An oxide of nitrogen contains 30.45 mass % N, if the molar mass is 90± 5 g/mol the molecular formula is N₂O₄.
<h3>What is molar mass?</h3>
The molar mass of a chemical compound is determined by dividing its mass by the quantity of that compound, expressed as the number of moles in the sample, measured in moles. A substance's molar mass is one of its properties. The compound's molar mass is an average over numerous samples, which frequently have different masses because of isotopes.
<h3>How to find the molecular formula?</h3>
The whole-number multiple is defined as follows.
Whole-number multiple = 
The empirical formula mass is shown below.
Mw of empirical formula = Mw of N+ 2 x (Mw of O)
= 14.01 g/mol + 2 x (16.00 g/mol)
= 46.01 g/mol
With the given molar mass or the molecular formula mass, we can get the whole-number multiple for the compound.
Whole-number multiple =
≈ 2
Multiplying the subscripts of NO2 by 2, the molecular formula is N(1x2)O(2x2)= N2O4.
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Lar mass of Ca<span> = 40.08 </span>grams/mole 77.4 g Ca<span> * ( 1 </span>mole Ca<span>/ 40.08 ... n = m / M 1mol </span>Ca<span>weights 40 gmol-1 n = 77,4 / 40 = 1.93 </span>mol<span>.</span>
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:
= 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>
Answer:As you can see from the very different numbers in the boxes above, your age changes (sometimes quite a lot) ... Mercury is the closest planet to the Sun and so has a smaller orbit path – it takes just 88 Earth days
Explanation:
Answer:
See the images below
Step-by-step explanation:
To draw a dot diagram of an atom, you locate the element in the Periodic Table and figure out how many valence electrons it has. Then you distribute the electrons as dots around the atom,
a. Silicon.
Si is in Group 14, so it has four valence electrons.
b. Xenon
Xenon is in Group 18, so it has eight valence electrons. We group them as four pairs around the xenon atom.
c. Calcium
Calcium is in Group 2, so it has two valence electrons. They are in a single subshell, so we write them as a pair on the calcium atom.
d. Water
Oxygen is in Group 16, so it has six valence electrons. The hydrogen atoms each contribute one electron, so there are eight valence electrons.
Chemists often use a dash to represent a pair of electrons in a bond.