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4 years ago

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

Physics

2 answers:

sattari [20]4 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]4 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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Alex777 [14]

Answer:

10.75 A

The current is in opposite direction since it causes a repulsion force between the wires

Explanation:

Force per unit length on the wires = 4.30×10^−5 N/m

distance between wires = 2.6 cm = 0.026 m

current through one wire = 0.52 A

current on the other wire = ?

Recall that the force per unit length of two wires conducting and lying parallel and close to each other is given as

$F/l$ = $\frac{u_{0}I_{1} I_{2} }{2\pi r }$

where $F/l$ is the force per unit length on the wires

$u_{0}$ = permeability of vacuum = 4π × 10^−7 T-m/A

$I_{1}$ = current on the first wire = 0.520 A

$I_{2}$ = current on the other wire = ?

r = the distance between the two wire = 0.026 m

substituting the value into the equation, we have

4.30×10^−5 = $\frac{4\pi *10^{-7}*0.520*I_{2} }{2\pi *0.026}$ = $\frac{ 2*10^{-7}*0.520*I_{2} }{0.026}$

4.30×10^−5 = 4 x 10^-6 $I_{2}$

$I_{2}$ = (4.30×10^-5)/(4 x 10^-6) = 10.75 A

The current is in opposite direction since it causes a repulsion force between the wires.

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