Answer:
- <u>The pointer will not be lined up with the zero mark; it will be above the zero mark.</u>
Explanation:
Althoug the list of statements is not provided, you can assure that the pointer will not be lined up with the zero mark, but above the zero mark.
A <em>triple beam balance</em> is very precise instrument, of common use in most labs, used to measure masses. It has three beams. Each beam works for different precisions and mass increments. One beam measures 1 g increments, other beam measures 10 g increments, and the other one 100 g increments.
If you place a mass of 250 grams on the measurement tray, then the weight will be balanced in the beams with 200 grams in the beam that reads increments of 100 g and 50 grams in the beam that reads increments of 10 g. When this is the situation, the weight is completely balanced, and the pointer will be lined up with the zero mark. But this is not the case.
When the rider on the 500 gram beam is set to the 200 gram mark, and the other riders are set to 0 grams, the weight of the mass (250 g) on the tray is greater than the force exerted by the rider on the beam (200 g) and so the tray will be below the level of balance, and the beam will be above the level of balance. The level of balance is indicated when the pointer is lined up with the zero mark.
Answer:
Conduction is the transfer of thermal energy through direct contact. Convection is the transfer of thermal energy through the movement of a liquid or gas. Radiation is the transfer of thermal energy through thermal emission.
Answer:
pH = 1.32
Explanation:
H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺
This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:
So first calculate the moles reacted and produced:
n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M
54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH
it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.
moles H₂M left = 0.074 - 0.015 = 0.059
moles HM⁻ produced = 0.015
Using the Henderson - Hasselbach equation to solve for pH:
ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325
Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.
For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.
Answer:
Explanation:
(a) Hybridization of orbitals
See the Lewis structure of propyne in the first diagram below.
C1 is directly bonded to two other atoms (H and C2) so it is sp hybridized.
C2 is directly bonded to two other atoms (C1 and C3) so it is sp hybridized.
C3 is directly bonded to four other atoms (C2 and 3 H) so it is sp³ hybridized.
(b) C2-C3 Sigma Bond
See the atomic and molecular orbitals in the second picture below.
C3 is using its hybrid atomic orbitals to form sigma molecular bonds. The C2-C3 sigma bond is formed by the overlap of the C2 sp atomic orbital with the C3 sp³ atomic orbital to make a σ(sp-sp³) molecular orbital.
(c) Bond angles
C1 and C2 are sp hybridized. Since the angle between sp orbitals is 180°, all atoms directly attached the C1 and C2 must be in a straight line. The C-C-C bond angle is 180°.
