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3 years ago

What mass of sucrose c12h22o11 is needed to make 300 ml of a 0.50m solution?

1 answer:

5 0

Use the concentration to obtain the moles. I am assuming you mean to write capital M. because little m means molality.

So, first convert the ml into Liters and then into moles, then moles to grams using the molar mass (just adding the values of each atom from the periodic table. )

Molar mass= 12 (12.0) + 22 (1.01)+ 11 (16.0)= 342 grams/mole

300 ml (1 liter/ 1000 mL) x (0.50 moles/ 1 Liter) x (342 grams/ 1 mole)= 51.3 grams

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2 years ago

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2 years ago

Calculate the amount of heat released when 27.0 g H2O is cooled from a liquid at 314 K to a solid at 263 K. The melting point of

BlackZzzverrR [31]

Answer : The amount of heat released, 45.89 KJ

Solution :

Process involved in the calculation of heat released :

$(1):H_2O(l)(314K)\rightarrow H_2O(l)(273K)\\\\(2):H_2O(l)(273K)\rightarrow H_2O(s)(273K)\\\\(3):H_2O(s)(273K)\rightarrow H_2O(s)(263K)$

Now we have to calculate the amount of heat released.

$Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+\Delta H_{fusion}+[m\times c_{p,s}\times (T_{final}-T_{initial})]$

where,

Q = amount of heat released = ?

m = mass of water = 27 g

$c_{p,l}$ = specific heat of liquid water = 4.184 J/gk

$c_{p,s}$ = specific heat of solid water = 2.093 J/gk

$\Delta H_{fusion}$ = enthalpy change for fusion = 40.7 KJ/mole = 40700 J/mole

conversion : $0^oC=273k$

Now put all the given values in the above expression, we get

$Q=[27g\times 4.184J/gK\times (314-273)k]+40700J+[27g\times 2.093J/gK\times (273-263)k]$

$Q=45896.798J=45.89KJ$ (1 KJ = 1000 J)

Therefore, the amount of heat released, 45.89 KJ

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3 years ago

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Evolution can not be tested by experiment given that it depends largely on observations that fit into a wider scope and point to a common conclusion.

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1 year ago