The balanced equation of the reaction between hydrogen and nitrogen to produce ammonia would be:
Ammonia is produced when nitrogen reacts with hydrogen.
Stoichiometrically, 1 mole of nitrogen is required to react with 3 moles of hydrogen with 2 moles of ammonia being the product.
Thus, the balanced chemical equation for the reaction would become:
More on balancedequations can be found here: brainly.com/question/7181548?referrer=searchResults
Answer:
Balanced chemical equation:
N₂ + 3H₂ → 2NH₃
Explanation:
Chemical equation:
N₂ + H₂ → NH₃
Balanced chemical equation:
N₂ + 3H₂ → 2NH₃
Step1:
Left side of equation Right side of equation
N = 2 N = 1
H = 2 H = 3
Step 2:
N₂ + H₂ → 2NH₃
Left side of equation Right side of equation
N = 2 N = 2
H = 2 H = 6
Step 3:
N₂ + 3H₂ → 2NH₃
Left side of equation Right side of equation
N = 2 N = 2
H = 6 H = 6
b. beta decay
c. alpha decay
d. isotope number
FALSE
Answer:
FALSE
Explanation:
Critical temperature can be defined as the temperature at which a substance can exist as a liquid. Below the critical temperature, the substance will freeze and above the critical temperature, the substance will evaporate.
Note: Liquid cannot evaporate below it's critical temperature
Answer:the standard reduction potential (E°red) for the reduction half-reaction of Pb^4+(aq) to Pb^2+(aq) is 1.50 V.
Explanation:The given chemical reaction is:
Pb^4+(aq) + 2 Ce^3+(aq) -> Pb^2+(aq) + 2 Ce^4+(aq)
The standard cell potential (E°cell) for this electrochemical reaction is 0.06 V.
The standard cell potential for a galvanic cell can be calculated using the Nernst equation:
E°cell = E°cathode - E°anode
In this reaction, Pb^4+(aq) is being reduced to Pb^2+(aq), so it is the reduction half-reaction, and Ce^3+(aq) is being oxidized to Ce^4+(aq), so it is the oxidation half-reaction.
The standard reduction potentials (E°) for the half-reactions are as follows:
For the reduction half-reaction:
Pb^4+(aq) + 2 e^- -> Pb^2+(aq) E°red = x (we'll solve for x)
For the oxidation half-reaction:
2 Ce^3+(aq) -> 2 Ce^4+(aq) + 2 e^- E°red = 1.44 V (This value is usually given)
Now, plug these values into the Nernst equation:
E°cell = E°cathode - E°anode
0.06 V = x - 1.44 V
Now, solve for x:
x = 0.06 V + 1.44 V
x = 1.50 V
So, the standard reduction potential (E°red) for the reduction half-reaction of Pb^4+(aq) to Pb^2+(aq) is 1.50 V.
Answer:
grams per milliliter.
Explanation:
While any units of mass and volume can be used to calculate density the most common are grams (g) and milliliters (ml). This gives density the units of grams per milliliter (g/ml).