Answer:
Each student will need;
1 red jelly bean, 1 white jelly bean, 1 black jelly bean and 3 red jelly beans.
Explanation:
Sodium bicarbonate molecule, NaHCO3, or baking soda is composed of the following:
1 atom of sodium, Na;
1 atom of hydrogen, H;
1 atom of carbon, C, and
3 atoms of oxygen.
For each of the models to be built by the two students, these atoms are to be represented accordingly.
Since Red jelly beans represent sodium atoms (Na), white jelly beans represent hydrogen atoms (H), black jelly beans represent carbon atoms (C), and blue jelly beans represent oxygen atoms (O), each student will need;
1 red jelly bean, 1 white jelly bean, 1 black jelly bean and 3 red jelly beans.
A sample model is found in the attachment below:
Answer: the answer is option (D). k[P]²[Q]
Explanation:
first of all, let us consider the reaction from the question;
2P + Q → 2R + S
and the reaction mechanism for the above reaction given thus,
P + P ⇄ T (fast)
Q + T → R + U (slow)
U → R + S (fast)
we would be applying the Rate law to determine the mechanism.
The mechanism above is a three step process where the slowest step seen is the rate determining step. From this, we can see that this slow step involves an intermediate T as reactant and is expressed in terms of a starting substance P.
It is important to understand that laws based on experiment do not allow for intermediate concentration.
The mechanism steps for the reactions in the question are given below when we add them by cancelling the intermediates on the opposite side of the equations then we get the overall reaction equation.
adding this steps gives a final overall reaction reaction.
2P + Q ------------˃ 2R + S
Thus the rate equation is given as
Rate (R) = K[P]²[Q]
cheers, i hope this helps
Answer:
it would definitely be wienerballs1977
Explanation:
fossil fuels x (3x1017kJ/yr) equals out to be wienerballs1977.
thx for the challenge !
Answer:
Cu(OH)₂ will precipitate first, with [OH⁻] = 2.97x10⁻¹⁰ M
Explanation:
The equilibriums that take place are:
Cu⁺² + 2OH⁻ ↔ Cu(OH)₂(s) ksp = 2.2x10⁻²⁰ = [Cu⁺²]*[OH⁻]²
Co⁺² + 2OH⁻ ↔ Co(OH)₂(s) ksp = 1.3x10⁻¹⁵ = [Co⁺²]*[OH⁻]²
Keep in mind that <em>the concentration of each ion is halved </em>because of the dilution when mixing the solutions.
For Cu⁺²:
2.2x10⁻²⁰ = [Cu⁺²]*[OH⁻]²
2.2x10⁻²⁰ = 0.25 M*[OH⁻]²
[OH⁻] = 2.97x10⁻¹⁰ M
For Co⁺²:
1.3x10⁻¹⁵ = [Co⁺²]*[OH⁻]²
1.3x10⁻¹⁵ = 0.25 M*[OH⁻]²
[OH⁻] = 7.21x10⁻⁸ M
<u>Because Copper requires less concentration of OH⁻ than Cobalt</u>, Cu(OH)₂ will precipitate first, with [OH⁻] = 2.97x10⁻¹⁰ M
