The accepted concentration of chlorine is 1.00 ppm that is 1 gram of chlorine per million of water.
The volume of water is 
.
Since, 1 gal= 3785.41 mL
Thus, 
Density of water is 1 g/mL thus, mass of water will be 
.
Since, 1 grams of chlorine →
grams of water.
1 g of water →
g of chlorine and,
of water →86.6 g of chlorine
Since, the solution is 9% chlorine by mass, the volume of solution will be:
Thus, volume of chlorine solution is 9.62\times 10^{2} mL.
Answer:
Volume of ammonia produced = 398.7 dm³
Explanation:
Given data:
Volume of N₂ = 200 dm³
Pressure and temperature = standard
Volume of ammonia produced = ?
Solution:
Chemical equation:
N₂ + 3H₂ → 2NH₃
Number of moles of N₂:
PV = nRT
1 atm× 200 L = n× 0.0821 atm.L/mol.K × 273 K
n = 200 atm.L /22.41 atm.L/mol
n = 8.9 mol
Now we will compare the moles of ammonia and nitrogen.
N₂ : NH₃
1 : 2
8.9 : 2/1×8.9 = 17.8 mol
Volume of ammonia:
1 mole of any gas occupy 22.4 dm³ volume
17.8 mol ×22.4 dm³/1 mol = 398.7 dm³
Answer:
<u>C) 4</u>
Explanation:
<u>The reaction</u> :
- C (s) + 2H₂ (g) ⇒ CH₄ (g)
12g 4g 16g
Hence, based on this we can say that : <u>2 moles of hydrogen gas are needed to produce 16g of methane.</u>
<u />
<u>For 32g of methane</u>
- Number of moles of H₂ = 32/16 × 2
- Number of moles of H₂ = <u>4</u>
A. It does not retain the properties of the substances that make it up
Answer:
5×10⁵ L of ammonia (NH3)
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
N2 + 3H2 —> 2NH3
From the balanced equation above, we can say that:
3 L of H2 reacted to produce 2 L of NH3.
Finally, we shall determine the volume of ammonia (NH3) produced by the reaction of 7.5×10⁵ L of H2. This can be obtained as illustrated below:
From the balanced equation above,
3 L of H2 reacted to produce 2 L of NH3.
Therefore, 7.5×10⁵ L of H2 will react to produce = (7.5×10⁵ × 2)/3 = 5×10⁵ L of NH3.
Thus, 5×10⁵ L of ammonia (NH3) is produced from the reaction.
