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Art [367]
3 years ago
15

A scientist conducted an experiment to investigate the effect of radon cellular mutations. Why is it important for another scien

tist to conduct the same scientific investigation independently? (1 point)
to verify the data to make the results stronger and more reliable
to ensure the popularity of the results so they will be accepted by the scientific community
to keep the scientific community competitive and motivated to continue finding new facts
To prove that the results are always different
Chemistry
1 answer:
sveticcg [70]3 years ago
3 0
To verify the data to make the results stronger and more reliable.
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Prepare a project report on various samples of
san4es73 [151]
<h2>Project Reports</h2>

<h3>A) Mixture</h3>

This refers to the material that is made when different substances mix up physically and causes a reaction.

You can make on the project of Mixture -

  • 1) Alloys
  • 2)Colloids
  • 3) Suspension
  • 4) Solution

<h3>B) Compound</h3>

This refers to the chemical bond that holds different atoms tightly

You can make on the project of Compound -

  • 1)Water
  • 2)Methane
  • 3)Carbon Dioxide
  • 4) Sulfuric Acid

<h3>C) Elements</h3>

Based on the fact that the atom is the smallest indivisible part of an element, elements like phosphorous cannot be further broken down.

You can make on the project of Elements -

  • 1) Mercury
  • 2) Iron
  • 3) Copper
  • 4)Carbon

Read more about mixtures and compounds here:

brainly.com/question/491220

#SPJ1

8 0
1 year ago
Propane has a normal boiling point of -42.04 °C and a heat of vaporization of 24.54 kJ/mole. What is the vapor pressure of propa
Mademuasel [1]

Answer : The vapor pressure of propane at 25.0^oC is 17.73 atm.

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of propane at 25.0^oC = ?

P_2 = vapor pressure of propane at normal boiling point = 1 atm

T_1 = temperature of propane = 25.0^oC=273+25.0=298.0K

T_2 = normal boiling point of propane = -42.04^oC=230.96K

\Delta H_{vap} = heat of vaporization = 24.54 kJ/mole = 24540 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

\ln (\frac{1atm}{P_1})=\frac{24540J/mole}{8.314J/K.mole}\times (\frac{1}{298.0K}-\frac{1}{230.96K})

P_1=17.73atm

Hence, the vapor pressure of propane at 25.0^oC is 17.73 atm.

3 0
3 years ago
If a solution is given to you, how will you determine whether it is acid,base or salt​
12345 [234]

Answer:

If your lab has litmus paper, you can use it to determine your solution's pH. When you place a drop of a solution on the litmus paper, the paper changes color based on the pH of the solution. Once the color changes, you can compare it to the color chart on the paper's package to find the pH.

Explanation:

A solution's pH will be a number between 0 and 14. A solution with a pH of 7 is classified as neutral. If the pH is lower than 7, the solution is acidic. When pH is higher than 7, the solution is basic. These numbers describe the concentration of hydrogen ions in the solution and increase on a negative logarithmic scale.

For example, If Solution A has a pH of 3 and Solution B has a pH of 1, then Solution B has 100 times as many hydrogen ions than A and is therefore 100 times more acidic.

5 0
3 years ago
Helpppppppppp!!!!!!!!!​
professor190 [17]

Answer:

a

Explanation:

the solution is extremely acidic and will dissolve the zinc rod

3 0
3 years ago
If 8.500 g CH is burned and the heat produced from the burning is added to 5691 g of water at 21 °C, what is the final
mojhsa [17]

The final temperature = 36 °C

<h3>Further explanation</h3>

The balanced combustion reaction for C₆H₆

2C₆H₆(l)+15O₂(g)⇒ 12CO₂(g)+6H₂O(l)  +6542 kJ

MW C₆H₆ : 78.11 g/mol

mol C₆H₆ :

\tt \dfrac{8.5}{78.11}=0.109

Heat released for 2 mol C₆H₆ =6542 kJ, so for 1 mol

\tt \dfrac{0.109}{2}\times 6542=356.539~kJ/mol

Heat transferred to water :

Q=m.c.ΔT

\tt 356.539=5.691~kg\times 4.18~kj/kg^oC\times (t_2-21)\\\\t_2-21=15\rightarrow t_2=36^oC

3 0
2 years ago
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