Answer:
Callie expect 600 molecules of CO2 to have been released as a waste during the same amount of time.
Explanation:
During cellular respiration 1 molecule of glucose undergoes oxidation to form 6 molecules of CO2 as a waste product.
According to the question callie determined that the germinating corn seed had utilized 100 molecules of glucose.
So 100 molecules of glucose will release 100×6=600 molecules of CO2 as a waste product.
Answer:
In our Sun, as in other stars, roughly 99.9% or so of all light emitted is emitted in a thin layer known as the photosphere, or light sphere. This is explained as follows. Interior to the photosphere the gas is ever denser and becomes far too opaque for any photon to emerge directly from that layer.
Explanation:
The value of Kc for the equilibrium is 0.150 mole² / litre ²
<u>Explanation:</u>
<u>Given:</u>
An equilibrium mixture in an 1.00 L vessel contains 5.30 moles of
Mg(OH )₂ 0.800 moles of Mg²⁺ and 0.0010 moles OH₋
We have to find the value of Kc
- Step 1: Find the equilibrium Concentration.
- Step 2: Substitute the values in the equation.
- Step 3: Find the value of Kc.
- I have attached the document for the detailed explanation
The value of Kc for the equilibrium is 0.150 mole² / litre ²
Answer:
C.If one light in the string burns out in a parallel circuit, the rest of the lights will continue to shine.
Solubility
product constants are values to describe the saturation of ionic compounds with
low solubility. A saturated solution is when there is a dynamic equilibrium
between the solute dissolved, the dissociated ions, the undissolved and the
compound. It is calculated from the product of the ion concentration in the
solution. For the base, Ca(OH)2, the dissociation would be as
follows:<span>
Ca(OH)2 = Ca2+ + 2OH-
So, the expression for the solubility product constant would be as follows:
Ksp = [Ca2+] [OH-]^2
let x be the concentration of the Ca2+. So,
</span>
Ksp = [x] [2x]^2
<span>Ksp = 4x^3
You have to substitute the value of the concentration of the calcium hydroxide in the final expression which is not given in the problem statement in order to evaluate Ksp.
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