Answer:
The alkaline hydrolysis of ester is known as saponification. When ester is heated with aqueous NaOH, sodium salt of acid and alcohol are formed.
Answer:
See the explanation
Explanation:
In this case, we have to keep in mind that in the monosubstituted product we only have to replace 1 hydrogen with another group. In this case, we are going to use the methyl group
.
In the axial position, we have a more steric hindrance because we have two hydrogens near to the
group. If we have <u>more steric hindrance</u> the molecule would be <u>more unstable</u>. In the equatorial positions, we don't <u>any interactions</u> because the
group is pointing out. If we don't have <u>any steric hindrance</u> the molecule will be <u>more stable</u>, that's why the molecule will <u>the equatorial position.</u>
See figure 1
I hope it helps!
Answer:
![\large \boxed{\text{D. 710 g}}](https://tex.z-dn.net/?f=%5Clarge%20%5Cboxed%7B%5Ctext%7BD.%20710%20g%7D%7D)
Explanation:
1. Calculate the molar mass of Na₂SO₄
![\begin{array}{ccc}\textbf{Atoms} &\textbf{M}_{\textbf{r}} & \textbf{Mass/u}\\\text{2Na} & 23 & 46\\\text{1S} & 32 & 32\\\text{4O}&16 & 64\\&\text{TOTAL =} & \mathbf{142}\\\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bccc%7D%5Ctextbf%7BAtoms%7D%20%26%5Ctextbf%7BM%7D_%7B%5Ctextbf%7Br%7D%7D%20%26%20%5Ctextbf%7BMass%2Fu%7D%5C%5C%5Ctext%7B2Na%7D%20%26%2023%20%26%2046%5C%5C%5Ctext%7B1S%7D%20%26%2032%20%26%2032%5C%5C%5Ctext%7B4O%7D%2616%20%26%2064%5C%5C%26%5Ctext%7BTOTAL%20%3D%7D%20%26%20%5Cmathbf%7B142%7D%5C%5C%5Cend%7Barray%7D)
The molar mass of Na₂SO₄ is 142 g/mol.
2. Calculate the moles of Na₂SO₄
![\text{Moles of Na$_{2}$SO}_{4} = \text{2.5 L solution} \times \dfrac{\text{2.0 mol Na$_{2}$SO}_{4}}{\text{1 L solution}} = \text{5.0 mol Na$_{2}$SO}_{4}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20Na%24_%7B2%7D%24SO%7D_%7B4%7D%20%3D%20%5Ctext%7B2.5%20L%20solution%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B2.0%20mol%20Na%24_%7B2%7D%24SO%7D_%7B4%7D%7D%7B%5Ctext%7B1%20L%20solution%7D%7D%20%3D%20%5Ctext%7B5.0%20mol%20Na%24_%7B2%7D%24SO%7D_%7B4%7D)
3. Calculate the mass of Na₂SO₄
![\text{Mass of Na$_{2}$SO}_{4} = \text{5.0 mol Na$_{2}$SO}_{4} \times \dfrac{\text{142 g Na$_{2}$SO}_{4}}{\text{1 mol Na$_{2}$SO}_{4}} = \text{710 g Na$_{2}$SO}_{4}\\\\\text{You need } \large \boxed{\textbf{710 g}} \text{ of Na$_{2}$SO}_{4}](https://tex.z-dn.net/?f=%5Ctext%7BMass%20of%20Na%24_%7B2%7D%24SO%7D_%7B4%7D%20%3D%20%5Ctext%7B5.0%20mol%20Na%24_%7B2%7D%24SO%7D_%7B4%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B142%20g%20Na%24_%7B2%7D%24SO%7D_%7B4%7D%7D%7B%5Ctext%7B1%20mol%20Na%24_%7B2%7D%24SO%7D_%7B4%7D%7D%20%3D%20%5Ctext%7B710%20g%20Na%24_%7B2%7D%24SO%7D_%7B4%7D%5C%5C%5C%5C%5Ctext%7BYou%20need%20%7D%20%5Clarge%20%5Cboxed%7B%5Ctextbf%7B710%20g%7D%7D%20%5Ctext%7B%20of%20Na%24_%7B2%7D%24SO%7D_%7B4%7D)
Answer:
![m_{Fe_2O_3}=549gFe_2O_3](https://tex.z-dn.net/?f=m_%7BFe_2O_3%7D%3D549gFe_2O_3)
Explanation:
Hello there!
In this case, according to the given chemical reaction for this problem about stoichiometry:
![4Fe+3O_2\rightarrow 2Fe_2O_3](https://tex.z-dn.net/?f=4Fe%2B3O_2%5Crightarrow%202Fe_2O_3)
Whereas there is a 3:2 mole ratio of oxygen (molar mass = 32.0 g/mol) to iron (III) oxide (molar mass = 159.69 g/mol) and therefore, the correct stoichiometric setup is:
![m_{Fe_2O_3}=165gO_2*\frac{1molO_2}{32.00gO_2}*\frac{2molFe_2O_3}{3molO_2} *\frac{159.69gFe_2O_3}{1molFe_2O_3} \\\\m_{Fe_2O_3}=549gFe_2O_3](https://tex.z-dn.net/?f=m_%7BFe_2O_3%7D%3D165gO_2%2A%5Cfrac%7B1molO_2%7D%7B32.00gO_2%7D%2A%5Cfrac%7B2molFe_2O_3%7D%7B3molO_2%7D%20%2A%5Cfrac%7B159.69gFe_2O_3%7D%7B1molFe_2O_3%7D%20%20%5C%5C%5C%5Cm_%7BFe_2O_3%7D%3D549gFe_2O_3)
Regards!
It has many different uses like being used to make the tube of a vacuum cleaner and also can be used to make oil in space crafts.