Answer : (C) Hafnium is the most likely identity of the given substance.
Solution : Given,
Mass of given substance (m) = 46.9 g
Volume of given substance (V) = 3.5 
First, find the Density of given substance.
Formula used :
Now,put all the values in this formula, we get
= 13.4 g/
So, we conclude that the density of given substance (13.4 g/) is approximately equal to the density of Mercury and Hafnium (13.53 and 13.31 g/ respectively).
According to the question the substance is solid at room temperature but Mercury is liquid at room temperature. So, Mercury is not identical to the given substance.
Another element i.e, Hafnium is the element whose density is approximately equal to the given substance and also solid at room temperature. And we know that the melting point of solid is high.
So, Hafnium is the most likely element which is the identity of the given substance.
Elements that exceed the octet rule must have an unoccupied d orbital. I believe.
The greater the energy, the larger the frequency and the shorter (smaller) the wavelength. Given the relationship between wavelength and frequency — the higher the frequency, the shorter the wavelength — it follows that short wavelengths are more energetic<span> than long wavelengths.</span>
Answer:
a) 
b) 1657 €
Explanation:
Hola,
a) En este problema, vamos a considerar el millón de litros de agua anuales, ya que con ellos podemos calcular el calor requerido para dicho calentamiento, sabiendo que la densidad del agua es de 1 kg/L:
Luego, usamos la entalpía de combustión del metano para calcular su requerimiento en kilogramos, sabiendo que la energía ganada por el agua, es perdida por el metano:
b) En este caso, consideramos que a condiciones normales de 1 bar y 273 K, 1 metro cúbico de metano cuesta 0,45 €, con esto, calculamos las moles de metano a dichas condiciones:
Con ello, los kilogramos de metano que cuestan 0,45 €:
Luego, aplicamos la regla de tres:
0.715 kg ⇒ 0.45 €
2630 kg ⇒ X
X = (2630 kg x 0.45 €) / 0.715 kg
X = 1657 €
Regards.
