The answer to this question would be: too low
Molar mass would be determined by the number of mol and the mass of the object. Mass wouldn't be influenced by the temperature, but number of mol is. Using ideal gas formula of PV=nRT you can conlude that the amount of mol(n) is inversely related to the temperature (T).
If the temperature is higher than it supposed to be, then the amount of mol would be lower than it supposed to be.
Answer:
Mg₁₂ = 1s² 2s² 2p⁶ 3s²
Explanation:
Abbreviated and unabbreviated electronic configuration:
The abbreviated electronic configuration uses the noble gas configuration i.e complete electronic shells. For example, the atomic number of neon is ten and magnesium is twelve. The abbreviated electronic configuration of magnesium is written by using the neon abbreviation in following way:
The electronic configuration of neon is given below:
Ne₁₀ = 1s² 2s² 2p⁶
The abbreviated electronic configuration of magnesium:
Mg₁₂ = [Ne] 3s²
While the unabbreviated electronic configuration is written without using noble gas electronic configuration.
Unabbreviated electronic configuration of magnesium:
Mg₁₂ = 1s² 2s² 2p⁶ 3s²
The particles are atoms, his theory is the atomic theory
Moles of H₂ are needed to produce 9.33 moles of NH₃ : 13.995
<h3>Further explanation</h3>
A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.
The reaction coefficient in a chemical equation shows the mole ratio of the reactants and products
Reaction for the synthesis of ammonia :
N₂+3H₂⇒2NH₃
moles of NH₃ = 9.33
From equation, mol ratio of H₂ : NH₃ = 3 : 2, so mol H₂ :
Answer:
The concentration of I at equilibrium = 3.3166×10⁻² M
Explanation:
For the equilibrium reaction,
I₂ (g) ⇄ 2I (g)
The expression for Kc for the reaction is:
![K_c=\frac {\left[I_{Equilibrium} \right]^2}{\left[I_2_{Equilibrium} \right]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%20%7B%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%7D%7B%5Cleft%5BI_2_%7BEquilibrium%7D%20%5Cright%5D%7D)
Given:
![\left[I_2_{Equilibrium} \right]](https://tex.z-dn.net/?f=%5Cleft%5BI_2_%7BEquilibrium%7D%20%5Cright%5D)
= 0.10 M
Kc = 0.011
Applying in the above formula to find the equilibrium concentration of I as:
![0.011=\frac {\left[I_{Equilibrium} \right]^2}{0.10}](https://tex.z-dn.net/?f=0.011%3D%5Cfrac%20%7B%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%7D%7B0.10%7D)
So,
![\left[I_{Equilibrium} \right]^2=0.011\times 0.10](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%3D0.011%5Ctimes%200.10)
![\left[I_{Equilibrium} \right]^2=0.0011](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%3D0.0011)
![\left[I_{Equilibrium} \right]=3.3166\times 10^{-2}\ M](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%3D3.3166%5Ctimes%2010%5E%7B-2%7D%5C%20M)
<u>Thus, The concentration of I at equilibrium = 3.3166×10⁻² M</u>
