Because it is more dense than the water or tea.
Answer:
0.625 g
Explanation:
Given data:
Half life of radon-222 = 3.8 days
Total mass of sample = 10 g
Mass left after 15.2 days = ?
Solution:
Number of half lives = T elapsed / Half life
Number of half lives = 15.2 / 3.8
Number of half lives = 4
At time zero = 10 g
At first half life = 10 g/2 = 5 g
At 2nd half life = 5 g/ 2= 2.5 g
At third half life = 2.5 g/2 = 1.25 g
At 4th half life = 1.25 g/2 = 0.625 g
Answer:
4.48 grams is the mass of potassium hydroxide that the chemist must weigh out in the second step.
Explanation:
The pH of the solution = 13.00
pH + pOH = 14
pOH = 14 - pH = 14 - 13.00 = 1.00
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
![1.00=-\log[OH^-]](https://tex.z-dn.net/?f=1.00%3D-%5Clog%5BOH%5E-%5D)
![[OH^-]=10^{-1.00} M=0.100 M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E%7B-1.00%7D%20M%3D0.100%20M)
![[KOH]=[OH^-]=[K^+]=0.100 M](https://tex.z-dn.net/?f=%5BKOH%5D%3D%5BOH%5E-%5D%3D%5BK%5E%2B%5D%3D0.100%20M)
Molariy of the KOH = 0.100 M
Volume of the KOH solution = 800 mL= 0.800 L
1 mL = 0.001 L
Moles of KOH = n
n = 0.0800 mol
Mass of 0.0800 moles of KOH :
0.0800 mol × 56 g/mol = 4.48 g
4.48 grams is the mass of potassium hydroxide that the chemist must weigh out in the second step.
Answer:
1.99 x 10⁻¹⁸J
Explanation:
Given parameters:
Frequency of the wave = 3 x 10¹⁵Hz
Unknown:
Energy of the photon = ?
Solution:
To solve this problem, we use the expression below;
E = hf
Where E is the energy, h is the Planck's constant and f is the frequency
Now insert the parameters and solve for E;
E = 6.63 x 10⁻³⁴ x 3 x 10¹⁵ = 19.9 x 10⁻¹⁹J or 1.99 x 10⁻¹⁸J
