Answer:
a) H₂SO₄ + Ba(OH)₂ ⇄ BaSO₄(s) + 2 H₂O(l)
b) H₂SO₄, H⁺, HSO₄⁻, SO₄²⁻. H₂O, H⁺, OH⁻.
c) H⁺, HSO₄⁻, SO₄²⁻
d) As the titration takes place, reaction [1] proceeds to the right. The conductivity of the solution decreases because the amount of H⁺, HSO₄⁻, SO₄²⁻ decreases. The formed solid is barium sulfate BaSO₄. Since BaSO₄ is very insoluble, the main responsible for conductivity are still H⁺, HSO₄⁻ and SO₄²⁻,
e) At the equivalence point equivalent amounts of H₂SO₄ and Ba(OH)₂ react. The conducting species are Ba²⁺, SO₄²⁻, H⁺ and OH⁻.
f) After the equivalence point there is an excess of Ba(OH)₂. The ions Ba²⁺ and OH⁻ are responsible for the increase in the conductivity, being the major conducting species.
Explanation:
a) Write an equation (including states of matter) for the reaction between H₂SO₄ and Ba(OH)₂.
The balanced equation is:
H₂SO₄ + Ba(OH)₂ ⇄ BaSO₄(s) + 2 H₂O(l) [1]
b) At the very start of the titration, before any titrant has been added to the beaker, what is present in the solution?
In the beginning there is H₂SO₄ and the ions that come from its dissociation reactions: H⁺, HSO₄⁻, SO₄²⁻. There is also H₂O and a very small amount of H⁺ and OH⁻ coming from its ionization.
H₂SO₄(aq) ⇄ H⁺(aq) + HSO₄⁻(aq)
HSO₄⁻(aq) ⇄ H⁺(aq) + SO₄²⁻(aq)
H₂O(l) ⇄ H⁺(aq) + OH⁻(aq)
c) What is the conducting species in this initial solution?
The main responsible for conductivity are the ions coming from H₂SO₄: H⁺, HSO₄⁻, SO₄²⁻.
d) Describe what happens as titrant is added to the beaker. Why does the conductivity of the solution decrease? What is the identity of the solid formed? What is the conducting species present in the beaker?
As the titration takes place, reaction [1] proceeds to the right. The conductivity of the solution decreases because the amount of H⁺, HSO₄⁻, SO₄²⁻ decreases. The formed solid is barium sulfate BaSO₄. Since BaSO₄ is very insoluble, the main responsible for conductivity are still H⁺, HSO₄⁻ and SO₄²⁻,
e) What happens when the conductivity value reaches its minimum value (which is designated as the equivalence point for this type of titration)? What is the conducting species in the beaker?
At the equivalence point equivalent amounts of H₂SO₄ and Ba(OH)₂ react. Only BaSO₄ and H₂O are present, and since they are weak electrolytes, there is a small amount of ions to conduct electricity. The conducting species are Ba²⁺ and SO₄²⁻ coming from BaSO₄ and H⁺ and OH⁻ coming from H₂O.
f) Describe what happens at additional titrant is added past the equivalence point. Why does the conductivity of the solution increase? What is the conducting species present in the beaker?
After the equivalence point there is an excess of Ba(OH)₂. The ions Ba²⁺ and OH⁻ are responsible for the increase in the conductivity, being the major conducting species.
The chemical reaction between H2SO4 and Ba(OH)2 forms BaSO4 and water, reducing conductivity by reducing the number of free ions. Beyond the equivalence point, the conductivity increases due to the dissociated ions from the excess Ba(OH)2 in the solution.
Firstly, the equation representing the reaction between sulfuric acid (H2SO4) and barium hydroxide (Ba(OH)2) is:
Ba(OH)2 (aq) + H2SO4 (aq) → BaSO4 (s) + 2H2O (l)
In the beginning, the solution only contains H2SO4 with its dissociated ions serving as the conducting species. As titrant (Ba(OH)2) is added, they react to form BaSO4, a solid precipitate reducing the number of free ions in the solution, thus decreasing conductivity. At the equivalence point, all H2SO4 has reacted, and conductivity reaches its minimum as there are lesser free ions for conduction. If additional titrant is added past the equivalence point, conductivity increases due to excess Ba(OH)2's dissociated ions that increase ion concentration in solution.
