A little boy is playing with a plastic water bottle. He blows air into the bottle to make a whistling sound. Then, he wedges a plastic ball into the bottle to make a rattle. But the ball is too big—it gets stuck in the mouth of the bottle! The little boy panics and runs to his big brother for help.

Luckily, his older brother just studied the principle of Boyle's law. Based on what he learned, how will he get the ball out of the bottle?

by placing the bottle in the freezer to decrease the mass of the air inside the bottle

by pushing down on the bottle to decrease the temperature of the air in the bottle

by placing the bottle in the freezer to increase the air pressure inside the bottle

by pushing down on the bottle to increase the air pressure inside the bottle?

Explanation:

You will need to heat the bottom of the bottle to make gas build up than the ball will pop out.

4 0

4 years ago

What is the formula antimony trichloride

Zielflug [23.3K]

Answer and hopefully this helps.

5 0

3 years ago

How long must a constant current of 50.0 A be passed through an electrolytic cell containing aqueous Cu2+ ions to produce 5.00 m

frozen [14]

Answer:

D) 5.36 hours

Explanation:

According to mole concept:

1 mole of an atom contains $6.022\times 10^{23}$ number of particles.

We know that:

Charge on 1 electron = $1.6\times 10^{-19}C$

Also, copper will produce 2 electrons. So, out of 5 moles of copper, 10 moles of electrons will be produced.

So,

Charge on 10 mole of electrons = $10\times 1.6\times 10^{-19}\times 6.022\times 10^{23}=9.6352\times 10^5C$

To calculate the time required, we use the equation:

$I=\frac{q}{t}$

where,

I = current passed = 50.0 A

q = total charge = $9.6352\times 10^5C$

t = time required = ?

Putting values in above equation, we get:

$50.0A=\frac{9.6352\times 10^5C}{t}\\\\t=\frac{9.6352\times 10^5C}{50.0A}=19270.4s$

Converting this into hours, we use the conversion factor:

1 hr = 3600 seconds

So, $19270.4s\times \frac{1hr}{3600s}=5.36hr$

Hence, the amount of time needed is 5.36 hrs.

8 0

3 years ago

A chemistry graduate student is given of a pyridine solution. Pyridine is a weak base with . What mass of should the student dis

Molodets [167]

Answer:

34.1g of C₅H₅NHCl the student need to dissolve to the solution.

Explanation:

Full question is:

A chemistry graduate student is given 500.mL of a 0.20M pyridine C5H5N solution. Pyridine is a weak base with =Kb×1.7x10−9 . What mass of C5H5NHCl should the student dissolve in the C5H5N solution to turn it into a buffer with pH =4.76 ?

Using H-H expression for weak bases, it is possible to find pH of a buffer thus:

pOH = pKb + log [BH⁺] / [B]

<em>Where pKb is -log Kb = 8.77, </em>[BH⁺] concentration of C₅H₅NHCl and [B] concentration of C₅H₅N (It is possible to take the moles of both compounds and not its concentration.

As pH the student wants is 4.76, pOH is:

pOH = 14 - pH = 14 - 4.76 = 9.24

Replacing:

9.24 = 8.77 + log [C₅H₅NHCl] / [C₅H₅N]

Moles of C₅H₅N are:

0.500L × (0.20mol / L) = 0.10mol C₅H₅N

Replacing again:

9.24 = 8.77 + log [C₅H₅NHCl] / [0.10mol]

2.9512 = [C₅H₅NHCl] / [0.10mol]

0.29512 moles = [C₅H₅NHCl].

As molar mass of C₅H₅NHCl is 115.56g/mol, mass of 0.29512 moles are:

0.29512 moles C₅H₅NHCl × (115.56g / mol) =

<h3>34.1g of C₅H₅NHCl the student need to dissolve to the solution.</h3>

3 0

4 years ago