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solong [7]
2 years ago
10

In science, observations are objective, but intereroes are subjective. True or False?

Chemistry
1 answer:
Fed [463]2 years ago
8 0

Answer:

the answer is False I hope this helps

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B.cell membrane is the answer
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In which natural cycle must an important gas in Earth's atmosphere be fixed before plants can use it?
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The answer is C . nitrogen cycle
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Which of the following properties of silver does not explain why this metal is often used in jewelry?
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The high conductivity of silver does not promote its use in jewelry. 
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3 years ago
A gas sample has a temperature of 22c with an unknown volume. The same gas has a volume of 456 mL when the temperature is 86c wi
tester [92]

Answer:

V₁  = 374.71  mL

Explanation:

Given data:

Initial volume of gas= ?

Initial temperature = 22°C

Final temperature = 86°C

Final volume = 456 mL

Solution:

Initial temperature = 22°C (22+273 = 295 k)

Final temperature = 86°C (86+273 = 359 k)

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₁ = V₂T₁ /T₂

V₁  = 456 mL × 295 K / 359 k

V₁  = 134520 mL.K /  359 k

V₁  = 374.71  mL

3 0
3 years ago
Th e molar absorption coeffi cient of a substance dissolved in water is known to be 855 dm3 mol−1 cm−1 at 270 nm. To determine t
Olegator [25]

Answer : The percentage reduction in intensity is 79.80 %

Explanation :

Using Beer-Lambert's law :

A=\epsilon \times C\times l

A=\log \frac{I_o}{I}

\log \frac{I_o}{I}=\epsilon \times C\times l

where,

A = absorbance of solution

C = concentration of solution = 3.25mmol.dm^{3-}=3.25\times 10^{-3}mol.dm^{-3}

l = path length = 2.5 mm = 0.25 cm

I_o = incident light

I = transmitted light

\epsilon = molar absorptivity coefficient = 855dm^3mol^{-1}cm^{-1}

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

\log \frac{I_o}{I}=(855dm^3mol^{-1}cm^{-1})\times (3.25\times 10^{-3}mol.dm^{-3})\times (0.25cm)

\log \frac{I_o}{I}=0.6947

\frac{I_o}{I}=10^{0.6947}=4.951

If we consider I_o = 100

then, I=\frac{100}{4.951}=20.198

Here 'I' intensity of transmitted light = 20.198

Thus, the intensity of absorbed light I_A = 100 - 20.198 = 79.80

Now we have to calculate the percentage reduction in intensity.

\% \text{reduction in intensity}=\frac{I_A}{I_o}\times 100

\% \text{reduction in intensity}=\frac{79.80}{100}\times 100=79.80\%

Therefore, the percentage reduction in intensity is 79.80 %

3 0
3 years ago
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