Answer:
0.93g/mL
Explanation:
Given parameters
Mass of wood = 3.16g
Initial volume of water = 20mL
Volume of water + wood = 23.4mL
Unknown:
Density of the wood = ?
Solution:
Density is the mass per volume of a substance.
It is mathematically expressed as;
Density =
Volume of wood = 23.4mL - 20mL = 3.4mL
So,
Density =
= 0.93g/mL
80Nm I'm pretty sure this is the answer if I'm wrong I'm sorry but I'm positive it the answer
Answer:
The foam board has a larger gravitational force but it also has a very large air resistance force. The net (total) force on the foam board will give it a smaller acceleration than paper.
Explanation:
The air molecules get knocked away and take some of the kinetic energy away from the object, which makes the object slow down
Answer:
Look at the properties of Oxygen and Silicon - the two most abundant elements in the Earth's crust - by clicking on their symbols on the Periodic Table.
Explanation:
Answer:
Isopropyl propionate
Explanation:
1. Information from formula
The formula is C₆H₁₂O₂. A six-carbon alkane would have the formula C₆H₁₄. The deficiency of two H atoms indicates the presence of either a ring or a double bond.
2. Information from the spectrum
(a) Triplet-quartet
A 3H triplet and a 2H quartet is the classic pattern for a CH₃CH₂- (ethyl) group
(b) Septet-doublet
A 1H septet and a 6H doublet is the classic pattern for a -CH(CH₃)₂ (isopropyl) group
(c) The rest of the molecule
The ethyl and isopropyl groups together add up to C₇H₁₂.
The rest of the molecule must have the formula CO₂ and one unit of unsaturation. That must be a C=O group.
The compound is either
CH₃CH₂-COO-CH(CH₃)₂ or (CH₃)₂CH-COO-CH₂CH₃.
(d) Well, which is it?
The O atom of the ester function should have the greatest effect on the H atom on the adjacent carbon atom.
The CH of an isopropyl is normally at 1.7. The adjacent O atom should pull it down perhaps 3.2 units to 4.9.
The CH₂ of an ethyl group is normally at 1.2. The adjacent O atom should pull it down to about 4.4.
We see a signal at 5.0 but none near 4.4. The compound is isopropyl propionate.
3. Summary
My peak assignments are shown in the diagram below.