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2 years ago

How does cold milk cool hot Cocoa

1 answer:

5 0

They will eventually reach equilibrium. But, before that, the thermal energy particles of the COLD milk will slow down the hot cocoa.

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A 4.0 L balloon has a pressure of 406 kPa. When the pressure increases to 2,030 kPa, what is the volume? *

elena-14-01-66 [18.8K]

The new volume when pressure increases to 2,030 kPa is 0.8L

BOYLE'S LAW:

The new volume of a gas can be calculated using Boyle's law equation:

P1V1 = P2V2

Where;

P1 = initial pressure (kPa)
P2 = final pressure (kPa)
V1 = initial volume (L)
V2 = final volume (L)

According to this question, a 4.0 L balloon has a pressure of 406 kPa. When the pressure increases to 2,030 kPa, the volume is calculated as:

406 × 4 = 2030 × V2

1624 = 2030V2

V2 = 1624 ÷ 2030

V2 = 0.8L

Therefore, the new volume when pressure increases to 2,030 kPa is 0.8L.

Learn more about Boyle's law calculations at: brainly.com/question/1437490?referrer=searchResults

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2 years ago

Which of these is an indication that a chemical reaction has occurred?

nevsk [136]

Formation Of A Gas.

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3 years ago

What planet is characterized by rings of ice particles?

iris [78.8K]

Saturn has the most extensive ring system in our solar system & most of the particles floating around in the ring are ice...

Answer=Saturn

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2 years ago

Read 2 more answers

25 mL of 0.10 M aqueous acetic acid is titrated with 0.10 M NaOH(aq). What is the pH after 30 mL of NaOH have been added?

weeeeeb [17]

Answer:

pH = 11.95≈12

Explanation:

Remember the reaction among aqueous acetic acid ( $CH_3COOH$ ) and aqueous sodium hydroxide (NaOH)

$CH_3COOH + NaOH -> CH_3COONa + H_2O$

First step. Need to know how much moles of the substances are present

$0.1 \frac{mol NaOH}{L} * 0.030 L = 0.003 mol NaOH[tex]0.1 \frac{mol CH_3COOH}{L} * 0.025 L= 0.0025 mol CH_3COOHSecond step. Know wich substance is in excess.0.0025 mol CH_3COOH * [tex]1 mol NaOH/ 1 mol CH_3COOH$ = 0.0025 mol NaOH

0.003 mol NaOH * $1 mol CH_3COOH$ / 1 mol NaOH = 0.003 mol CH_3COOH[/tex]

NaOH is in excess. Now, how much?

0.003 mole NaOH - 0.0025 mole NaOH = 0.0005 mole NaOH

Then, that amount in excess would be responsable for the pH.

Third step. Know the pH

Remember that pH= -log[H+]

According to the dissociation of water equilibrium

Kw=[H+]*[OH-]= 10^(-14)

The dissociation of NaOH is

NaOH -> $Na^{+} + OH^{-}$

Now, concentration of OH^{-}[/tex] would be given for the excess of NaOH.

[OH-]= 0.0005 mole / 0.055 L = 0.00909 M

Careful: we have to use the total volumen

Les us to calculate pH

$pH= -log [H+]\\pH= -log \frac{K_w}{[OH-]} \\pH= 11.95$

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2 years ago

What happens if I overshoot the endpoint in titration?

xxTIMURxx [149]

Your solution in the conical flask will be acidic since you will have gone beyond the amount of acid that should completely neutralise the base completely. thax

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2 years ago