We have Kc = 4.2 x 10^-2 (given but missing in the question)
and When the balanced equation for this reaction is:
PCl5(g) ↔ PCl3(g) + Cl2(g)
so, according to the Kc formula:
Kc = the concentration of products / the concentration of the reactants
so, to get the concentration of the reactants in equilibrium, the concentration of the products / the concentration of the reactants should equal the Kc value which is given in the question (missing in your question).
So by substitution in Kc formula:
Kc = [PCl3]*[Cl2] / [PCl5]
4.2 x 10^-2 = 0.18 * 0.25 /[PCl5]
∴[PCl5] = 0.18*0.25 / 4.2x10^-2 = 1.07
So the concentration of the reactants in equilibrim = 1.07
The fact that CO2 is released from oceans due to further rise in temperature is an example of a negative feedback.
A negative feedback is one in which the process that produces the feedback is interrupted. That is, the process is stopped as a result of the feedback received.
In this case, CO2 which leads to global warming dissolves in the ocean which serves a large sink for the gas. However, as the increase in ocean temperatures causes decrease in solubility of CO2, more CO2 is released leading to further temperature rise. This is an example of a negative feedback loop.
Learn more: brainly.com/question/13440572
Answer:
Number of moles = 0.042 mol
Explanation:
Given data:
Number of moles = ?
Mass of calcium carbonate = ?
Solution:
Formula:
Number of moles = mass/ molar mass
now we will calculate the molar mass of calcium carbonate.
atomic mass of Ca = 40 amu
atomic mass of C = 12 amu
atomic mass of O = 16 amu
CaCO₃ = 40 + 12+ 3×16
CaCO₃ = 40 + 12+48
CaCO₃ = 100 g/mol
Now we will calculate the number of moles.
Number of moles = 4.15 g / 100 g/mol
Number of moles = 0.042 mol
Answer:
Explanation:
Not Many
1 mol of CO has a mass of
C = 12
O = 16
1 mol = 28 grams.
1 mol of molecules = 6.02 * 10^23
x mol of molecules = 3.14 * 10^15 Cross multiply
6.02*10^23 x = 1 * 3.14 * 10^15 Divide by 6.02*10^23
x = 3.14*10^15 / 6.02*10^23
x = 0.000000005 mols
x = 5*10^-9
1 mol of CO has a mass of 28
5*10^-9 mol of CO has a mass of x Cross Multiply
x = 5 * 10^-9 * 28
x = 1.46 * 10^-7 grams
Answer: there are 1.46 * 10-7 grams of CO if only 3.14 * 10^15 molecules are in the sample
