Answer:
Boron and Aluminum
Explanation:
If you write the electron configuration for boron and aluminum, you get:
for boron and 
for aluminum. Both have 3 valance electrons and has 2 electrons in a s-orbital and 1 in a p-orbital. These valance electron similarities are based on the column/group the elements are. Therefore, Boron and Aluminum have similar chemical behaviours and similar arrangement of outer/valance electrons.
Once for the water and once for the copper. Set up a table that accounts for each of the variables you know, and then identify the ones you need to obtain. Give me a moment or two and I will work this out for you.
Okay, so like I said before, you will need to use the equation twice. Now, keep in mind that when the copper is placed in the water (the hot into the cold), there is a transfer of heat. This heat transfer is measured in Joules (J). So, the energy that the water gains is the same energy that the copper loses. This means that for your two equations, they can be set equal to each other, but the copper equation will have a negative sign in front to account for the energy it's losing to the water.
When set equal to each other, the equations should resemble something like this:
(cmΔt)H20 = -(cmΔt)Cu
(Cu is copper).
Remember, Δt is the final temperature minus the initial temperature (T2-T1). We are trying to find T2. Since we are submerging the copper into the water, we can assume that the final temperature at equilibrium is the same for both the copper and the water. At a thermodynamic equilibrium, there is no heat transfer because both materials are at the same temperature.
T2Cu = T2H20
Now, the algebra for this part of the problem is a bit confusing, so make sure you keep track of your variables. If done right, the algebra should work out so you have this:
T2 = ((cmT1)Cu + (cmT1)H20) / ((cm)H20 + (cm)Cu)
Insert the values for the variables. Once you plug and chug, your final answer should be
26.8 degrees Celsius.
The balanced chemical equation is given as:
2CH3CH2OH(l) → CH3CH2OCH2CH3(l) + H2O(l)
We are given the yield of CH3CH2OCH2CH3 and the amount of ethanol to be used for the reaction. These values will be the starting point for the calculations.
Theoretical amount of product produced:
329 g CH3CH2OH ( 1 mol / 46.07 g ) ( 1 mol CH3CH2OCH2CH3 / 2 mol CH3CH2OH ) (74.12 g / mol ) = 264.66 g CH3CH2OCH2CH3
% yield = .775 = actual yield / 264.66
actual yield = 205.11 g CH3CH2OCH2CH3
Answer:
pH =3.8
Explanation:
Lets call the monoprotic weak acid HA, the dissociation equilibria in water will be:
HA + H₂O ⇄ H₃O⁺ + A⁻ with Ka = [ H₃O⁺] x [A⁻]/ [HA]
The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.
In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:
HA H₃O⁺ A⁻
Initial, M 0.40 0 0
Change , M -x +x +x
Equilibrium, M 0.40 - x x x
Lets express these concentrations in terms of the equilibrium constant:
Ka = x² / (0.40 - x )
Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,
7.3 x 10⁻⁶ = x² / 0.40 ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³
[H₃O⁺] = 1.71 x 10⁻³
Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.
pH = - log ( 1.71 x 10⁻³ ) = 3.8
Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.
Answer: Nuclei combine to form a heavier nucleus, releasing energy.
Explanation: e.g two deuterium nucleus (Hydrogen-2 isotopes) forms an He nucleus and energy is released.
