Answer:
1. 75.82 m^3 (cubic meters)
2. 75820 L (Liters)
Explanation:
Given data
L = 21 FT
W = 17 ft
h = 7 ft 6 in
Calculation
Formula to calculate volume is:
V= L×W×h
put the values in formula
V = 21 ft × 17 ft × 7 ft 6 in
V = 2677.5 ft
Now we will convert Cubic ft to cubic meters, as we know that 1 cubic meter is equal to 35.3147 cubic feet
V (m^3) = 2677.5 ÷ 35.3147
V (m^3) = 75.82 m^3
Now we will calculate liters
as we know that 1 cubic meter is equal to 1000 liters
so multiply the ans. with 1000
L = 75.82 × 1000
L = 75820 l
Answer:

Explanation:
Hello there!
In this case, according to the given information it will be firstly necessary to set up the chemical equation taking place:

We infer we need to calculate the moles of NH3 by using both of the moles of N2 and H2 at the beginning, in order to identify the limiting reactant:

Thus, since hydrogen yields the fewest moles of ammonia, we conclude that we are just able to yield 4 moles of NH3.
Regards!
Answer:
- <em>The net ionic equation is: </em><u>Ag⁺ (aq) + Cl ⁻ (aq) → AgCl (s)</u>
Explanation:
<u>1) Start by writing the total ionic equation:</u>
The total ionic equation shows each aqueous substance in its ionized form, while the solid or liquid substances are shown with their chemical formula.
These are the ionic species:
- AgF (aq) → Ag⁺ (aq) + F⁻ (aq)
- NH₄Cl (aq) → NH₄⁺ (aq) + Cl ⁻ (aq)
- NH₄F(aq) → NH₄⁺ (aq) + F⁻ (aq)
Then, replace each chemical formula in the chemical equation by those ionic forms:
- Ag⁺ (aq) + F⁻ (aq) + NH₄⁺ (aq) + Cl ⁻ (aq) → AgCl (s) + NH₄⁺ (aq) + F⁻ (aq)
That is the total ionic equation.
<u>2) Spectator ions:</u>
The ions that appear in both the reactant side and the product side are considered spectator ions (they do not change), and so they are canceled.
In our total ionic equation they are F⁻ (aq) and NH₄⁺ (aq).
After canceling them, you get the net ionic equation:
<u>3) Net ionic equation:</u>
- Ag⁺ (aq) + Cl ⁻ (aq) → AgCl (s) ← answer
<u>Answer:</u> The density of phosphorus is 
<u>Explanation:</u>
To calculate the density of phosphorus, we use the equation:

where,
= density
Z = number of atom in unit cell = 1 (CCP)
M = atomic mass of phosphorus = 31 g/mol
= Avogadro's number = 
a = edge length of unit cell =
(Conversion factor:
)
Putting values in above equation, we get:

Hence, the density of phosphorus is 