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Answer:
THE NEW FREEZING POINT IS -4.196 °C
Explanation:
ΔTf = 1 Kf m
molarity of MgCl2:
Molar mass = (24 + 35.5 *2) g/mol
molar mass = 95 g/mol
7.15 g of MgCl2 in 100 g of water
7.15 g = 100 g
(7.15 * 100 / 1000) = 1000 g or 1 L or 1 dm3
= 0.715 g /dm3
Molarity in mol/dm3 = molarity in g/dm3 / molar mass
= 0.715 g /dm3 / 95 g/mol
m = 0.00752 mol/ dm3
So therefore:
ΔTf = i Kf m
1 = 3 (1 Mg and 2 Cl)
Kf = 1.86 °C/m
M = 0.752 moles
So we have:
ΔTf = 3 * 1.86 * 0.752
ΔTf = 4.196 °C
The new freezing point therefore will be 0 °C - 4.196 °C which is equals to - 4.196 °C
B. a change in
c. not related to
D. only an increase in
Answer:
option B is correct
Explanation:
impulse is a change in momentum
Answer: C2H4S5
Explanation:
Since the total mass is 5.000g
Mass of sulphur = 5.000-(0.6357+0.1070)
Mass of sulphur = 4.2573g
Using Empirical relation
C= 0.6357 H= 0.1070 S= 4.2573
Divide through by their molar mass to obtain the smallest ratio
C= 0.6357/12 H=0.1070/1 S=4.2573/32
C= 0.053 H= 0.1070 S= 0.133
Divide through by the smallest ratio (0.053)
C=0.053/0.053 H=0.1070/0.053 S=0.133/0.053
C=1 H=2 S=2.5
1:2:2.5 ,multiply through by 2 ,to obtain whole numbera
2:4:5
Therefore the empirical formula is C2H4S5. Thus only gives the ratio
Molecular formula is the chemical formula .
(Empirical formula) n = molecular formula
(C2H4S5)n = molar mass
[(12×2) + ( 1×4) +(32×5)]n = 188.4
188n=188.4
n= 1
Molecular formula = (C2H4S5)×1
Therefore the chemical formula of
lenthionine is C2H4S5
Answer:
0,12 μmol/L of MgF₂
Explanation:
Preparation of solutions is a common work in chemist's life.
In this porblem says that you measure 0,00598 μmol of MgF₂ in 50,0 mL of water and you must calculate concentration in μmol/L
You have 0,00598 μmol but not Liters.
To obtain liters you sholud convert mL to L, knowing 1000mL are 1 L, thus:
50,0 mL (1L/1000mL) = 0,05 L of water.
Thus, concentration in μmol/L is:
0,00598 μmol / 0,05 L = 0,12 μmol/L -The problem request answer with two significant digits-
I hope it helps!
Answer: The lone pair of electron on nitrogen is accommodated in a 2p orbital hence it interacts with the pi system in aniline.
Explanation:
Aniline is less basic than amines. This is because, the nitrogen atom in aniline is not purely sp3 hybridized. Its actual hybrization state is closer to sp2 because the lone pair on nitrogen is accommodated in a 2p orbital. The nitrogen atim in aniline is planar and its
lonely pair interacts with the pi electron system of aniline. This makes the lone pair unavailable for protonation hence aniline is less basic than amines.
The calculated hybridization of the N-atom lone pair in aniline is affected by electron-electron repulsions, resonance, and steric effects from substituents on the aromatic ring.
The calculated hybridization of the N-atom lone pair in aniline is different from the predicted sp3 hybridization due to a combination of factors:
Overall, these factors contribute to the observed hybridization of the N-atom lone pair in aniline, deviating from the predicted tetrahedral electron geometry.
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The minimum space cushion defines the required amount of space which vehicles should maintain in other to afford them the time and space to gain control in emergency scenarios. Hence, the minimum space cushion required in the scenario is 4 seconds.
In cases of mishaps or accidents, the space cushion might just afford other cars the space to maneuver their way to safety rather than being caught up in the collison or accident.
The required space cushion in most scenario is usually between 2 - 5 seconds, with additional space afforded depending on the length and type of the vehicle.
Therefore, to ensure safety, the required minimum spacecushion to be left when driving being a cargo van traveling at a speed of 25mph is 4 seconds.
Learn more :brainly.com/question/24535523
In order to ensure safety while driving a cargo van at 25 MPH, the driver should maintain a space cushion of about 3-4 van lengths, which accounts for speed, reaction time, and distance needed to apply brakes and avert a collision.
The subject of your question revolves around optimal space cushion required for safety while driving a cargo van at the speed of 25 MPH, adhering to REPS (Reference point, Eye lead time, Posting and Scanning) and Checks (Check side mirrors and Rearview mirror every 5-8 seconds). This question falls under the domain of physics, as it involves velocity (speed of the vehicle), distance (space cushion), and time.
As a general rule of thumb, for every 10 miles per hour, a driver should ideally stay approximately one car length away from the car in front of them. Therefore, at 25 MPH, the driver should maintain a distance of at least 2.5 car lengths. In the case of a cargo van, which is typically larger than a regular car, this distance should ideally be increased to 3-4 van lengths to ensure safe stopping distance and reaction time in case of any sudden stoppage by the vehicle ahead.
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