<u>Answer:</u> The new concentration of lemonade is 3.90 M
<u>Explanation:</u>
To calculate the number of moles for given molarity, we use the equation:
.....(1)
Molarity of lemonade solution = 2.66 M
Volume of solution = 473 mL
Putting values in equation 1, we get:

Now, calculating the new concentration of lemonade by using equation 1:
Moles of lemonade = 1.26 moles
Volume of solution = (473 - 150) mL = 323 mL
Putting values in equation 1, we get:

Hence, the new concentration of lemonade is 3.90 M
A volumetric flask is used to contain a predetermined volume of substance and only measures that volume, for example 250 ml.
Conical flasks can be used to measure the volume of substances but the accuracy they provide is usually up to 10ml. Conical flasks are used in titrations, reactions where the liquid may boil, and reactions which involve stirring.
Pippettes are of two types, volumetric and graduated. Pippettes are used where high accuracy is required and volumetric pippettes come in as little as 1 ml. Pippettes are usually used in titrations.
Graduated cylinders come in a wide variety of sizes and their accuracy can be down to as much as 1 ml. They are used to contain liquids.
Answer:
Density = Mass / Volume. so, x = 90.5 g / 96 mL ... The Density would be 0.942 g/mL
Given parameters:
Mass of sucrose = 5g
Density of sucrose = 1.12g/mL
Percentage of sucrose per liter of cane juice = 12%
Unknown:
Volume of cane juice needed = ?
We need to establish the volume - density relationship. Density is the mass of a substance per unit volume.
Mathematically;
Density =
Now solve for the volume of sucrose;
1.12g/mL =
Volume =
= 4.46mL = 4.46 x 10⁻³L since 1000mL = 1L
Since 12% of 1 liter of cane juice is sucrose;
12% of x liter of cane juice = 4.46 x 10⁻³L
Volume of cane juice = 4.46 x 10⁻³ x
= 0.037L
Volume of cane juice is 0.037L
Vanillin is the common name for 4-hydroxy-3-methoxy-benzaldehyde.
See attached figure for the structure.
Vanillin have 3 functional groups:
1) aldehyde group: R-HC=O, in which the carbon is double bonded to oxygen
2) phenolic hydroxide group: R-OH, were the hydroxyl group is bounded to a carbon from the benzene ring
3) ether group: R-O-R, were hydrogen is bounded through sigma bonds to carbons
Now for the hybridization we have:
The carbon atoms involved in the benzene ring and the red carbon atom (from the aldehyde group) have a <u>sp²</u> hybridization because they are involved in double bonds.
The carbon atom from the methoxy group (R-O-CH₃) and the blue oxygen's have a <u>sp³</u> hybridization because they are involved only in single bonds.