15.5% by mass is
equivalent 15.5 g urea in 100 g solution or 155 g urea in 1 kg solution. <span>
<span>we know that molality = moles solute / kg solvent
<span>moles solute = 155 g / 60 g/mol = 2.58 moles urea
</span></span></span>
Since there are 155 g
urea in 1000g solution, hence the solvent is 845 g or 0.845 kg
So:<span>
<span>molality = 2.58 / 0.845 = 3.06 m</span></span>
Answer:
% Mass of sand = 35.95%
% Mass of KBr = 64.05%
Explanation:
From the question given, we obtained the following:
Mass of mixture( KBr + sand) = 10g
Mass of sand = 3.595g
Mass of KBr = 10 — 3.595 = 6.405g
% Mass of sand = ( Mass of sand / mass of mixture) x 100
% Mass of sand = (3.595 / 10) x100
% Mass of sand = 35.95%
We can obtain the percentage of KBr by subtracting the percentage of sand from 100. This is illustrated below:
% Mass of KBr = 100 — 35.95
% Mass of KBr = 64.05%
6H2O represents 6 molecules of water
Your answer is a, boron. Boron has 5 protons, 6 neutrons, and 5 electrons. Hope this helps!
Answer:
810 pm
Explanation:
Step 1: Given and required data
- Velocity of the atom (v): 490 m/s
- Mass of a hydrogen atom (m): 1.67 × 10⁻²⁷ kg
- Planck's constant (h): 6.63 × 10⁻³⁴ J.s
Step 2: Calculate the de Broglie wavelength of the hydrogen atom
We will use de Broglie's equation.
λ = h / m × v
λ = 6.63 × 10⁻³⁴ J.s / 1.67 × 10⁻²⁷ kg × 490 m/s = 8.10 × 10⁻¹⁰ m
Step 3: Convert 8.10 × 10⁻¹⁰ m to picometers
We will use the conversion factor 1 m = 10¹² pm.
8.10 × 10⁻¹⁰ m × 10¹² pm/1 m = 810 pm
