The balanced equation for the reaction is as follows;
Li₂O + H₂O ---> 2LiOH
Stoichiometry of Li₂O to H₂O is 1:1
Mass of Li₂O reacted - 18.9 g
Number of Li₂O moles reacted - 18.9 g / 30 g/mol = 0.63 mol
An equivalent amount of moles of water have reacted - 0.63 mol
mass of water required - 0.63 mol x 18 g/mol = 11.34 g
A mass of 11.34 g of water is required
"The inert gases are obtained by fractional distillation of air, with the exception of helium which is separated from a few natural gas sources rich in this element, through cryogenic distillation or membrane separation. For specialized applications, purified inert gas shall be produced by specialized generators on-site. They are often used by chemical tankers and product carriers (smaller not a big as well as the tendency of inert gases vesselshtop specialized generators are also available for laboratories."
Answer:
first one is a second one is e
Explanation:
Answer:most of the positively charge particles should be bounce back at a range of angles as they collide with the atoms in the foil; only a few should pass straight through the foil
Explanation:
These are two questions and two answers
Question 1.
Answer:
Explanation:
<u>1) Data:</u>
a) m = 9.11 × 10⁻³¹ kg
b) λ = 3.31 × 10⁻¹⁰ m
c) c = 3.00 10⁸ m/s
d) s = ?
<u>2) Formula:</u>
The wavelength (λ), the speed (s), and the mass (m) of the particles are reltated by the Einstein-Planck's equation:
- h is Planck's constant: h= 6.626×10⁻³⁴J.s
<u>3) Solution:</u>
Solve for s:
Substitute:
- s = 6.626×10⁻³⁴J.s / ( 9.11 × 10⁻³¹ kg × 3.31 × 10⁻¹⁰ m) = 2.20 × 10 ⁶ m/s
To express the speed relative to the speed of light, divide by c = 3.00 10⁸ m/s
- s = 2.20 × 10 ⁶ m/s / 3.00 10⁸ m/s = 7.33 × 10 ⁻³
Answer: s = 7.33 × 10 ⁻³ c
Question 2.
Answer:
Explanation:
<u>1) Data:</u>
a) m = 45.9 g (0.0459 kg)
b) s = 70.0 m/s
b) λ = ?
<u>2) Formula:</u>
Macroscopic matter follows the same Einstein-Planck's equation, but the wavelength is so small that cannot be detected:
- h is Planck's constant: h= 6.626×10⁻³⁴J.s
<u>3) Solution:</u>
Substitute:
- λ = 6.626×10⁻³⁴J.s / ( 0.0459 kg × 70.0 m/s) = 2.06 × 10 ⁻³⁴ m
As you see, that is tiny number and explains why the wave nature of the golf ball is undetectable.
Answer: 2.06 × 10 ⁻³⁴ m.