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dem82 [27]
3 years ago
15

Air pollution may have a greater effect on health than on atmospheric temperature.

Chemistry
1 answer:
worty [1.4K]3 years ago
8 0
<span>Yes it is true, Because we are breath in air in the form of smoke and very small practical plagued our lungs, Which are very danger for our body. Computerized tomography imaging studies on the bodies of ancient mummies have detected evidence of pneumonia, emphysema and many types of asthma problem occur in our body .</span>
You might be interested in
A sample of quartz is put into a calorimeter (see sketch at right) that contains of water. The quartz sample starts off at and t
pashok25 [27]

Answer:

0.71 J/g°C

Explanation:

Here is the complete question

thermometer A 51.9 g sample of quartz is put into a calorimeter (see sketch at right) that contains 300.0 g of water. The quartz sample starts off at 97.8 °C and the temperature of the water starts off at 17.0 °C. When the temperature of the water stops changing it's 19.3 °C. The pressure remains constant at 1 atm. insulated container water sample Calculate the specific heat capacity of quartz according to this experiment. Be sure your answer is rounded to 2 significant digits. a calorimeter g °C

Solution

Since the temperature of the water increases from 17.0 °C to 19.3 °C, it means that it loses heat. Also, the final temperature of the quartz equals the final temperature of the water 19.3 °C. Since the quartz temperature decreases from 97.8 °C to 19.3 °C it loses heat.

So, heat lost by quartz, Q = heat gained by water, Q'

-Q = Q'

-mc(θ₂ - θ₁) = m'c'(θ₂ - θ₃) where m = mass of quartz = 51.9 g, c = specific heat capacity of quartz, θ₁ = initial temperature of quartz = 97.8 °C, θ₂ = final temperature of quartz = 19.3 °C, m' = mass of water = 300 g, c = specific heat capacity of water = 4.2 J/g °C , θ₃ = initial temperature of water = 17.0 °C, θ₂ = final temperature of water = 19.3 °C

Making c subject of the formula, we have

c = -m'c'(θ₂ - θ₃)/m(θ₂ - θ₁)

Substituting the values of the variables into the equation, we have

c = -300 g × 4.2 J/g °C(19.3 °C - 17.0 °C)/51.9 g(19.3 °C - 97.8 °C)

c = -1260 J/°C(2.3 °C)/51.9 g(-78.5 °C)

c = -2898 J/-4074.15 g°C

c = 0.711 J/g°C

c ≅ 0.71 J/g°C to 2 significant digits

5 0
3 years ago
A gas with a volume of 4.0 L at a pressure of 2.02 atm is allowed to expand to a volume of 12.0
Maurinko [17]
<h3>Answer:</h3>

P₂ = 0.67 atm

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right  

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality<u> </u>

<u>Chemistry</u>

<u>Gas Laws</u>

Boyle's Law: P₁V₁ = P₂V₂

  • P₁ is pressure 1
  • V₁ is volume 1
  • P₂ is pressure 2
  • V₂ is volume 2
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] P₁ = 2.02 atm

[Given] V₁ = 4.0 L

[Given] V₂ = 12.0 L

[Solve] P₂

<u>Step 2: Solve</u>

  1. Substitute in variables [Boyle's Law]:                                                              (2.02 atm)(4.0 L) = P₂(12.0 L)
  2. [Pressure] Multiply:                                                                                           8.08 atm · L = P₂(12.0 L)
  3. [Pressure] [Division Property of Equality] Isolate unknown:                          0.673333 atm = P₂
  4. [Pressure] Rewrite:                                                                                           P₂ = 0.673333 atm

<u>Step 3: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs as our smallest.</em>

0.673333 atm ≈ 0.67 atm

4 0
3 years ago
Copper is formed when aluminum reacts with copper (II) sulfate in a single-replacement reaction. How many moles of copper can be
serg [7]

Answer:

The answer to your question is the limiting reactant is CuSO₄ and 0.975 moles of Cu were obtained

Explanation:

moles of Copper = ?

mass of Aluminum = 29 g

mass of CuSO₄ = 156 g

Limiting reactant = ?

Balanced Chemical reaction

                  3 CuSO₄   +  2 Al   ⇒   3 Cu   +  Al₂(SO₄)₃

Calculate the moles of reactants

CuSO₄ = 64 + 32 + (16 x 4) = 160g

Al = 27 g

                160 g of CuSO₄  ----------------- 1 mol

                156 g                   -----------------  x

                      x = (156 x 1) / 160

                      x = 0.975 moles

               27 g of Al -------------------------- 1 mol

               29 g of Al -------------------------- x

                x = (29 x 1)/27

                x = 1.07 moles

Calculate proportions to find the limiting reactant

Theoretical     3 moles CuSO₄/2 moles Al = 1.5 moles

Experimental  0.975 moles CuSO₄/1.07 moles = 0.91

The experimental proportion was lower than the theoretical proportion that means that the limiting reactant is CuSO₄.      

                    3 moles of CuSO₄ ------------------ 3 moles of Cu

                 0.975 moles of CuSO₄ ---------------  x

                         x = (0.975 x 3)/3

                        x = 0.975 moles of Cu were obtained.        

3 0
2 years ago
Read 2 more answers
Approximately what percentage of a human's DNA is used for coding Information for the synthesis of proteins?
skelet666 [1.2K]
I’m pretty sure it is 25 % :))
4 0
3 years ago
What ion is oxygen likely to form?
Leto [7]
It's most natural state has a charge of -2. So, a negative ion with two more electrons than is normal.
7 0
2 years ago
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