Answer:
Acceleration
Explanation:
Acceleration =
liters of the 40% solution and
liters of the 25% solution must be mixed to obtain a 28% solution of H2SO4.
(Round to the nearest tenth, if necessary.)
Answer:
16 liters of the solution with 40% concentration must be mixed with 62 liters of the solution with 25% concentration in order to obtain 78 liters, 25% concentration solution.
Explanation:
Let the required volume of solution 1 be represented by x.
The required volume of solution 2 would then be 78-x.
The number of moles of solution 1 that would be required = 0.4x
The number of moles of solution 2 that would be required = 0.25(78-x)
The number of moles of the final mixture = 78 x 0.28 = 21.84
moles of solution 1 + moles of solution 2 = moles of final mixture
0.4x + 0.25(78 - x) = 21.84
0.4x + 19.5 - 0.25x = 21.84
0.4x - 0.25x = 21.84 - 19.5
0.15x = 2.34
x = 15.6 liters
To the nearest tenth = 16 liters
Liters of 40% solution needed = 16 liters
Liters of 25% solution needed = 78 - 16 = 62 liters.
Hence, 16 liters of the solution with 40% concentration must be mixed with 62 liters of the solution with 25% concentration in order to obtain 78 liters, 25% concentration solution.
B. 100 mL volumetric flask
C. 10 mL volumetric pipet
D. 25 mL volumetric pipet
E. 10 mL graduated pipet
F. 25 mL graduated pipet
Answer:
C
Explanation:
The volumetric pipet is designed for a specific volume of solution. In this case, a 10 mL of volumetric pipet is required to take exactly 10 mL of the solution 1.00 M and add into a volumetric flask of 100 mL. Then water or solvent(s) can be added to reach the ring of the volumetric flask. This volumetric flask is also a laboratory equipment for precis dilutions beside the volumetric pipet
Answer:
(0,653±0,002) M of HNO₃
Explanation:
The reaction of standarization of HNO₃ with Na₂CO₃ is:
2 HNO₃ + Na₂CO₃ ⇒ 2 Na⁺ + H₂O + CO₂ + 2NO₃⁻
To obtain molarity of HNO₃ we need to know both moles and volume of this acid. The volume is (27,71±0,05) mL and to calculate the moles it is necessary to obtain the Na₂CO₃ moles and then convert these to HNO₃ moles, thus:
0,9585 g of Na₂CO₃ × ( 1 mole / 105,988 g) =
9,043×10⁻³ mol Na₂CO₃ × ( 2 moles of HNO₃ / 1 mole of Na₂CO₃) = 1,809×10⁻² moles of HNO₃
Molarity is moles divide liters, thus, molarity of HNO₃ is:
1,809×10⁻² moles / 0,02771 L = 0,6527 M of HNO₃
The absolute uncertainty of multiplication is the sum of relative uncertainty, thus:
ΔM = 0,6527M× (0,0007/0,9585 + 0,001/105,988 + 0,05/27,71) =
0,6527 M× 2,54×10⁻³ = 1,7×10⁻³ M
Thus, molarity of HNO₃ solution and its absolute uncertainty is:
(0,653±0,002) M of HNO₃
I hope it helps!
Answer:
The concentration of HCl is - 0.2675 M
Explanation:
At equivalence point,
Moles of = Moles of NaOH
Considering:-
Given that:
So,
The concentration of HCl is - 0.2675 M
Answer:
An apple.
A person.
A table.
Air.
Water.
A computer.
Paper.
Iron.
Hope this helps you
Answer:
your boddy is made of mater and a clock too it is still a mater of time.
Explanation:
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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