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3 years ago

Which compound is most likely a powerful and dangerous acid? H2S HBr Li2O LiBR

Physics

2 answers:

sattari [20]3 years ago

6 0

HBr is the most powerful and dangerous acid .

Explanation:

A strong acid is one that instantly disunites or grants its protons in suspension. HBr is a strong acid. HBr is powerful than HCl or HF because the overlapping of a 1s-orbital and a 4p-orbital is surprisingly small, thus the binding is weak so splitting is very easy..

BARSIC [14]3 years ago

5 0

Answer: HBr is a powerful and dangerous acid

Explanation:

HBr gives hydronium ions ( $H^+$ ) in its aqueous solution and presence of large amount of ions in the solution makes it corrosive in nature. This is because when these ions present in the solution comes in contact with any of the objects they tends to neutralize their charge and due to this action of ions we observe corrosive nature of an acid.

Chemical equation for HBr dissociating into ions is given as:

$HBr(aq)\rightarrow H^+(aq)+Br^-(aq)$

Hence ,HBr is a powerful and dangerous acid. Where as $H_2S,Li_2O\text{ and }LiBr$ are gas , basic oxide and salt respectively.

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3 years ago

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3 years ago

Maggie walks to a friends house which is exactly 1500 meters due South. It takes Maggie 45 minutes for the walk and Maggie has t

Sindrei [870]

Answer:

<em>Explanation below</em>

Explanation:

<u>Speed vs Velocity </u>

These are two similar physical concepts. They only differ in the fact that the velocity is vectorial, i.e. having magnitude and direction, and the speed is scalar, just the magnitude regardless of the direction. They are strongly related to the concepts of displacement and distance, which are the vectorial and scalar versions of the space traveled by a moving object. The velocity can be computed as

$\displaystyle \vec v=\frac{\vec r}{t}$

Where $\vec r$ is the position vector and t is the time. The speed is

$\displaystyle v=\frac{d}{t}$

To compute $\vec r$ , we only need to know the initial and final positions and subtract them. To compute d, we need to add all the distances traveled by the object, regardless of their directions.

Maggie walks to a friend's house, located 1500 meters from her place. The initial position is 0 and the final position is 1500 m. The displacement is

$\vec r=1500\ m \text{ to the south}$

and the velocity is

$\displaystyle \vec v=\frac{1500}{45}=33.33\ m/s\text{ to the south}$

Now, we know Maggie had to make three different turns of direction to finally get there. This means her distance is more than 1500 m. Let's say she walked 500 m in all the turns, then the distance is

$d=1500+500=2000\ m$

If she took the same time to reach her destiny, she would have to run faster, because her average speed is

$\displaystyle v=\frac{2000}{45}=44.44\ m/s$

5 0

3 years ago

The 26-kg sphere c is released from rest when θ = 0∘ and the tension in the spring is f = 100 n

aleksandr82 [10.1K]

You are given the mass of a sphere that is 26 kg sphere and it is released from rest when θ = 0°. You are also given the force of the spring that is F = 100 N. You are asked to find the tension of the spring. Imagine that the sphere is connected to a spring. The spring exerts a tension and the spring exerts gravitational pull. This will follow the second law of newton.

T - F = ma
T = ma + F
T = 26kg (9.81m/s²) + 100 N
T = 355.06 N

5 0

3 years ago

Each driver has mass 79.0 kg. Including the masses of the drivers, the total masses of the vehicles are 800 kg for the car and 4

Mademuasel [1]

Answer:

Force exerted on the car driver by the seatbelt = 8139.4 N = 8.14 kN

Force exerted on the truck driver by the seatbelt = 1628.2 N = 1.63 kN

It is evident that the driver of the smaller vehicle has it worse. The car driver is in way more danger in this perfectly inelastic head-on collision with a bigger vehicle (the truck).

Explanation:

First of, we calculate the velocity of the vehicles after collision using the law of conservation of Momentum

Momentum before collision = Momentum after collision

Since the collision of the two vehicles was described as a head-on collision, for the sake of consistent convention, we will take the direction of the velocity of the bigger vehicle (the truck) as the positive direction and the direction of the car's velocity automatically is the negative direction.

Velocity of the truck before collision = 6.80 m/s

Velocity of the car before collision = -6.80 m/s

Let the velocity of the inelastic unit of vehicles after collision be v

Momentum before collision = (4000)(6.80) + (800)(-6.80) = 27200 - 5440 = 21,760 kgm/s

Momentum after collision = (4000 + 800)(v) = (4800v) kgm/s

Momentum before collision = Momentum after collision

21760 = 4800v

v = (21760/4800)

v = 4.533 m/s (in the direction of the big vehicle (the truck)

So, we then apply Newton's second law of motion which explains that the magnitude change in momentum is equal to the magnitude of impulse.

|Impulse| = |Change in momentum|

But Impulse = (Force exerted on each driver by the seatbelt) × (collision time) = (F×t)

Change in momentum = (Momentum after collision) - (Momentum before collision)

So, for the driver of the truck

Initial velocity = 6.80 m/s (the driver moves with the velocity of the truck)

Final velocity = 4.533 m/s

Change in momentum of the truck driver = (79)(6.80) - (79)(4.533) = 179.1 kgm/s

(F×t) = 179.1

F × 0.110 = 179.1

F = (179.1/0.11)

F = 1628.2 N = 1.63 kN

So, for the driver of the car

Initial velocity = -6.80 m/s (the driver moves with the velocity of the car)

Final velocity = 4.533 m/s

Change in momentum of the car driver = (79)(-6.80) - (79)(4.533) = -895.3 kgm/s

(F×t) = |-895.3|

F × 0.110 = 895.3

F = (895.3/0.11)

F = 8139.4 N = 8.14 kN

Hope this Helps!!!

3 0

3 years ago