<u>Answer:</u> The molality of potassium hydroxide solution is 0.608 m
<u>Explanation:</u>
We are given:
3.301 mass % of potassium hydroxide solution.
This means that 3.301 grams of potassium hydroxide is present in 100 grams of solution
Mass of solvent = Mass of solution - Mass of solute (KOH)
Mass of solvent = (100 - 3.301) g = 96.699 g
To calculate the molality of solution, we use the equation:
Where,
= Given mass of solute (KOH) = 3.301 g
= Molar mass of solute (KOH) = 56.1 g/mol
= Mass of solvent = 96.699 g
Putting values in above equation, we get:
Hence, the molality of potassium hydroxide solution is 0.608 m
<span>35.0 mL of 0.210 M
KOH
molarity = moles/volume
find moles of OH
do the same thing for: 50.0 mL of 0.210 M HClO(aq) but for H+
they will cancel out: H+ + OH- -> H2O
but you'll have some left over,
pH=-log[H+]
pOH
=-log[OH-]
pH+pOH
=14</span>
Landslides and slumps are forms of mass movements. Types of erosion both move, in different ways. In slumps, land breaks off in one piece, or chunk. Landslides are when rocks and dirt rapidly move down a slide.
Answer:
Evaporation
Explanation:
Heat makes molecules move and eventually evaporate.
Answer:
(a) 7.11 x 10⁻³⁷ m
(b) 1.11 x 10⁻³⁵ m
Explanation:
(a) The de Broglie wavelength is given by the expression:
λ = h/p = h/mv
where h is plancks constant, p is momentum which is equal to mass times velocity.
We have all the data required to calculate the wavelength, but first we will have to convert the velocity to m/s, and the mass to kilograms to work in metric system.
v = 19.8 mi/h x ( 1609.34 m/s ) x ( 1 h / 3600 s ) = 8.85 m/s
m = 232 lb x ( 0.454 kg/ lb ) = 105.33 kg
λ = h/ mv = 6.626 x 10⁻³⁴ J·s / ( 105.33 kg x 8.85 m/s ) = 7.11 x 10⁻³⁷ m
(b) For this part we have to use the uncertainty principle associated with wave-matter:
ΔpΔx > = h/4π
mΔvΔx > = h/4π
Δx = h/ (4π m Δv )
Again to utilize this equation we will have to convert the uncertainty in velocity to m/s for unit consistency.
Δv = 0.1 mi/h x ( 1609.34 m/mi ) x ( 1 h/ 3600 s )
= 0.045 m/s
Δx = h/ (4π m Δv ) = 6.626 x 10⁻³⁴ J·s / (4π x 105.33 kg x 0.045 m/s )
= 1.11 x 10⁻³⁵ m
This calculation shows us why we should not be talking of wavelengths associatiated with everyday macroscopic objects for we are obtaining an uncertainty of 1.11 x 10⁻³⁵ m for the position of the fullback.
