Answer:
If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia
Explanation:
Step 1: Data given
Mass of nitrogen gas (N2) = 13.4 grams
Molar mass of N2 = 28 g/mol
Molar mass of NH3 = 17.03 g/mol
Step 2: The balanced equation
N2 + 3H2 → 2NH3
Step 3: Calculate moles of N2
Moles N2 = Mass N2 / molar mass N2
Moles N2 = 13.4 grams / 28.00 g/mol
Moles N2 = 0.479 moles
Step 4: Calculate moles of NH3
For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3
For 0.479 moles N2 we'll produce 2*0.479 = 0.958 moles
Step 5: Calculate mass of NH3
Mass of NH3 = moles NH3 * molar mass NH3
Mass NH3 = 0.958 moles * 17.03 g/mol
Mass NH3 = 16.3 grams
If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia
Answer: The coefficients for the given reaction species are 1, 6, 2, 3.
Explanation:
The given reaction equation is as follows.
Now, the two half-reactions can be written as follows.
Reduction half-reaction: 
This will be balanced as follows.
... (1)
Oxidation half-reaction: 
This will be balanced as follows.
... (2)
Adding both equation (1) and (2) we will get the resulting equation as follows.
Thus, we can conclude that coefficients for the given reaction species are 1, 6, 2, 3.
Answer:
3.2 × 10⁻⁸
Explanation:
Let's consider the solution of magnesium carbonate.
MgCO₃ ⇄ Mg²⁺(aq) + CO₃²⁻(aq)
We can relate the molar solubility (S) with the solubility product (Ksp) using an ICE chart.
MgCO₃ ⇄ Mg²⁺(aq) + CO₃²⁻(aq)
I 0 0
C +S +S
E S S
The Ksp is:
Ksp = [Mg²⁺] × [CO₃²⁻] = S × S = S² = (1.8 × 10⁻⁴)² = 3.2 × 10⁻⁸
The answer is B mark me as Brainliest while your at it
