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3 years ago

What will you observe when you open a bottle of concentrated hydrochloric acid?

Chemistry

1 answer:

Soloha48 [4]3 years ago

7 0

Answer:

Strong affinity of HCl gas for moisture in air results in forming of droplets of liquid solution which appears like a cloudy smoke.

Explanation:

#followformore

Hope this helps :)

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A salt bridge: Question 3 options: provide a pathway through which the electrons travel from the cathode to the anode. provide a

Vesnalui [34]

Answer:

The correct option is: provide a source of counterions to prevent the build-up of charge at both the cathode to the anode.

Explanation:

A salt bridge is a U-shaped glass tube that is used in a voltaic cell or galvanic cell to connect the oxidation and reduction half-cells and complete the electric circuit.

It allows the ions to pass through it, thus preventing the accumulation of charge on the anode and cathode as the chemical reaction proceeds.

Therefore, the correct option is: provide a source of counterions to prevent the build-up of charge at both the cathode to the anode.

5 0

4 years ago

You walk out to your car on a cold January day, and you notice that your tires appear to be flat. You take your car to the shop

Naddik [55]

The answer to this question would be the rate of change

4 0

3 years ago

Convert 55 m3 to liters

luda_lava [24]

Explanation:

<h3>hope it help you thanku byy </h3>

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4 0

2 years ago

Someone please help!!!!

Blababa [14]

I'm not 100% sure on this one, but I think it's C.

Sorry, I'm not really good with Chemistry.

Hope this helps :)

8 0

3 years ago

Benzoyl peroxide, the substance most widely used against acne, has a half-life of 9.8 × 103 days when refrigerated. How long wil

prohojiy [21]

Answer:

$7.3\times 10^2\ days$

Explanation:

Given that:

Half life = $9.8\times 10^3$ days

$t_{1/2}=\frac{\ln2}{k}$

Where, k is rate constant

So,

$k=\frac{\ln2}{t_{1/2}}$

$k=\frac{\ln2}{9.8\times 10^3}\ days^{-1}$

The rate constant, k = 0.00007 days⁻¹

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given:

5 % is lost which means that 0.05 of $[A_0]$ is decomposed. So,

$\frac {[A_t]}{[A_0]}$ = 1 - 0.05 = 0.95

t = ?

$\frac {[A_t]}{[A_0]}=e^{-k\times t}$

$0.95=e^{-0.00007\times t}$

t = 732.76 days = $7.3\times 10^2\ days$

7 0

3 years ago