Answer:
Si hay algo que sucedió entre ustedes raro o un comportamiento que tuviste con esa persona o que te vio hacer asi como le pudieron decir algo de ti que no le gustara a tu amigo hace que se comporte raro contigo
Explanation:
Answer:
Explanation:
Use Dalton's law and the vapor pressure of water at 23.0 o C to correct the pressure to units of atmoshperes.
PT = Poxygen +Pwater
At 23.0 o C the vapor pressure of water is 21.1 mmHg. (This can be found on a vapor pressure table.)
762 mmHg = Poxygen + 21.1 mmHg
Poxygen = 762 mmHg - 21.1 mmHg
Poxygen =741 mmHg
Convert the corrected pressure to atmospheres.
(741 mmHg) (1 atm / 760 mmHg) = 0.975 atm
Use the ideal gas law to find out how many moles of gas were produced:
PV = nRT (remember to put volume in liters and temperature in Kelvin)
(0.975 atm) (.193 L) = n (.0821 L atm / mol K) (298 K)
n = (0.975 atm) (.193 L) / (.0821 L atm / mol K) (298 K)
n = 7.69 X 10-4 mol
Use the number of moles and the molecular weight of oxygen to find out how many grams of oxygen were collected.
(7.69 X 10-4 mol) (32.0 g / 1 mol) = 2.46 X 10-2 g
Answer:
pA = 0.095 atm
pB = 0.303 atm
Explanation:
Step 1: the reaction
AB(s) ⇔ A(g) + B(g)
Kp = pA * pB
⇒ with Kp = equilibrium constant
Kp = 0.126 * 0.23 ⇒ Kp = 0.02898
Since the container will be compressed to half of its original volume, means that he pressure will be doubled.
⇒pA = 0.252
⇒pB =0.46
To establish this equilibrium, each pressure has to be lowered by x
⇒pA = 0.252 - x
⇒pB = 0.46 - x
Kp = 0.02898 = (0.252 - x)(0.46-x)
0.02898 = 0.11592 - 0.252x -0.46x + x²
-x² + 0.712x - 0.08694 = 0
D= b² - 4ac
⇒ D = 0.712² -4*(-1) *(-0.08694) = 0.506944 -0.34776 =0.159184
x = (-b ± √D)/2a
x = (-0.712 ± √0.159184)/(2*-1) = (-0.712 ± 0.398978696)/-2
x = 0.156510652 or x= 0.555489348
x = 0.555489348 is impossble or the pressure would be negative
x=0.156510652
pA =0.252 - 0.156510652 = 0.095489348 atm
pB = 0.46 - 0.156510652 = 0.303489348 atm
Answer:
250cm³ de disolución se deben preparar
Explanation:
El porcentaje volumen volumen (%v/v) se define como el volumen de soluto -En este caso, H2O2- en 100cm³ de disolución.
Se desea preparar una solución que contenga 20cm³ de H2O2 por cada 100cm³
Como solamente se cuenta con 50cm³ de H2O2, el volumen que se debe preparar es:
50cm³ H2O2 * (100cm³ Disolución / 20cm³ H2O2) =
<h3>250cm³ de disolución se deben preparar</h3>
