PLS HELP omg im gonna fail :')

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

igomit [66]

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

+ because the heat is absorbed, the reaction is endothermic-

7 0

3 years ago

A gas stream containing n-hexane in nitrogen with a relative saturation of 0.58 (as a fraction, multiply by 100% if you prefer %

kondor19780726 [428]

This problem is describing a gas mixture whose mole fraction of hexane in nitrogen is 0.58 and which is being fed to a condenser at 75 °C and 3.0 atm, obtaining a product at 3.0 atm and 20 °C, so that the removed heat from the system is required.

In this case, it is recommended to write the enthalpy for each substance as follows:

$H_{C-6}=y_{C-6}C_v(T_b-Ti)+\Delta _vH+C_v(T_f-Tb)\\\\H_{N_2}=y_{N_2}C_v(T_f-Ti)$

Whereas the specific heat of liquid and gaseous n-hexane are about 200 J/(mol*K) and 160 J/(mol*K) respectively, its condensation enthalpy is 31.5 kJ/mol, boiling point is 69 °C and the specific heat of gaseous nitrogen is about 29.1 J/(mol*K) according to the NIST data tables and $y_{C-6}$ and $y_{N_2}$ are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:

$H_{C-6}=0.58*200(69-75)+(-31500)+160(20-69)=-40036J/mol\\\\H_{N_2}=0.42*29.1(20-75)=-672.21J/mol$

It is seen that the heat released by the nitrogen is neglectable in comparison to n-hexanes, however, a rigorous calculation is being presented. Then, we add the previously calculated enthalpies to compute the amount of heat that is removed by the condenser:

$Q=-40036+(-672.21)=-40708.21J$

Finally we convert this result to kJ:

$Q=-40708.21J*\frac{1kJ}{1000J}\\\\Q=-40.7kJ$

Learn more:

6 0

2 years ago

For the purposes of determining electron configuration of ions, when electrons are added to a neutral atom, they will inhabit or

irga5000 [103]

Answer: Aufbau principle

Explanation: Aufbau Principle suggests that in filling electrons into orbitals, lower energy orbitals must be filled before higher energy orbitals.

8 0

3 years ago

Read 2 more answers

The mole is defined as the amount of a substance containing the same number of particles as exactly 12 g of C-12. The amu is def

Pavlova-9 [17]

Answer:

Because it wouldn't make any sense

Explanation:

First of all, I think it's important to highlight the definition of isotope.

Isotope (Wikipedia): Variants of a particular chemical element which differ in neutron number...

This means that two isotopes are the same element but have different net mass per atom, due to the different number of neutrons.

Therefore, it's important to make the definition on the same isotope so that the proportion is equal. If the definition would be made on different isotopes, the proportion wouldn't have any sense. Let me be clear with this example:

mass in grams of a C-12 atom = 1.9944235 × 10 ^ -23 g --> this is the mass of a single C atom.

By definition --> 1 mol of anything = 6.02 x 10 ^ 23 particles of anything

Therefore, we know how much a single C atom weights. How many grams do you think that 6.02 x 10 ^ 23 atoms of C (i.e a mol of C) could weight??

1 single C atom ----------------------------- 1.9944235 × 10 ^ -23 g

6.02 x 10 ^ 23 atoms of C ------------ 12.006 g !!

These 12 g is the same quantity than above! Therefore, 1 mol of C weights 12 g. If the definition were made with 13 g of C-13 (the other C isotope), these numbers will not be the same --> There would be a contradiction.

Regarding the second question, we need to search Ne-20 atomic mass in grams --> 3,3509177 × 10 ^ -23 g

Hence, if we follow the same rule, the amu would be 1/12 of Ne-20.

[ 3,3509177 × 10 ^ -23 g ] / 12 = 2.79 ^ -24 g

3 0

3 years ago

Mendeleev put elements in groups while creating the periodic table. What important role do groups play in the structure of the p

NikAS [45]

Answer:

Groups show similar chemical properties about the elements

Explanation:

The periodic table is divided into groups and periods. Groups are vertical while periods are horizontal.

All elements in the same group possess the same number of outermost shell electrons. This number of outermost shell electrons controls the reactivity of the elements in the group.

Hence, groups show similar chemical properties of the elements in them, Groups separate elements into families which assist in the speedy study of elements.

3 0

3 years ago