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VikaD [51]
4 years ago
14

A certain molecular compound M has a solubility in acetone of 0.737g/mL at 10.°CCalculate the mass of M that's dissolved in 9.0L

of a saturated solution of M in acetone at this temperature.Be sure your answer has the correct unit symbol and number of significant digits.
Chemistry
1 answer:
Paha777 [63]4 years ago
5 0

Answer:

6633 grams of M are dissolved in 9L

Explanation:

If the compound M has a solubility in acetone of 0.737 g/mL, means that in 1 mLof solution, 0.737 grams of solute is dissolved.

Let's make a rule of three:

9 L = 9000 mL

In 1 mL ___ 0.737 g of M is dissolved

In 9000 mL ___ (9000 . 0.737) /1 = 6633 grams

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Calculate the standard entropy of vaporization of ethanol at its boiling point, 352 K. The standard molar enthalpy of vaporizati
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Answer : The correct option is, (b) +115 J/mol.K

Explanation :

Formula used :

\Delta S=\frac{\Delta H_{vap}}{T_b}

where,

\Delta S = change in entropy

\Delta H_{vap} = change in enthalpy of vaporization = 40.5 kJ/mol

T_b = boiling point temperature = 352 K

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

\Delta S=\frac{\Delta H_{vap}}{T_b}

\Delta S=\frac{40.5kJ/mol}{352K}

\Delta S=115J/mol.K

Therefore, the standard entropy of vaporization of ethanol at its boiling point is +115 J/mol.K

8 0
3 years ago
"46.7 g of water at 80.6 oC is added to a calorimeter that contains 45.33 g of water at 40.6 oC. If the final temperature of the
soldier1979 [14.2K]

<u>Answer:</u> The specific heat of calorimeter is 30.68 J/g°C

<u>Explanation:</u>

When hot water is added to the calorimeter, the amount of heat released by the hot water will be equal to the amount of heat absorbed by cold water and calorimeter.

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[(m_2\times c_2)+c_3](T_{final}-T_2)       ......(1)

where,

q = heat absorbed or released

m_1 = mass of hot water = 46.7 g

m_2 = mass of cold water = 45.33 g

T_{final} = final temperature = 59.4°C

T_1 = initial temperature of hot water = 80.6°C

T_2 = initial temperature of cold water = 40.6°C

c_1 = specific heat of hot water = 4.184 J/g°C

c_2 = specific heat of cold water = 4.184 J/g°C

c_3 = specific heat of calorimeter = ? J/g°C

Putting values in equation 1, we get:

46.7\times 4.184\times (59.4-80.6)=-[(45.33\times 4.184)+c_3](59.4-40.6)

c_3=30.68J/g^oC

Hence, the specific heat of calorimeter is 30.68 J/g°C

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3 years ago
Help (also tell me what the phenotype and genotypes are)
ad-work [718]

Answer:

genotype are the organism's hereditary information for example DNA

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A sample of gas has an initial volume of 15 L and an initial pressure of 4.5 atm. If the pressure changes to 1.8 atm, what is th
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Answer:

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3 years ago
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