The question is incomplete, here is the complete question:
What volume (mL) of the partially neutralized stomach acid having concentration 2 M was neutralized by 0.1 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)
<u>Answer:</u> The volume of HCl neutralized is 1.25 mL
<u>Explanation:</u>
To calculate the volume of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of stomach acid which is HCl
are the n-factor, molarity and volume of base which is NaOH.
We are given:
Putting values in above equation, we get:
Hence, the volume of HCl neutralized is 1.25 mL
Answer:
1. 
2. 
3. 
Explanation:
¡Hola!
En este caso, dada la información para estos problemas, procedemos de la siguiente manera, basado en las leyes de los gases ideales:
1. Una masa de aire ocupa un volumen de 5 litros a una temperatura de 120 °C Cual será el nuevo volumen si la temperatura se reduce a la mitad:
Aqui, utilizamos la ley de Charles, asegurándonos que la temperatura está en Kelvin:
2. Un gas ideal ocupa un volumen de 4000 ml a una presión absoluta de 1500 kilo pascal Cual será la presión si el gas es comprimido lentamente hasta 750 kilo pascal a temperatura constante?
Aquí, utilizamos la ley de Boyle, dado que la temperatura se mantiene constante, calculando el volumen, ya que lo que se da es la presión final:
3. Un gas ocupa un volumen de 200 litros a 95°C y 782 mmHg Cual será el volumen ocupado por dicho gas a 65°C y 815 mmHg:
Aquí, utilizamos la ley combinada de los gases ideales, asegurándonos que las temperaturas están en Kelvin:
¡Saludos!
1. 12 L = 12 dm³
2. 3.18 g
<h3>Further explanation</h3>
Given
1. Reaction
K₂CO₃+2HNO₃⇒ 2KNO₃+H₂O+CO₂
69 g K₂CO₃
2. 0.03 mol/L Na₂CO₃
Required
1. volume of CO₂
2. mass Na₂CO₃
Solution
1. mol K₂CO₃(MW=138 g/mol) :
= 69 : 138
= 0.5
mol ratio of K₂CO₃ : CO₂ = 1 : 1, so mol CO₂ = 0.5
Assume at RTP(25 C, 1 atm) 1 mol gas = 24 L, so volume CO₂ :
= 0.5 x 24 L
= 12 L
2. M Na₂CO₃ = 0.03 M
Volume = 1 L
mol Na₂CO₃ :
= M x V
= 0.03 x 1
= 0.03 moles
Mass Na₂CO₃(MW=106 g/mol) :
= mol x MW
= 0.03 x 106
= 3.18 g
Answer:
Copper carbonate decomposes to copper oxide and carbon dioxide
Explanation:
the coefficients are equal is NOT a sign that a double displacement reaction has occurred
<u>Explanation:</u>
Usually during solutions of two ionic compounds are combined a precipitate (solid) is created. The mix may become cloudy, or may also separate, with the solids falling at the seat of the receptacle. Infrequently, the mix may bubble as a gas or water is produced.
A precipitation reaction is at two aqueous ionic compounds that make a unique ionic compound that is not soluble in water. The insoluble product compound is called the precipitate. The reaction, in this case, is not balanced, which are evaluated by summing coefficients.
