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Nutka1998 [239]
1 year ago
13

The enthalpy of combustion of lactose, C12H22O11, is -5652 kJ/mol. A 2.50 g sample of lactose was burned in a calorimeter that c

ontained 1350 g of water. The heat capacity of the calorimeter is 1630 J/oC, and the initial temperature was 24.58oC. What was the final temperature(oC)?
Chemistry
1 answer:
Katena32 [7]1 year ago
4 0

Answer:

30.25°C

Explanation:

The calorimeter is an equipment used to measure the combustion enthalpy of a substance. The heat loss in the reaction is used to heat the water and the equipment. By the conservation of energy:

Qcombustion + Qcalorimeter + Qwater = 0

Because there is no phase change:

Qcalorimeter = C*ΔT, where C is the heat capacity, and ΔT the variation in temperature (final - initial)

Qwater = m*c*ΔT, where m is the mass, and c is the specific heat (4.184 J/g°C).

The molar mass of lactose is 342.3 g/mol, so the number of moles in 2.50 g is:

n = mass/molar mass

n = 2.50/342.3

n = 0.0073 mol

Qcombustion = -5652 kJ/mol * 0.0073 mol

Qcombustion = -41.28 kJ

Qcombustion = - 41280 J

Thus,

-41280 + 1630*(T - 24.58) + 1350*4.184*(T - 24.58) = 0

(T - 24.58) * (1630 + 5648.4) = 41280

7278.4(T - 24.58) = 41280

T - 24.58 = 5.67

T = 30.25°C

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C₂H₇F₂P

Explanation:

Given parameters:

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                H = 7%

                 F  = 38%

                 P  = 31%

Unknown:

Empirical formula of compound;

Solution :

The empirical formula is the simplest formula of a compound. To solve for this, follow the process below;

                                   C                          H                         F                   P

% composition

by mass                     24                          7                        38                  31

Molar mass                 12                           1                         19                  31

Number of

moles                       24/12                          7/1                    38/19           31/31

                                     2                               7                       2                   1

Dividing

by the

smallest                      2/1                             7/1                       2/1                1/1

                                     2                                7                        2                   1

           Empirical formula        C₂H₇F₂P

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The optimal pH for sulfate‑based buffers is when pka=pH, that means that optimal pH is <em>2.</em>

<em />

I hope it helps!

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