The balanced chemical equation for the formation ammonia is
N2(g) + 3H2(g) ----> 2NH3(g) .
The balanced chemical equations explains that the same number of each element exist as reactants and products. The coefficients in a balanced equation must be the simplest whole number ratio. Mass is always conserved in chemical reactions.
For the formation of ammonia, the chemical equation is
N2(g) + H2(g) ----> NH3(g)
Balancing the chemical reaction, we can write,
N2(g) + 3H2(g) ----> 2NH3(g) .
This equation shows two nitrogen entering the reaction together and two hydrogens entering the reaction together. Since NH3 is multiplied by a coefficient of 2 there are now 2 nitrogen and 6 hydrogens. The 6 hydrogens come from the 2 multiplied by the subscript of 3. This is the balanced chemical reaction.
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➷ The formula for Potassium oxide is K2O
This is because potassium has a 1+ charge and the oxygen has a 2- charge. Two potassium atoms are required to balance the charge.
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Answer:
the third option
Explanation:
The circuit is incomplete and broken, and the lights in the circuit do not shine.
An open circuit means an incomplete circuit.
The first option is wrong since the lights do not shine in an open circuit.
The second option is wrong since an open circuit is incomplete.
The fourth option is also wrong since an open circuit is broken or incomplete.
∴ The third option is correct.
MgCl2(s) + H2O(l) → MgO(s) + 2 HCl(g)
Using the standard enthalpies of formation given in the source below:
(−601.24 kJ) + (2 x −92.30 kJ) − (−641.8 kJ) − (−285.8 kJ) = +141.76 kJ
So:
MgCl2(s) + H2O(l) → MgO(s) + 2 HCl(g), ΔH = +141.76 kJ
Answer:
pH = 12.61
Explanation:
First of all, we determine, the milimoles of base:
0.120 M = mmoles / 300 mL
mmoles = 300 mL . 0120 M = 36 mmoles
Now, we determine the milimoles of acid:
0.200 M = mmoles / 100 mL
mmoles = 100 mL . 0.200M = 20 mmoles
This is the neutralization:
HCOOH + OH⁻ ⇄ HCOO⁻ + H₂O
20 mmol 36 mmol 20 mmol
16 mmol
We have an excess of OH⁻, the ones from the NaOH and the ones that formed the salt NaHCOO, because this salt has this hydrolisis:
NaHCOO → Na⁺ + HCOO⁻
HCOO⁻ + H₂O ⇄ HCOOH + OH⁻ Kb → Kw / Ka = 5.55×10⁻¹¹
These contribution of OH⁻ to the solution is insignificant because the Kb is very small
So: [OH⁻] = 16 mmol / 400 mL → 0.04 M
- log [OH⁻] = pOH → 1.39
pH = 14 - pOH → 12.61
