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marissa [1.9K]
3 years ago
11

Why is observational evidence important in an experiment?

Chemistry
2 answers:
irga5000 [103]3 years ago
7 0

Answer:

Observational evidence is essential for investigating the way disease affects populations, the patterns and distribution of risk within them, and the emergence of trends in health and disease over time.

const2013 [10]3 years ago
3 0

Answer:

It tests a prediction It supports the results. It asks a testable question It predicts what will happen

Explanation:

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What is the definition of pendulum
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Answer:

weight hung from a fixed point so that it can swing freely backward and forward.

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3 years ago
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: Citric acid, H3C6H5O7, is a triprotic acid. Consider a buffer system comprising H2C6H5O7 - and HC6H5O7 2- ions. What is the ne
olchik [2.2K]

Answer:

The Net reaction is    

  1.    H_3C_6H_5O_7 + 0H^- ----> H_2C_6H_5O_7^- +H_2O \ \ \ \ \ \ \ Main \ Reaction
  2.    H_2C_6H_5O_7^ - + OH^- ----> H_C_6H_5O_7^{ 2-} + H_2O \ \ \ \ Add \ NaOH
  3.    HC_6H_5O_7 ^{2-} + OH^- ---> C_6H_5O_7^{3-} +H_2O \ \ \ \ \ Add \ NaOH  

Explanation:

From the Question we are told that the buffers are

                H_2C_6H_5O_7^ - and  HC_6H_5O_7 ^{ 2-}

When NaOH is added the Net ionic reaction would be

  1.    H_3C_6H_5O_7 + 0H^- ----> H_2C_6H_5O_7^- +H_2O \ \ \ \ \ \ \ Main \ Reaction
  2.    H_2C_6H_5O_7^ - + OH^- ----> H_C_6H_5O_7^{ 2-} + H_2O \ \ \ \ Add \ NaOH
  3.    HC_6H_5O_7 ^{2-} + OH^- ---> C_6H_5O_7^{3-} +H_2O \ \ \ \ \ Add \ NaOH  

             

8 0
3 years ago
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If velocity is positive, which would most likely yield a negative acceleration?
Nataliya [291]

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An initial velocity that is faster than a final velocity

Explanation:

Edg 2020

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3 years ago
While in science class a student measured a cube. She
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7cm

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3 years ago
A 110.0-mL sample of a solution that is 3.0×10−3 M in AgNO3 is mixed with a 230.0- mL sample of a solution that is 0.10 M in NaC
almond37 [142]

Answer:

[Ag⁺] = 0.0666M

Explanation:

For the addition of Ag⁺ and CN⁻, the (Ag(CN)₂⁻ is produced, thus:

Ag⁺ + 2CN⁻  ⇄  Ag(CN)₂⁻

Kf = 1x10²¹ = [Ag(CN)₂⁻] / [CN⁻]² [Ag⁺]

As initial concentrations of Ag⁺ and CN⁻ are:

[Ag⁺] = 0.110L × (3.0x10⁻³mol / L) = 3.3x10⁻⁴mol / (0.110L + 0.230L) = 9.7x10⁻⁴M

[CN⁻] = 0.230L × (0.1mol / L) = 0.023mol / (0.110L + 0.230L) = 0.0676M

The equilibrium concentrations of each compound are:

[CN⁻] = 9.7x10⁻⁴M - x

[Ag⁺] = 0.0676M - x

[Ag(CN)₂⁻] = x

<em>Where x is reaction coordinate</em>

Replacing in Kf formula:

1x10²¹ = [x] / [9.7x10⁻⁴M - x]² [0.0676M - x]

1x10²¹ = [x] / 6.36048×10⁻⁸ - 0.000132085 x + 0.06954 x² - x³

-1x10²¹x³ + 6.954x10¹⁹x² - 1.32085x10¹⁷ x + 6.36x10¹³ = x

-1x10²¹x³ + 6.954x10¹⁹x² - 1.32085x10¹⁷ x + 6.36x10¹³ = 0

Solving for x:

X = 9.614x10⁻⁴M

Thus, equilibrium concentration of Ag⁺ is:

[Ag⁺] = 0.0676M - 9.614x10⁻⁴M = <em>0.0666M</em>

6 0
3 years ago
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