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Lesechka [4]
3 years ago
13

What is the highest percentage of water vapor that could be in the air hurry help

Chemistry
1 answer:
Inessa05 [86]3 years ago
4 0

Answer:

4%

Explanation:

At 30 °C (86 °F), for example, a volume of air can contain up to 4 percent water vapor.

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Sedimentary and metamorphic :)


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8 0
2 years ago
Some fruits and vegetables are preserved by pickling them. Nandini got confused
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2 years ago
Can the pH scale be utilized for all acids (Arrhenius, Bronsted-Lowry, and Lewis)? Give examples of substances from each definit
CaHeK987 [17]

Answer:

No

Explanation:

The pH scale is a scale graduated from 0-14 which shows the degree of acidity of alkalinity of a substance. The pH scale is graduated in such a way that 0-6.9 indicates acidity, 7.0 indicate a neutral substance, while a pH of 8-14 indicates alkalinity respectively.

There are three main definitions of acids/bases

- Arrhenius definition

-Brownstead-Lowry definition

-Lewis definition

Arrhenius explains acids as any substance that produces hydrogen ions as its only positive ion in solution while a base produces hydroxide ions as its only negative ion in solution. The pH scale is based on corresponding values of pH derived from aqueous solutions of these substances.

However, not all acids/bases produces hydrogen or hydroxide ions in solution. Brownstead-Lowry definition of acids and Lewis definition of acids could be extended to nonaqueous media where the pH can not be measured as there are no hydrogen or hydroxide ions present in the solution.

This implies that pH measurement may not apply to acids/bases in the all the categories of acids/bases hence it can not be utilized for all acids and bases.

Arrhenius - sodium carbonate

Brownstead-Lowry - concentrated HF

Lewis acid - AlCl3

6 0
3 years ago
Ayuda es urgente!! el tronco de madera y las cenizas pesan lo mismo ?​
skad [1K]

Answer:

Spanish

Explanation:

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3 0
2 years ago
A 20g piece of lead absorbs 566 joules of heat and its temperature changes from 35 oC to 195 oC. Calculate the specific heat.
Alja [10]

Answer:

  • <u>Question 1: 0.2J/(gºC)</u>
  • <u>Question 2: 6,000J</u>
  • <u>Question 3: 300J</u>
  • <u>Question 4: 80g</u>
  • <u>Question 5: 74ºC</u>
  • <u>Question 6: 50g</u>

<u></u>

Explanation:

Question 1.<em> A 20g piece of lead absorbs 566 joules of heat and its temperature changes from 35º oC to 195º C. Calculate the specific heat.</em>

<em />

The thermal energy equation is:

  • Q = m × C × ΔT

<em />

Substitute and solve for C:

  • 566J = 20g × C × (195ºC - 35ºC)
  • C = 566J / (20g × 160ºC)
  • C = 0.177 J/(gºC) ≈ 0.2J/(gºC)

<em />

You must round to one significant figure because one factor has one significant figure).

<em />

<em />

Qustion 2.<em> 40g of water is heat at 40ºC and the temperature rise to 75ºC. What is the amount of heat needed for the temperature to rise? (specific heat of water is 4.184 J/gºC)</em>

<em />

Use the thermal energy equation again:

  • Q = m × C × ΔT

<em />

Substitute and compute:

  • Q = 40g × 4.184 J/gºC × (75ºC - 40ºC)
  • Q = 5,857.6J

Round to one significant figure: 6,000J

<em />

Question 3. <em>Graphite has a mass of 50g and a specific heat of 0.420 J/gºC. If graphite is cooled from 50ºC to 35ºC, how much energy was lost?</em>

  • Q = m × C × ΔT
  • Q = 50g × 0.420J/gºC × (35ºC - 50ºC)
  • Q = 315J

Round to one significant figure (because 50g has one significant figure)

  • Q = 300J

<em />

Question 4.<em> </em><em>Iron has a specific heat of 0.712 J/gºC. A piece of iron absorbs 3000J of energy and undergoes a temperature change totaling 50ºC, What is the mass of iron?</em>

<em />

  • Q = m × C × ΔT

Solve for m:

  • m = Q / (C × ΔT)

Substitute and compute:

  • m = 3,000J / (0.712J/gºC × 50ºC)
  • m = 84.26 g ≈ 80 g (rounded to one significant figure, because the factor 3,000J has one significant figure).

Question 5. <em>If 400g of an unknown solution at 70ºC loses 7500 J of heat, what is the final temperature of the unknown solution. The unknown solution has a specific heat of 4.184 J/gºC.</em>

<em />

  • Q = m × C × ΔT

<em />

Q is negative, since it is released.

Substitute and solve for T:

  • - 7,500J = 400g × 4.184J/gºC × (T - 70ºC)

  • T = - 7500J / 400g × 4.184J/gºC) + 70ºC

  • T = 74ºC

<em />

If you round to one significant figure you cannot tell the temperature difference, thus leave two significant figures.

<em />

Question 6. <em>How many grams of water would require 9500J of heat to raise the temperature from 50ºC to 100ºC</em>

  • Q = m × C × ΔT

Subsitute:

  • 9,500J = m × 4.184J/gºC × (100ºC - 50ºC)

Solve for m and compute:

  • m = 9,500J / (4.184J/gºC × 50ºC)

  • m = 45g

Since the temperatures indicate one singificant figure, the mass should be rounded to one significant figure:

  • m = 50g.
8 0
3 years ago
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