"Endothermic" is the way among the following choices given that you would <span>classify the following chemical reaction, in terms of energy. The correct option among all the options that are given in the question is the second option or option "B". I hope that the answer has come to your help.</span>
Answer:
612 K
Explanation:
From the question given above, the following data were obtained:
Initial temperature (T₁) = 306 K
Initial pressure (P₁) = 150 kPa
Final pressure (P₂) = 300 kPa
Volume = 4 L = constant
Final temperature (T₂) =?
Since the volume is constant, the final (i.e the new) temperature of the gas can be obtained as follow:
P₁ / T₁ = P₂ / T₂
150 / 306 = 300 / T₂
Cross multiply
150 × T₂ = 306 × 300
150 × T₂ = 91800
Divide both side by 150
T₂ = 91800 / 150
T₂ = 612 K
Thus, the new temperature of the gas is 612 K
A gas with a vapor density greater than that of air, would be most effectively displaced out off a vessel by ventilation.
The two following principles determine the type of ventilation: Considering the impact of the contaminant's vapour density and either positive or negative pressure is applied.
Consider a vertical tank that is filled with methane gas. Methane would leak out if we opened the top hatch since its vapour density is far lower than that of air. A second opening could be built at the bottom to greatly increase the process' efficiency.
A faster atmospheric turnover would follow from air being pulled in via the bottom while the methane was vented out the top. The rate of natural ventilation will increase with the difference in vapour density. Numerous gases that require ventilation are either present in fairly low concentrations or have vapor densities close to one.
Answer:
Average atomic mass = 17.5 amu.
Explanation:
Given data:
X-17 isotope = atomic mass17.2 amu, abundance:78.99%
X-18isotope = atomic mass 18.1 amu, abundance 10.00%
X-19isotope = atomic mass:19.1 amu, abundance: 11.01%
Average atomic mass of X = ?
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass) / 100
Average atomic mass = (78.99×17.2)+(10.00×18.1) +(11.01+ 19.1) /100
Average atomic mass = 1358.628 + 181 +210.291 / 100
Average atomic mass = 1749.919 / 100
Average atomic mass = 17.5 amu.
For this problem we use the wave equation. It is expressed as the speed (c) is equal to the product of frequency (f) and wavelength (v).
c = v x f
We know the wavelength of the an red light which is 6.5 x 10^-7 m. Now, we solve for the wavelength of the unknown wave to see the relation between the two waves.
2.998 X 10^8 = 5.3 X 10^15 X v
v = 2.998 X 10^8 / (5.3 X 10^15) = 5.657 X 10^-8 m
Therefore, the wavelength of the unknown wave is less than the wavelength of the red light.