The weight of your car will also affect its ____

Calculate the heat required when 2.50 mol of a reacts with excess b and a2b according to the reaction: 2a + b + a2b → 2ab + a2 g

Softa [2035]

Answer:

Q = 12.5 kJ

Explanation:

The formula used to compute heat is:

Q = H° * n

Where:

Q: heat (J or kJ)

H°: enthalpy of reaction (kJ/mol)

n: moles

Now, as noted in the comments, the question lacks completeness, here is the part that is missing:

Given:

2A + B  A2B (1)

ΔH° = – 25.0 kJ/mol

2A2B  2AB + A2 (2)

ΔH° = 35.0 kJ/mol

Using these two reactions, we can determine the heat change.

Using the above two reactions, we need to establish the overall reaction (the one presented in the question), so let’s combine (1) and (2):

2A + B -------> A2B H°1 = -25 kJ/mol

2A2B --------> 2AB + A2 H°2 = 35 kJ/mol

When we add these equations, one A2B cancels out with one A2B from reaction 2, thus, we have:

2A + B + 2A2B -------> 2AB + A2

So, for the enthalpy, the values are summed:

H°3 = -25 + 35 = 10 kJ/mol

Now we can calculate the heat:

Q = 10 * 2.5 = 25 kJ

However, as we have 2A in the reaction, it does not maintain a 1:1 mole ratio, instead, it is 1:2, which requires us to adjust; thus:

Q = 25 / 2 = 12.5 kJ

3 0

8 days ago

On a hot summer day, you decide to make some iced tea. First, you brew 1.50 LL of hot tea and leave it to steep until it has rea

Yuliya22 [2438]

Explanation:

Please refer to the attachment for the solution.

6 0

10 days ago

A closed system of mass 10 kg undergoes a process during which there is energy transfer by work from the system of 0.147 kJ per

Softa [2035]

Result: -50.005 kJ

Details:

Provided Data

mass of the system = 10 kg

work done = 0.147 kJ/kg

Elevation change $(\Delta h)=-50 m$

initial speed $(v_1)=15 m/s$

Final Speed $(v_2)=30 m/s$

Specific internal Energy $(\Delta U)=-5 kJ/kg$

according to the first Law of thermodynamics

$Q-W=\Delta H$

$\Delta H=\Delta KE+ \Delta PE +\Delta U$

where KE represents kinetic energy

PE indicates potential energy

U denotes internal Energy

$\Delta H=\frac{m(v_2^2-v_1^2)}{2}+mg(\Delta h)+\Delta U$

$Q=W+\Delta KE+ \Delta PE +\Delta U$

$Q=0.147\times 10+\frac{10\cdot (30^2-15^2)}{2}+10\cdot 9.81\cdot (-5)-5\times 10$

Q = 1.47 + 3.375 - 4.850 - 50

Q = -50.005 kJ

5 0

1 month ago

In a third class lever, the distance from the effort to the fulcrum is ____________ the distance from the load/resistance to the

serg [2593]

The full sentence states:
In a third class lever, the distance between the effort and the fulcrum is LESS than the distance between the load/resistance and the fulcrum.
In a third class lever, the fulcrum is positioned on one end of the effort, while the load/resistance is on the opposite side, placing the effort somewhere in between. Consequently, the distance from the effort to the fulcrum is less than that from the load to the fulcrum.

4 0

2 days ago

1. Susie wondered if the height of a hole punched in the side of a quart-size milk carton would affect how far from the containe

Sav [2230]

Hypothesis: The liquid will project far.
Independent Variable: Height of the hole.
Dependent Variable: Distance of the squirt.
Constant: All other factors aside from the independent variable, such as the liquid volume.
Control: None that I recognize.
Number of groups: 4
Trials per group: 4

7 0

22 days ago

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The weight of your car will also affect its _____.