Answer:
229,098.96 J
Explanation:
mass of water (m) = 456 g = 0.456 kg
initial temperature (T) = 25 degrees
final temperature (t) = - 10 degrees
specific heat of ice = 2090 J/kg
latent heat of fusion =33.5 x 10^(4) J/kg
specific heat of water = 4186 J/kg
for the water to be converted to ice it must undergo three stages:
- the water must cool from 25 degrees to 0 degrees, and the heat removed would be Q = m x specific heat of water x change in temp
Q = 0.456 x 4186 x (25 - (-10)) = 66808.56 J
- the water must freeze at 0 degrees, and the heat removed would be Q = m x specific heat of fusion x change in temp
Q = 0.456 x 33.5 x 10^(4) = 152760 J
- the water must cool further to -10 degrees from 0 degrees, and the heat removed would be Q = m x specific heat of ice x change in temp
Q = 0.456 x 2090 x (0 - (-10)) = 9530.4 J
The quantity of heat removed from all three stages would be added to get the total heat removed.
Q total = 66,808.56 + 152,760 + 9,530.4 = 229,098.96 J
Answer:
D. a cation that has a smaller radius than the atom.
Explanation:
When electrons are removed from the outermost shell of a calcium atom, the atom becomes a cation that has a smaller radius than the atom.
Answer:
Mc = 1920[lb*in]
Explanation:
Para poder solucionar este problema debemos realizar un análisis estático, por tal motivo lo primero es realizar un diagrama de cuerpo libre con las respectivas fuerzas actuando sobre la barra ABC. DE igual manera calcular la geometría de la configuración mostrada.
El diagrama de cuerpo libre se puede ver en la imagen adjunta, con la solución de este problema.
Lo primero es determinar el angulo t, el cual por medio de las propiedades del triangulo rectángulo se puede determinar.
Con este angulo (t) ya determinado, fijamos la atención en el triangulo BCD, este triangulo no es rectángulo, pero por medio de la ley de senos podemos determinar el angulo omega.
Después de determinar el angulo omega, restamos el angulo (t) para poder determinar el angulo (a).
Seguidamente realizamos una sumatoria de momentos alrededor del punto C, utilizado las respectivas fuerzas con los ángulos descompuestos.
El momento en el punto C es de 1920 [Lb*in].
Nota: ya que no se menciona la fuerza en el punto A, esta se desprecia y no se tiene en cuenta en los calculos. En la imagen adjunta se puede ver el procedimiento desarrollado.
Answer:
Because the wavelengths of macroscopic objects are too short for them to be detectable.
Explanation:
Wavelength of an object is given by de Broglie wavelength as:

Where, 'h' is Planck's constant, 'm' is mass of object and 'v' is its velocity.
So, for macroscopic objects, the mass is very large compared to microscopic objects. As we can observe from the above formula, there is an inverse relationship between the mass and wavelength of the object.
So, for vary larger masses, the wavelength would be too short and one will find it undetectable. Therefore, we don't observe wave properties in macroscopic objects.
Answer: C. unknown element he predicted would have properties similar to those of aluminum.
Explanation: Mendeleev arranged the elements in order of increasing atomic masses. He left spaces for the elements that were predicted to exist, but were not discovered.
One space was left below Aluminum where an element of mass 70 was predicted to exist with properties similar to Aluminum. It was referred to as eka-Aluminum.
Later it was named as gallium.