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Romashka-Z-Leto [24]
3 years ago
6

In a coffee cup calorimeter, 1.60 g of NH4NO3 is mixed with 75.0 g of water at an initial temperature of 25.00 degrees C. After

dissolution of the salt, the final temperature of the calorimeter contents is 23.34 degrees C. Assuming the solution has a heat capacity of 4.18 J/Cg and assuming no heat loss to the calorimeter, calculate the enthalpy change for the dissolution of NH4NO3 in units of kJ/mol.
Chemistry
1 answer:
igomit [66]3 years ago
7 0

Answer:

+26.6kJ/mol

Explanation:

The enthalpy of dissolution of NH₄NO₃ is:

NH₄NO₃(aq) + ΔH  → NH₄⁺ + NO₃⁻

Where ΔH is the heat of reaction that is absorbed per mole of NH₄NO₃,

The moles that reacts in 1.60g are (Molar mass NH₄NO₃:80g/mol):

1.60g * * (1mol / 80g) = 0.02 moles reacts

To find the heat released in the coffee cup calorimeter, we must use the equation:

Q = m×ΔT×C

Where Q is heat released,

m is mass of the solution

ΔT is change in temperature (Final temperature - Initial temperature)

C is specific heat of the solution (4.18J/g°C)

Mass of the solution is:

1.60g + 75g = 76.60g

Change in temperature is:

25.00°C - 23.34°C = 1.66°C

Replacing:

Q = m×ΔT×C

Q = 76.60g×1.66°C×4.18J/g°C

Q = 531.5J

This is the heat released per 0.02mol. The heat released per mole (Enthalpy change for the dissolution of NH₄NO₃) is:

531.5J / 0.02mol = 26576J/ mol =

+26.6kJ/mol

<em>+ because the heat is absorbed, the reaction is endothermic-</em>

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4 years ago
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The compound known as butylated hydroxytoluene, abbreviated as BHT, contains carbon, hydrogen, and oxygen. A 3.001 g sample of B
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The compound known as butylated hydroxytoluene, abbreviated as BHT, contains carbon, hydrogen, and oxygen. A 3.001 g sample of BHT was combusted in an oxygen rich environment to produce 8.990 g of CO2(g) and 2.944 g of H2O(g). Insert subscripts below to appropriately display the empirical formula of BHT

Answer:

C_{15}H_{24}0

Explanation:

A 3.001 g sample of BHT was combusted in an oxygen rich environment to produce 8.990 g of CO2(g) and 2.994 g of H2O(g).

If all the carbon in BHT is present in CO_2 and also, all the hydrogen in BHT is  present in H_2O, Then we can determine for the corresponding numbers of moles of Carbon(C) and Hydrogen (H) respectively as:

moles of  CO_2 = 8.990 g*(\frac{1mole}{44.01g})

                       =  0.2043 moles

∴ moles of C =  0.2043 moles

moles of H_2O = 2.944 g *(\frac{1mole}{18.01g} )

                       = 0.1635 moles

∴ moles of H = 2 × 0.1635 moles

                      = 0.327 moles

Since number of moles= \frac{mass}{molarmass}

number of moles of H =  0.327 moles

molar mass of H = 1.008 g/mol

∴  mass of H in the sample = 0.327 moles × 1.008 g/mol

                                             = 0.329616g

                                             

mass of C in the sample can be calculated as = 0.2043 moles × (\frac{12.01g}{1 mole} )

= 2.453643 g

mass of C+H in the sample = 2.453643g + 0.329616g

= 2.783259 g

mass of O can be calculated as = 3.001 g - 2.783259 g

= 0.217741 g

∴ moles of O = 0.217741g × (\frac{1mole}{16.0g})

= 0.0136 moles

Now, since; we've gotten our data, we can now proceed to calculate for the empirical formula.

C                                          H                                    O

0.2043                                0.327                             0.0136

Dividing by the least number (0.0136) , we have :

\frac{0.2043}{0.0136}                                     \frac{0.327}{0.0136}                               \frac{0.0136}{0.0136}

15.02                                      24.04                             1

≅

15                                             24                                  1

Therefore, the empirical formula would be : C_{15}H_{24}0

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Calculating Moles of H₂:
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             32 g (1 mole) of CH₃OH is produced by  =  2 Mole of H₂
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Solving for X,
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