<span>the element that requires the least amount of energy to remove the outer-most electron from a gaseous atom in the ground state is s</span><span>odium</span>
You can split the process in two parts:
1) heating the liquid water from 10.1 °C to 25.0 °C , and
2) vaporization of liquid water at constant temperature of 25.0 °C.
For the first part, you use the formula ΔH = m*Cs*ΔT
ΔH = 30.1g * 4.18 j/(g°C)*(25.0°C - 10.1°C) = 1,874 J
For the second part, you use the formula ΔH = n*ΔHvap
Where n is the number of moles, which is calculated using the mass and the molar mass of the water:
n = mass / [molar mass] = 30.1 g / 18.0 g/mol = 1.67 mol
=> ΔH = 1.67 mol * 44,000 J / mol = 73,480 J
3) The enthalpy change of the process is the sum of both changes:
ΔH total = 1,874 J + 73,480 J = 75,354 J
Answer: 75,354 J
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.
Answer:
20 neutrons
Explanation:
(not really any just look at a periodic table)
