LiOH+HCL LiCL+H2O represents a double replacement reaction.
Millimeters are an SI unit of length that =0.0001 m
<u>Given:</u>
Moles of He = 15
Moles of N2 = 5
Pressure (P) = 1.01 atm
Temperature (T) = 300 K
<u>To determine:</u>
The volume (V) of the balloon
<u>Explanation:</u>
From the ideal gas law:
PV = nRT
where P = pressure of the gas
V = volume
n = number of moles of the gas
T = temperature
R = gas constant = 0.0821 L-atm/mol-K
In this case we have:-
n(total) = 15 + 5 = 20 moles
P = 1.01 atm and T = 300K
V = nRT/P = 20 moles * 0.0821 L-atm/mol-K * 300 K/1.01 atm = 487.7 L
Ans: Volume of the balloon is around 488 L
Answer:
The answer is 465.6 mg of MgI₂ to be added.
Explanation:
We find the mole of ion I⁻ in the final solution
C = n/V -> n = C x V = 0.2577 (L) x 0.1 (mol/L) = 0.02577 mol
But in the initial solution, there was 0.087 M KI, which can be converted into mole same as above calculation, equal to 0.02242 mol.
So we need to add an addition amount of 0.02577 - 0.02242 = 0.00335 mol of I⁻. But each molecule of MgI₂ yields two ions of I⁻, so we need to divide 0.00335 by 2 to find the mole of MgI₂, which then is 0.001675 mol.
Hence, the weight of MgI₂ must be added is
Weight of MgI₂ = 0.001675 mol x 278 g/mol = 0.4656 g = 465.6 mg
Answer:
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Explanation:
