Answer:
9.25
Explanation:
Let first find the moles of
and 
number of moles of
= 0.40 mol/L × 200 × 10⁻³L
= 0.08 mole
number of moles of
= 0.80 mol/L × 50 × 10⁻³L
= 0.04 mole
The equation for the reaction is expressed as:

The ICE Table is shown below as follows:

Initial (M) 0.08 0.04 0
Change (M) - 0.04 -0.04 + 0.04
Equilibrium (M) 0.04 0 0.04







for buffer solutions
since they are in the same solution


Answer is: B.) Yes, if work is done, this transfer process can take place.
For example, air conditioner involves a cyclic process that transfers heat from a cold reservoir to a hot reservoir, but with use of electricity.
Thermal conductuction is the transfer of heat through physical contact. Thermal conduction is the transfer of heat by microscopic collisions of particles. Heat spontaneously flows from a hotter to a colder body.
The process of heat conduction depends on four basic factors: the temperature gradient, the cross section of the materials involved, their path length and the properties of those materials.
<span>A scientific question is like a hypothesis. It's the question that you're trying to answer throughout the experiment. So, a scientific question in this case could be: If the car has bigger wheels, will it travel faster? This is something you can test in the experiment, by having different cars with different sized wheels. In this way, you can track how fast each car goes, and determine whether or not the wheel size increases speed, decreases speed, or has no effect on speed.</span>
Answer:
All of the statements above are true.
Explanation:
Ice is solid water. Ice consists of an array of water molecules arranged into a crystal lattice. Ice has spaces between the water molecules so it is less dense than liquid water. Ice is about 9% less dense than liquid water. This accounts for the fact that it floats on water.
Ice contains more hydrogen bonds per water molecule when compared to liquid water.
Answer : The energy required to melt 58.3 g of solid n-butane is, 4.66 kJ
Explanation :
First we have to calculate the moles of n-butane.

Given:
Molar mass of n-butane = 58.12 g/mole
Mass of n-butane = 58.3 g
Now put all the given values in the above expression, we get:

Now we have to calculate the energy required.

where,
Q = energy required
= enthalpy of fusion of solid n-butane = 4.66 kJ/mol
n = moles = 1.00 mol
Now put all the given values in the above expression, we get:

Thus, the energy required to melt 58.3 g of solid n-butane is, 4.66 kJ