Answer: The empirical formula for the given compound is 
Explanation : Given,
Mass of O = 0.370 g
Mass of N = 0.130 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.
Moles of Oxygen = 
Moles of Nitrogen = 
Step 2: Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00928 moles.
For Oxygen = 
For Nitrogen = 
Step 3: Taking the mole ratio as their subscripts.
The ratio of O : N = 2 : 1
Hence, the empirical formula for the given compound is 
Answer:
0.44 moles
Explanation:
Given that :
A mixture of water and graphite is heated to 600 K in a 1 L container. When the system comes to equilibrium it contains 0.17 mol of H2, 0.17 mol of CO, 0.74 mol of H2O, and some graphite.
The equilibrium constant ![K_c= \dfrac{[CO][H_2]}{[H_2O]}](https://tex.z-dn.net/?f=K_c%3D%20%20%5Cdfrac%7B%5BCO%5D%5BH_2%5D%7D%7B%5BH_2O%5D%7D)
The equilibrium constant 
The equilibrium constant 
Some O2 is added to the system and a spark is applied so that the H2 reacts completely with the O2.
The equation for the reaction is :

Total mole of water now = 0.74+0.17
Total mole of water now = 0.91 moles
Again:
![K_c= \dfrac{[CO][H_2]}{[H_2O]}](https://tex.z-dn.net/?f=K_c%3D%20%20%5Cdfrac%7B%5BCO%5D%5BH_2%5D%7D%7B%5BH_2O%5D%7D)
![0.03905 = \dfrac{[0.17+x][x]}{[0.91 -x]}](https://tex.z-dn.net/?f=0.03905%20%3D%20%20%5Cdfrac%7B%5B0.17%2Bx%5D%5Bx%5D%7D%7B%5B0.91%20-x%5D%7D)
0.03905(0.91 -x) = (0.17 +x)(x)
0.0355355 - 0.03905x = 0.17x + x²
0.0355355 +0.13095
x -x²
x² - 0.13095
x - 0.0355355 = 0
By using quadratic formula
x = 0.265 or x = -0.134
Going by the value with the positive integer; x = 0.265 moles
Total moles of CO in the flask when the system returns to equilibrium is :
= 0.17 + x
= 0.17 + 0.265
= 0.435 moles
=0.44 moles (to two significant figures)
Answer:
The order will be:
CCH > CHCH₂ > CH₂CH₃> CH₃
Explanation:
According to Cahn-Ingold-Prelog system we rank the groups based on the atomic number of directly attached atom with the chiral carbon.
For example: between C and H, we rank Carbon first.
If the same atoms are attached for different groups then we prioritized based on the second element with highest atomic number.
For example:
Among CH₃ and C₂H₅, the priority will be given to C₂H₅.
If an atom is double or triple bonded to the directly attached atom then each pi bond is considered to be a new atom.
Hence CH=CH₂ means, that there are two carbons attached to CH carbon.
So the order based on above selection rules will be:
CCH > CHCH₂ > CH₂CH₃> CH₃
Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane
Answer:
sulfur
Explanation:
In oxygen family sulfur has yellow color and also having stinky smell. Thus given statements are about sulfur.
It is present in oxygen family.
It has six valance electrons.
Its atomic number is 16.
Its atomic weight is 32 amu.
The electronic configuration of sulfur is given below,
S₁₆ = 1s² 2s² 2p⁶ 3s² 3p⁴
We can see the valance shell is third shell and it have six electrons thus sulfur have six valance electrons. (3s² 3p⁴ )
Sulfur is used in vulcanisation process.
It is used in bleach and also as a preservative for many food.
it is used to making gun powder.