Answer:
Static Electricity.
Explanation:
Static electricity is defined as 'an electric charge that has built up on an insulated body, often due to friction.' <u> It is an outcome of the disparity among the positive, as well as, negative charges residing in a body</u> or object and causes the charge to build up on the surface of the body. The accumulation of electric charges on the objects like wool, hair, silk, plastic, etc. causes them to possess static electricity. These charges stay on the surface until it is discharged or released through a source. Thus, <u>'static electricity</u>' is the correct answer.
Answer:
V = 65.81 L
Explanation:
En este caso, debemos usar la expresión para los gases ideales, la cual es la siguiente:
PV = nRT (1)
Donde:
P: Presion (atm)
V: Volumen (L)
n: moles
R: constante de gases (0.082 L atm / mol K)
T: Temperatura (K)
De ahí, despejando el volumen tenemos:
V = nRT / P (2)
Sin embargo como estamos hablando de condiciones normales de temperatura y presión, significa que estamos trabajando a 0° C (o 273 K) y 1 atm de presión. Lo que debemos hacer primero, es calcular los moles que hay en 50 g de amoníaco, usando su masa molar de 17 g/mol:
n = 50 / 17 = 2.94 moles
Con estos moles, reemplazamos en la expresión (2) y calculamos el volumen:
V = 2.94 * 0.082 * 273 / 1
<h2>
V = 65.81 L</h2>
Answer: Freezing point of a solution will be 
Explanation:
Depression in freezing point is given by:
= Depression in freezing point
i= vant hoff factor = 1 (for non electrolyte)
= freezing point constant = 
m= molality
Weight of solvent (benzene)= 1480 g =1.48 kg
Molar mass of solute (octane) = 114.0 g/mol
Mass of solute (octane) = 220 g
Thus the freezing point of a solution will be
Answer:
1.4952 grams of sodium bicarbonate she would need to ingest to neutralize this much HCl.
Explanation:
Moles of hydrochloric acid = n
Volume of hydrochloric acid solution = 200.0 mL = 0.200 L
Molarity of the hydrochloric acid = 0.089 M
of HCL
According to reaction, 1 mole of HCl is neutralized by 1 mole of sodium bicarbonate.
Then 0.0178 moles of HCl wil be neutralized by :
of sodium bicarbonate
Mass of 0.0178 moles of sodium bicarbonate:
0.0178 mol × 72 g/mol = 1.4952 g
1.4952 grams of sodium bicarbonate she would need to ingest to neutralize this much HCl.
