B - Atomic number. Dmitri Mendeleev organised the table according to atomic weight, however this caused problems with elements such as iodine and tellurium, Iodine has a higher mass, but a lower atomic number. And to make iodine in the same group as similar elements (halogens), Mendeleev had to break his own rules and put it before tellurium in the table. Moseley fixed this problem by ordering the elements according to atomic (proton) number.
Answer:
90.3 L
Explanation:
Given data:
Volume of water produced = 77.4 L
Volume of oxygen required = ?
Solution:
Chemical equation:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
It is known that,
1 mole = 22.414 L
There are 7 moles of oxygen = 7×22.414 = 156.9 L
There are 6 moles of water = 6×22.414 = 134.5 L
Now we will compare:
H₂O : O₂
134.5 : 156.9
77.4 : 156.9/134.5×77.4 =90.3 L
So for the production of 77.4 L water 90.3 L oxygen is required.
The second and first one but if it isn’t 2 choices then 1
The electron group arrangement of NO²⁻is trigonal planar. The molecular shape is bent, and the bond angle is 120°.
<h3>What is the molecular shape of a compound?</h3>
The molecular geometry of the compound shows the position of nuclei and the electron of the compound. It shows how the joining of electrons and nuclei makes the shape of the compound.
Like here, the shape of nitrite is bent with lone pair which is shown by Lewis's structure The bond angle will be the distance between the nuclei of the neighbor atoms.
Thus, the electron geometry arrangement of nitrite is trigonal planer with a bent shape and the bond angle will be 120°.
To learn more about molecular geometry, refer to the link:
brainly.com/question/7558603
#SPJ4
First, we have to remember the molarity formula:
![M=\text{ }\frac{moles\text{ of solute}}{L\text{ solution}}](https://tex.z-dn.net/?f=M%3D%5Ctext%7B%20%7D%5Cfrac%7Bmoles%5Ctext%7B%20of%20solute%7D%7D%7BL%5Ctext%7B%20solution%7D%7D)
Part 1:
In this case, our solute is sodium nitrate (NaNO3), and we have the mass dissolved in water, then we have to convert grams to moles. For that, we need the molecular weight:
![M.W_{NaNO_3}=\text{ 23+14+16*3= 85 g/mol}](https://tex.z-dn.net/?f=M.W_%7BNaNO_3%7D%3D%5Ctext%7B%2023%2B14%2B16%2A3%3D%2085%20g%2Fmol%7D)
Then, we calculate the moles present in the solution:
![3.976\text{ g NaNO}_3\text{ * }\frac{1\text{ mol}}{85\text{ g}}=\text{ 0.04678 mol NaNO}_3](https://tex.z-dn.net/?f=3.976%5Ctext%7B%20g%20NaNO%7D_3%5Ctext%7B%20%2A%20%7D%5Cfrac%7B1%5Ctext%7B%20mol%7D%7D%7B85%5Ctext%7B%20g%7D%7D%3D%5Ctext%7B%200.04678%20mol%20NaNO%7D_3)
Now, we have the necessary data to calculate the molarity (with the solution volume of 200 mL):
![M=\frac{0.04678\text{ mol}}{200\text{ mL*}\frac{1\text{ L}}{1000\text{ mL}}}=\text{ 0.2339 M}](https://tex.z-dn.net/?f=M%3D%5Cfrac%7B0.04678%5Ctext%7B%20mol%7D%7D%7B200%5Ctext%7B%20mL%2A%7D%5Cfrac%7B1%5Ctext%7B%20L%7D%7D%7B1000%5Ctext%7B%20mL%7D%7D%7D%3D%5Ctext%7B%200.2339%20M%7D)
The molarity of this solution equals 0.2339 M.
Part 2:
In this case, we have the same amount (in moles and mass) of sodium nitrate, but a different volume of solution, then we only have to change it:
![M=\text{ }\frac{0.04678\text{ mol}}{275\text{ mL *}\frac{1\text{ L}}{1000\text{ mL}}}=\text{ 0.1701 M}](https://tex.z-dn.net/?f=M%3D%5Ctext%7B%20%7D%5Cfrac%7B0.04678%5Ctext%7B%20mol%7D%7D%7B275%5Ctext%7B%20mL%20%2A%7D%5Cfrac%7B1%5Ctext%7B%20L%7D%7D%7B1000%5Ctext%7B%20mL%7D%7D%7D%3D%5Ctext%7B%200.1701%20M%7D)
So, the molarity of this solution is 0.1701 M.