D. Electrons orbit in a fixed shell
Answer:
7 and 11
Explanation:
The amount of sand on the beaches can be found using this formula:
volume (m3) = length (m) × width (m) × depth (m)
(6 × 108 m) × 60 m × 20 m = 7 × 1011 m3
Therefore, there would be a total of 7 × 1011 cubic meters of sand on the beaches.
<h3>Further explanation</h3>
Basic oxides ⇒ metal(usually alkali/alkaline earth) +O₂
L + O₂ ⇒ L₂O
L + O₂ ⇒ LO
Dissolve in water becomes = basic solution
L₂O+H₂O⇒ 2LOH
LO + H₂O⇒ L(OH)₂
So the basic oxides : Na₂O and MgO
Na₂O + H₂O⇒NaOH
MgO +H₂O⇒Mg(OH)₂
The aqueous solution of CO₂(dissolve in water)
CO₂ + +H₂O⇒ H₂CO₃(carbonic acid)
Answer:
643g of methane will there be in the room
Explanation:
To solve this question we must, as first, find the volume of methane after 1h = 3600s. With the volume we can find the moles of methane using PV = nRT -<em>Assuming STP-</em>. With the moles and the molar mass of methane (16g/mol) we can find the mass of methane gas after 1 hour as follows:
<em>Volume Methane:</em>
3600s * (0.25L / s) = 900L Methane
<em>Moles methane:</em>
PV = nRT; PV / RT = n
<em>Where P = 1atm at STP, V is volume = 900L; R is gas constant = 0.082atmL/molK; T is absolute temperature = 273.15K at sTP</em>
Replacing:
PV / RT = n
1atm*900L / 0.082atmL/molK*273.15 = n
n = 40.18mol methane
<em>Mass methane:</em>
40.18 moles * (16g/mol) =
<h3>643g of methane will there be in the room</h3>
Concentration and ion moles are equal only when volume is 1. Of 1.0 x 10-6 and 1.0 x 10-4, the larger number is 1.
