<span>Response:
A 1.00 L solution that includes 3.00x10^-4 M Cu(NO3)2 and 2.40x10^-3 M ethylenediamine (en).
Contains
0.000300 moles of Cu(NO3)2 and 0.00240 moles of ethylenediamine.
Using the formula Cu(en)2^2+
0.000300 moles of Cu(NO3)2 reacts with double that amount of en = 0.000600 mol of en.
Thus, 0.00240 moles of ethylenediamine - 0.000600 mol of en reacted leaves 0.00180 mol en unreacted.
According to the formula Cu(en)2^2+
0.000300 moles of Cu(NO3)2 reacts to yield an equivalent of 0.000300 moles of Cu(en)2^2+
The formation constant Kf for Cu(en)2^2+ is 1x10^20.
Therefore,
1 Cu+2 and 2 en --> Cu(en)2^2+
Kf = [Cu(en)2^2+] / [Cu+2] [en]^2
1x10^20 = [0.000300] / [Cu+2] [0.00180 ]^2
Solving for [Cu+2] gives [Cu+2] = [0.000300] / (1x10^20) (3.24 e-6)
Thus, Cu+2 = 9.26 e-19 Molar.
Since Kf only has 1 significant figure, round that to 9 X 10^-19 Molar Cu+2.</span>
Although multiple values are given, our focus is on HCl.
<span>We have 215 mL (0.215 L) of 0.300 M HCl fully consumed in the reaction. It's important to recall that the number of moles is found by multiplying volume by molarity:</span>
moles = 0.215 L × 0.300 M
<span>moles = 0.0645 moles of HCl</span>
Students dealing with ionic bonds comprehend better how to convey what the model should showcase.
Explanation:
- Upon dissolving ionic compounds in water, the compounds separate into their constituent ions via a process called dissociation.
- The ions become attracted to water molecules, which carry a polar charge.
- If the pull between the ions and the water molecules is strong enough to disband the ionic bonds, the compound dissolves.
- The ions disperse in the solution, each surrounded by water molecules to inhibit reattachment.
- The ionic solution forms an electrolyte, allowing it to conduct electricity.
- In contrast, while covalent compounds do dissolve in water, they separate into molecules, not individual atoms.
- Water acts as a polar solvent, yet covalent compounds are generally nonpolar.
- This implies that covalent compounds often do not dissolve in water and instead form a distinct layer on top of the water.
N₀ signifies the quantity of C-14 atoms per kg of carbon in the original sample at time = 0 seconds, when the carbon composition matched that in today’s atmosphere. As time progresses to ts, the number of C-14 atoms per kg declines to N, due to radioactive decay. λ indicates the decay constant.
Hence, we have N = N₀e - λt, which is the equation for radioactive decay. Rearranging gives us N₀/N = e λt, or In(N₀/N) = - λt, which becomes equation 1.
The sample contains mc kg of carbon, leading to an activity measured as A/mc decay per kg. The variable r represents the initial mass of C-14 in the sample at t=0 relative to the total mass of carbon which is calculated as [(total number of C-14 atoms at t = 0) × ma] / total mass of carbon. Thus, N₀ equates to r/ma, which becomes equation 2.
The activity of the radioactive element is directly related to the atom count at the moment. The activity equation A = dN/dt = λ(N) indicates that: A = λ₁(N × mc). Rearranging provides N = A / (λmc), represented in equation 3.
By integrating equations 2 and 3, we can solve for t yielding
t = (1/λ) In(rλmc/m₀A).
The result is 200 g. Given that the molar mass of CaCl2 is 110.98 g/mol, this indicates that there are 110.98 g in 1 L of a 1 M solution. Let's calculate the amount of CaCl2 in 0.720 M. Using the proportion 110.98 g: 1 M = x: 0.720 M, we find x to be 79.90 g. Therefore, in 1 L of a 0.720 M solution, there is 79.90 g. Next, we need to create ten beakers with 250 mL each, totaling 10 * 250 mL = 2500 mL or 2.5 L. Then, using the equation 79.90 g: 1 L = x: 2.5 L, we calculate x = 79.90 g * 2.5 L: 1 L, resulting in x = 199.75 g, approximately 200 g.
