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Alenkasestr [34]
2 years ago
14

How does physical properties differ from chemical properties? 1.Chemical properties are observed using your senses. 2.Chemical p

roperties can only be observed using your taste sense. 3.Physical properties can be observed and chemical properties cannot. 4.Physical properties cannot be observed and chemical properties can.
Chemistry
1 answer:
mel-nik [20]2 years ago
7 0

Answer:

3. Physical properties can be observed and chemical properties can't

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Escriba la formula de:
nevsk [136]

Answer:

<h2>Translate your language to English </h2>
4 0
1 year ago
Why do you think forensic scientists are so careful that the tests they do are sensitive, reproducible, and specific?
mario62 [17]

Critical Thinking Questions

1. Why do you think forensic scientists are so careful that the tests they do are sensitive, reproducible, and specific? What might happen if they were less careful about this?

They have to be careful to ensure as much accuracy as possible.

2.Which type of evidence do you think is most useful in an investigation? Why?

Physical evidence would probably be most important because it is the best way to connect someone directly with that crime.

3.Why do you think that forensic scientists continue to look for class characteristics given their limitations?

Class characteristics are good in court because it provides details of different aspects of the crime.


4 0
3 years ago
Read 2 more answers
Codeine C18H21NO3 is a weak organic base. A 5.0 x 10^-3 M solution of codeine has a pH of 9.95. Calculate the value of Kb for th
arsen [322]

Answer:

Kb = 4.45\times10^{-7}\ mol/L

p^{Kb}=6.35

Explanation:

For a weak organic base, the formula to find p^{OH} is given by:

p^{OH}=p^{K_b}+\log c

where c is the concentration of base.

Here c= 5\times10^{-3}\ M

p^{H}=9.95\\p^{OH}=14-p^{H}=14-9.95=4.05

Substituting the above values in the formula,we get:

p^{k_b}=p^{OH}-\log c\\p^{k_b}=4.05-\log (5\times10^{-3})\\p^{K_b}=6.35\\K_b=$antilog 6.35=4.45\times10^{-7}\ mol/L

Hence:

Kb = 4.45\times10^{-7}\ mol/L

p^{Kb}=6.35

7 0
2 years ago
CHEMISTRY
andrezito [222]
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8 0
2 years ago
Read 2 more answers
What is the pH of a buffer that consists of 0.254 M CH3CH2COONa and 0.329 M CH3CH2COOH? Ka of propanoic acid, CH3CH2COOH is 1.3
Shtirlitz [24]

Answer:

4.77 is the pH of the given buffer .

Explanation:

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[salt]}{[acid]})

pH=-\log[K_a]+\log(\frac{[salt]}{[acid]})

pH=-\log[K_a]+\log(\frac{[CH_3CH_2COONa]}{[CH_3CH_2COOH]})

We are given:

K_a = Dissociation constant of propanoic acid = 1.3\times 10^{-5}

[CH_3CH_2COONa]=0.254 M

[CH_3CH_2COOH]=0.329 M

pH = ?

Putting values in above equation, we get:

pH=-\log[1.3\times 10^{-5}]+\log(\frac{[0.254 M]}{[0.329]})

pH = 4.77

4.77 is the pH of the given buffer .

7 0
2 years ago
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