Answer:
Co
H20 somewhat ionic as well as covalent
SO2 is covalent
K2O is ionic
O2 ionic with somewhat covalent bonds
Answer:
A. Actual Yield
Since all of the others are calculated not measured outright.
<span>We can solve this problem by assuming that the decay of
cyclopropane follows a 1st order rate of reaction. So that the
equation for decay follows the expression:</span>
A = Ao e^(- k t)
Where,
A = amount remaining at
time t = unknown (what to solve for) <span>
Ao = amount at time zero = 0.00560
M </span><span>
<span>k = rate constant
t = time = 1.50 hours or 5400 s </span></span>
The rate constant should
be given in the problem which I think you forgot to include. For the sake of
calculation, I will assume a rate constant which I found in other sources:
k = 5.29× 10^–4 s–1 (plug in the correct k value)
<span>Plugging in the values
in the 1st equation:</span>
A = 0.00560 M * e^(-5.29 × 10^–4 s–1 * 5400 s )
A = 3.218 <span>× 10^–4 M (simplify
as necessary)</span>
1. C : Mg(CN)2
2. B : N2O5
3. D : Ti(ClO4)3
4. A : Ni(NO3)3
5. D : Cobalt (III) Acetate
6. B : Nickel(II) sulfate
7. C : Dinitrogen Tetrafluoride
8. A : Phosphorus pentachloride<em />
9. C : Metallic <em>(<!> This is the only one I'm not 100% sure of)</em>
10. A : Ionic
11. C : Metallic
12. B : Covalent
It can be explained in<span> simple </span>terms<span>: Oxidation is the loss of electrons or an increase </span>in<span> oxidation state by a molecule, atom, or ion. </span>Reduction<span> is the gain of electrons or a decrease </span>in<span> oxidation state by a molecule, atom, or ion.</span>
