thus, Delta H = -635 J/0.01 mol
= -63500 J/mol
= -63.5 kJ/mol.
The correct sequence would be B, A, E, D, C. I hope this information assists you!
