Balance Chemical equation is as follow,
<span> 3 H</span>₂ <span>(g) + N</span>₂ <span>(g) </span>→<span> 2 NH</span>₃ <span>(g)
According to balanced equation, 3 Molecules (3 moles) of Hydrogen reacts with 1 Molecule of N</span>₂ to produce 2 moles (2 Molecules) of NH₃.
Result:
2 Molecules of Ammonia are produced by reacting 3 molecules of Hydrogen and 1 molecule of Nitrogen.
Answer:
1.02 × 10⁶ g
Explanation:
Step 1: Given data
- Volume of the balloon (V): 5400 m³
- Absolute pressure (P): 1.10 × 10⁵ Pa
- Molar mass of He (M): 4.002 g/mol
Step 2: Convert "V" to L
We will use the conversion factor 1 m³ = 1000 L.
5400 m³ × 1000 L/1 m³ = 5.400 × 10⁶ L
Step 3: Convert "P" to atm
We will use the conversion factor 1 atm = 101325 Pa.
1.10 × 10⁵ Pa × 1 atm / 101325 Pa = 1.09 atm
Step 4: Calculate the moles of He (n)
We will use the ideal gas equation.
P × V = n × R × T
n = P × V / R × T
n = 1.09 atm × 5.400 × 10⁶ L / 0.08206 atm.L/mol.K × 280 K
n = 2.56 × 10⁵ mol
Step 5: Calculate the mass of He (m)
We will use the following expression.
m = n × M
m = 2.56 × 10⁵ mol × 4.002 g/mol
m = 1.02 × 10⁶ g
Answer: 9.3 x 10^ 18 g CO
Explanation:
Start by knowing that carbon monoxide is the compound CO. To convert molecules to grams, you first need to convert molecules to moles. This can be done using the conversion factor for Avogadro's Number:
(2.0 x 10^5 molecules CO) x 1 mol CO / 6.02 x 10^23 molecules CO
This cancels molecules CO.
Then, you can convert moles to grams, which is your desired quantity. You can find the number of grams for CO by looking at the periodic table and adding together their masses. C = 12 g and O = 16 g. Total of 28 g CO:
(1 mol CO) x 28 g CO / 1 mol CO
This cancels mol CO, which leaves grams CO.
Answer:
1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)
Explanation:
Main reaction: 2Ag⁺(aq) + Mn(s) ⇄ 2Ag(s) + Mn²⁺(aq)
In the oxidation half reaction, the oxidation number increases:
Mn changes from 0, in the ground state to Mn²⁺.
The reduction half reaction occurs where the element decrease the oxidation number, because it is gaining electrons.
Silver changes from Ag⁺ to Ag.
1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)
To balance the hole reaction, we need to multiply by 2, the second half reaction:
Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
(Ag⁺(aq) + 1e⁻ ⇄ Ag(s)) . 2
2Ag⁺(aq) + 2e⁻ ⇄ 2Ag(s)
Now we sum, and we can cancel the electrons:
2Ag⁺(aq) + Mn(s) + 2e⁻ ⇄ 2Ag(s) + Mn²⁺(aq) + 2e⁻
