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STatiana [176]
3 years ago
5

As the temperature of a liquid solvent increases, the amount of solute that can dissolve in it ________. decreases remains const

ant increases decreases by 1º celsius for every milliliter of solvent
Chemistry
1 answer:
Nookie1986 [14]3 years ago
6 0
Answer:
              
<span>As the temperature of a liquid solvent increases, the amount of solute that can dissolve in it <u>increases</u>.

Explanation:
                    The solubility of most solutes in a solvent increases with increase in temperature. This solubility is closely related to the heat of solution, (the heat evolved or absorbed when solute is dissolved in solvent). Hence, majority of solutes when dissolved in solvent absorbs heat and makes the overall heat of solution positive. Hence, in this case more heat provided will increase the rate of solubility.</span>
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Which law states that the volume and absolute temperature of a fixed quantity of gas are directly proportional under constant pr
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The mass of 0.10 mole of methane
saveliy_v [14]

Answer:

A methane molecule is made from one carbon atom and four hydrogen atoms. Carbon has a mass of 12.011 u and hydrogen has a mass of 1.008 u. This means that the mass of one methane molecule is 12.011 u + (4 × 1.008u), or 16.043 u. This means that one mole of methane has a mass of 16.043 grams.

メタン分子は、1つの炭素原子と4つの水素原子から作られています。炭素の質量は12.011uで、水素の質量は1.008uです。これは、1つのメタン分子の質量が12.011 u +(4×1.008u)、つまり16.043uであることを意味します。これは、1モルのメタンの質量が16.043グラムであることを意味します。^>^

5 0
3 years ago
Baking soda reacts with vinegar and forms a gas why is it chemical
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Read 2 more answers
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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