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vaieri [72.5K]
3 years ago
6

If constant heat is added to ice water, what changes will take place and at what temperatures?

Chemistry
1 answer:
EleoNora [17]3 years ago
6 0

Answer:

Explanation:

When you heat ice, its temperature rises, but as soon as the ice starts to melt, the temperature stays constant until all the ice has melted. This happens because all the heat energy goes into breaking the bonds of the ice's crystal lattice structure.   :)

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A. density
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g If 50.0 mL of a 0.75 M acetic acid solution is titrated with 1.0 M sodium hydroxide, what is the pH after 10.0 mL of NaOH have
V125BC [204]

Answer:

pH = 2.66

Explanation:

  • Acetic Acid + NaOH → Sodium Acetate + H₂O

First we <u>calculate the number of moles of each reactant</u>, using the <em>given volumes and concentrations</em>:

  • 0.75 M Acetic acid * 50.0 mL = 37.5 mmol acetic acid
  • 1.0 M NaOH * 10.0 mL = 10 mmol NaOH

We<u> calculate how many acetic acid moles remain after the reaction</u>:

  • 37.5 mmol - 10 mmol = 27.5 mmol acetic acid

We now <u>calculate the molar concentration of acetic acid after the reaction</u>:

27.5 mmol / (50.0 mL + 10.0 mL) = 0.458 M

Then we <u>calculate [H⁺]</u>, using the<em> following formula for weak acid solutions</em>:

  • [H⁺] = \sqrt{C*Ka}=\sqrt{0.458M*1.76x10^{-5}}
  • [H⁺] = 0.0028

Finally we <u>calculate the pH</u>:

  • pH = -log[H⁺]
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3 years ago
How will the concentration of H+ and OH− ions change when a substance with a pH 3.2 is added to water? Both H+ and OH− will incr
Lunna [17]
Also water H2O is made of H+ and OH- ions. so when an acidic substance is added to water the concentration of H+ ions increase.

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I'm stuck on this assignment, +20 PTS and BRAINLIEST for step by step:
dolphi86 [110]

Answer:

2. 2.74 L

3. 488 K

4. 7.47 L

5. 38.6 L

6. 2.85 mol

7. 319 K

8. 3.43 kPa

Explanation:

Generally speaking, you want to convert units to SI units, but in this case, we are working with ratios.  This makes up for using the units that wouldn't appropriate elsewhere.

2.  Use the equation P₁V₁ = P₂V₂.  Solve for V₂.  

(3.05 L)(870 kPa) = (969 kPa)(V₂)  

V₂ = 2.74 L.  

3.  Use the equation V₁/T₁ = V₂/T₂. Solve for T₂.  

(3.32 L)/(360 K) = (4.50 L)/(T₂)  

T₂ = 488 K.

4.  Do the same as above, but for V₂.  

(5.10 L)/(-56°C) = V₂/(-82°C)  

V₂ = 7.47 L

5.  Use the equation V₁/n₁ = V₂/n₂.  Solve for V₂.  

(37.2 L)/(0.750 mol) = (V₂)/(0.778 mol)  

V₂ = 38.6 L

6.  Do the same as above, but for n₂.

(86.0 L)/(2.65 mol) = (92.5 L)(n₂)  

n₂ = 2.85 mol

7.  Use the equation P₁/T₁ = P₂/T₂.  Solve for T₂.  

(3.00 atm)/(390 K) = (2.45 atm)/(T₂)  

T₂ = 319 K

8.  Do the same as above, but for P₂.  

In this specific case, however you will need to convert units.  Since both temperatures don't have the same sign, the ratio won't come out right.  Convert to Kelvin.  Add 273.15 to the temperature in Celsius to convert to Kelvin -12.3°C = 260.85 K  25°C = 298.15 K.

(3.00 kPa)/(260.85 K) = P₂/(298.15 K)

P₂ = 3.43 kPa

There is a lot in here... If you are confused about something, let me know!

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How is rock candy formed
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Answer:

A supersaturated solution made out of sugar and oil is crystallized on a surface suitable for crystal nucleation such as a stick.

I hope this helps.

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