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4 years ago

7

How can litmus paper and phenolphthalein be used to tell whether a solution is an acid or a base? Be specific.

1 answer:

MArishka [77]4 years ago

6 0

Both litmus paper and phenolphthalein are used as acid/base (pH) indicators.

Litmus paper:
Litmus paper will turn either blue or red when a drop of solution is added to the paper. Red indicates an acidic solution (pH is <4.5), and blue indicates a basic solution (pH >8.3).

Phenolphthalein:
Phenolphthalein is commonly used during acid/base titrations. When the solution is acidic and the indicator is added, it will remain colorless. If the solution is basic, the phenolphthalein will turn pink.

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Helium gas in a cylinder is under 1.12atm pressure at 25.0C. What will be the pressure if the temperature increases to 37.0C?

Irina18 [472]

Answer:

$p_2=1.17atm$

Explanation:

Hello!

In this case, considering that the Gay-Lussac's law allows us to relate the temperature-pressure problems as directly proportional relationships:

$\frac{p_2}{T_2} =\frac{p_1}{T_1} \\\\$

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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

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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

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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 %

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