D as it only has a negative effect when being used and it is small at that anyway.
Answer:- 171 g
Solution:- It asks to calculate the grams of sucrose required to make 1 L of 0.5 Molar solution of it.
We know that molarity is moles of solute per liter of solution.
If molarity and volume is given then, moles of solute is molarity times volume in liters.
moles of solute = molarity* liters of solution
moles of solute = 0.5*1 = 0.5 moles
To convert the moles to grams we multiply the moles by molar mass.
Molar mass of sucrose = 12(12) + 22(1) + 11(16)
= 144 + 22 + 176
= 342 grams per mol
grams of sucrose required = moles * molar mass
grams of sucrose required = 0.5*342 = 171 g
So, 171 g of sucrose are required to make 1 L of 0.5 molar solution.
Moles of Oxygen= 2.8075 moles
<h3>Further explanation</h3>
Given
29.2 grams of acetylene
Required
moles of Oxygen
Solution
Reaction(Combustion of Acetylene) :
2 C₂H₂ (g) + 5 O₂ (g) ⇒ 4CO₂ (g) + 2H₂O (g)
Mol of Acetylene :
= mass : MW Acetylene
= 29.2 g : 26 g/mol
= 1.123
From equation, mol ratio of Acetylene(C₂H₂) : O₂ = 2 : 5, so mol O₂ :
= 5/2 x mol C₂H₂
= 5/2 x 1.123
= 2.8075 moles
Answer:
![CH_3 CH_2 CH_2 CH_2 CH_2 CH_2 CHO + 2[Ag(NH_3)_2]+ 3 H_2 O](https://tex.z-dn.net/?f=CH_3%20CH_2%20CH_2%20CH_2%20CH_2%20CH_2%20CHO%20%2B%202%5BAg%28NH_3%29_2%5D%2B%20%203%20H_2%20O)
= 
Explanation:
Tollens test is carried out to perceive difference between aldehydes and ketones on the basis of their capability to oxidized easily.
when Tollens react with aldehyde (heptanal) , a silver mirror is form on inner side of container.
The reaction between tollens and heptanal is given as
![CH_3 CH_2 CH_2 CH_2 CH_2 CH_2 CHO + 2[Ag(NH_3)_2]+ 3 H_2 O](https://tex.z-dn.net/?f=CH_3%20CH_2%20CH_2%20CH_2%20CH_2%20CH_2%20CHO%20%2B%202%5BAg%28NH_3%29_2%5D%2B%203%20H_2%20O)
= 
