Answer:
the cold will get hot and the hot will get cold
Explanation:
Answer:
[EtOH] = 2.2M and Wt% EtOH = 10.1% (w/w)
Explanation:
1. Molarity = moles solute / Volume solution in Liters
=> moles solute = mass solute / formula weight of solute = 9.8g/46g·mol⁻¹ = 0.213mol EtOH
=> volume of solution (assuming density of final solution is 1.0g/ml) ...
volume solution = 9.81gEtOH + 87.5gH₂O = 97.31g solution x 1g/ml = 97.31ml = 0.09731 Liter solution
Concentration (Molarity) = moles/Liters = 0.213mol/0.09731L = 2.2M in EtOH
2. Weight Percent EtOH in solution (assuming density of final solution is 1.0g/ml)
From part 1 => [EtOH] = 2.2M in EtOH = 2.2moles EtOH/1.0L soln
= {(2.2mol)(46g/mol)]/1000g soln] x 100% = 10.1% (w/w) in EtOH.
Answer:
I can be a friend if u need
1) number of moles of N2 = n/2
2) Number of moles of CH4 = n/2
3) Total number of moles of the mixture = n/2 + n/2 = n
4) Kg of N2
mass in grams = number of moles * molar mass
molar mass of N2 = 2 * 14.0 g/mol = 28 g/mol
=> mass of N2 in grams = (n/2) * 28 = 14n
mass of N2 in Kg = mass of N2 in grams * [1 kg / 1000g] = 14n/1000 kg = 0.014n kg
Answer: mass of N2 in kg = 0.014n kg
Answer:
Explanation:
The given pH = 8.55
Unknown:
[H₃O⁺] = ?
[OH⁻] = ?
In order to find these unknowns we must first establish some relationship.
pH = -log[H₃O⁺]
8.55 = -log[H₃O⁺]
[H₃O⁺] = inverse log₁₀(-8.55) = 2.82 x 10⁻⁹moldm⁻³
To find the [OH⁻],
pH + pOH = 14
pOH = 14 - pH = 14 - 8.55
pOH = 5.45
pOH = -log[OH⁻]
[OH⁻] = inverse log₁₀ (-5.45) = 3.55 x 10⁻⁶moldm⁻³
The solution is basic because it has more concentration of OH⁻ ions compared to H⁺ ions.
