After dipping the litmus paper strips in the solutions, the answer to the question is that one is acidic because according to brad's observation on turned red and the others are basic or alkaline with the two papers turning blue.
Answer:

Explanation:
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In this case, according to the given chemical reaction, it is possible for us to calculate the produced grams of nitrogen monoxide by starting with 25.0 g of nitrogen via their 1:2 mole ratio and the molar masses of 30.1 g/mol and 28.02 g/mol, respectively and by some stoichiometry:

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The given molarity of sodium hydroxide solution = 2.0 M
The required concentration of sodium hydroxide is 65 mL of 0.6 M NaOH
Converting 65 mL to L:

Calculating the moles of NaOH in the final solution:

Finding out the volume of 2.0 M solution taken to prepare the final solution:

Therefore, 19.5 mL of 2.0 M NaOH solution and make it up to 65 mL to prepare 0.6 M NaOH solution.
Respuesta:
2 FeCl₃(aq) + 3 Na₂CO₃(s) ⇒ Fe₂(CO₃)₃(s) + 6 NaCl(aq)
Explicación:
Consideremos la ecuación no balanceada que ocurre cuando cloruro férrico acuoso reacciona con carbonato de sodio sólido para formar carbonato férrico sólido y cloruro de sodio acuoso. Esta es una reacción de doble desplazamiento.
FeCl₃(aq) + Na₂CO₃(s) ⇒ Fe₂(CO₃)₃(s) + NaCl(aq)
Vamos a usar el método de tanteo. Empezaremos balanceando los átomos de C, multiplicando Na₂CO₃ por 3.
FeCl₃(aq) + 3 Na₂CO₃(s) ⇒ Fe₂(CO₃)₃(s) + NaCl(aq)
Luego, balancearemos los átomos de Fe, multiplicando FeCl₃ por 2.
2 FeCl₃(aq) + 3 Na₂CO₃(s) ⇒ Fe₂(CO₃)₃(s) + NaCl(aq)
Finalmente, obtendremos la ecuación balanceada, multiplicando NaCl por 6.
2 FeCl₃(aq) + 3 Na₂CO₃(s) ⇒ Fe₂(CO₃)₃(s) + 6 NaCl(aq)