Answer:
True
Explanation:
Significant digits are numbers that helps to present the precision of measurements calculations.
Numbers that do not contribute to the precision of a reading should not be counted as significant.
There are rules of assigning significant numbers:
- Leading or trailing zeros are insignificant and should only be counted as a place holder.
- All non-zero digits are significant
- Zeroes between non-zero digits are significant.
- Leading zeros in a decimal are significant before the number.
- All the numbers in a scientific notation are significant.
Answer:
The mole fraction of NaOH in an aqueous solution that contain 22.9% NaOH by mass=0.882
Explanation:
We are given that
Aqueous solution that contains 22.9% NaOH by mass means
22.9 g NaOH in 100 g solution.
Mass of NaOH(WB)=22.9 g
Mass of water =100-22.9=77.1
Na=23
O=16
H=1.01
Molar mass of NaOH(MB)=23+16+1.01=40.01
Number of moles =
Using the formula
Number of moles of NaOH
Molar mass of water=16+2(1.01)=18.02g
Number of moles of water
Now, mole fraction of NaOH
=
=0.882
Hence, the mole fraction of NaOH in an aqueous solution that contain 22.9% NaOH by mass=0.882
Answer:
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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.
Answer:
The beaker holds 307.94 mL
Explanation:
As we know that the volume that beaker hold is the volume of water that occupied by it.
For this first we have to find mass of the water in the beaker
This can be calculated by the subtraction of beaker's weight from the weight of beaker and water.
weight of water (m) = total weight - weight of beaker
Empty weight of beaker = 25.91 g
Weight of beaker with water = 333.85 g
Weight of water = 333.85 - 25.91 = 307.94 g
Density of water = 1 g/mL
We have
Mass = Volume x density
307.94 = Volume x 1
Volume = 307.94 mL
The beaker holds 307.94 mL