3. The figure below show a bar of negligible weight pivot at point A attached to the
1 answer:
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Explanation:
The given data is as follows.
Length of beam, (L) = 5.50 m
Weight of the beam, () = 332 N
Weight of the Suki, () = 505 N
After crossing the left support of the beam by the suki then at some overhang distance the beam starts o tip. And, this is the maximum distance we need to calculate. Therefore, at the left support we will set up the moment and equate it to zero.
= 0
x =
=
= 0.986 m
Hence, the suki can come (2 - 0.986) m = 1.014 from the end before the beam begins to tip.
Thus, we can conclude that suki can come 1.014 m close to the end before the beam begins to tip.
Answer:
The volume of the final solution, V = 0.0305L
Explanation:
Number of moles = Concentration * volume
Concentration of HA = 1.00 * 10⁻⁴M
Volume of HA = 1000mL = 1 L
Number of moles of HA = 1.00 * 10⁻⁴ * 1
Number of moles of HA = 1.00 * 10⁻⁴ mols
Equation of reaction:
HA → H⁺ + A⁻
If 1 mol of HA produces 1 mol of H⁺ and A⁻, 1.00 * 10⁻⁴ mol of HA will produce 1.00 * 10⁻⁴ mol of H⁺ and A⁻.
Since only 16% dissociation occurs = 0.16
Number of moles of H⁺ produced = 0.16 * 1.00 * 10⁻⁴
Number of moles of H⁺ produced = 1.6 * 10⁻⁵mols
Number of moles of A⁻ produced = 0.16 * 1.00 * 10⁻⁴
Number of moles of A⁻ produced = 1.6 * 10⁻⁵mols
Since 16% of HA dissociated into H⁺ and A⁻, 84% of HA is left
Number of mols of HA left = 0.84 * 1.00 * 10⁻⁴
Number of mols of HA left = 8.4 * 10⁻⁵mols
Concentration = num of moles/volume
Let the volume of the final solution be V
Conc of HA = 8.4 * 10⁻⁵/V
Conc of H⁺ = 1.6 * 10⁻⁵/V
Conc of A⁻ = 1.6 * 10⁻⁵/V
To calculate the dissociation constant