The answer should be <span>enteropeptidase
</span>
When the reaction equation is:
CaSO3(s) → CaO(s) + SO2(g)
we can see that the molar ratio between CaSO3 & SO2 is 1:1 so, we need to find first the moles SO2.
to get the moles of SO2 we are going to use the ideal gas equation:
PV = nRT
when P is the pressure = 1.1 atm
and V is the volume = 14.5 L
n is the moles' number (which we need to calculate)
R ideal gas constant = 0.0821
and T is the temperature in Kelvin = 12.5 + 273 = 285.5 K
so, by substitution:
1.1 * 14.5 L = n * 0.0821 * 285.5
∴ n = 1.1 * 14.5 / (0.0821*285.5)
= 0.68 moles SO2
∴ moles CaSO3 = 0.68 moles
so we can easily get the mass of CaSO3:
when mass = moles * molar mass
and we know that the molar mass of CaSO3= 40 + 32 + 16 * 3 = 120 g/mol
∴ mass = 0.68 moles* 120 g/mol = 81.6 g
Answer:
The answer to your question is 50 moles of O₂
Explanation:
Balanced Chemical reactions
1.- N₂(g) + 3H₂ (g) ⇒ 2NH₃ (g)
2.- 4NH₃ (g) + 5O₂(g) ⇒ 4NO (g) + 6H₂O (l)
moles of N₂(g) = 20 moles
moles of O₂(g) = ?
Process
1.- Calculate the moles of NH₃
1 mol of N₂ ------------- 2 moles of NH₃
20 moles of N₂ --------- x
x = (20 x 2) / 1
x = 40 moles of NH₃
2.- Calculate the moles of O₂
4 moles of NH₃ -------------- 5 O₂
40 moles of NH₃ ------------ x
x = (40 x 5) / 4
x = 200 / 4
x = 50 moles of O₂
<u>Answer:</u> The value of 
for the reaction at 690 K is 0.05
<u>Explanation:</u>
We are given:
Initial pressure of 
= 1.0 atm
Total pressure at equilibrium = 1.2 atm
The chemical equation for the decomposition of phosgene follows:
Initial: 1 - -
At eqllm: 1-x x x
We are given:
Total pressure at equilibrium = [(1 - x) + x+ x]
So, the equation becomes:
![[(1 - x) + x+ x]=1.2\\\\x=0.2atm](https://tex.z-dn.net/?f=%5B%281%20-%20x%29%20%2B%20x%2B%20x%5D%3D1.2%5C%5C%5C%5Cx%3D0.2atm)
The expression for for above equation follows:
Putting values in above equation, we get:
Hence, the value of for the reaction at 690 K is 0.05
The noble gas notation is the short or abbreviated form of the electron configuration.
It means that you use the symbol of the previous noble gas as part of the electron configuration of an element.
The gas noble previous to antimony is Kr, so you do not use Xe to write the electron configuration of Sb.
The gas noble previous to radium is Rn, so you do not use Xe to wirte the electron configuration of Ra.
The gas noble previous to uranium is Rn, so you do not use Xe to write the electron configuration of U.
The gas noble previous to cesium is Xe, so you use Xe to write the noble notation for Sb. This is it: Cs: [Xe] 6s.
Answer: cesium
The ga
