Behavioral Adaptation is something that is natural in animals that are used for survival. Example: Birds Migrating north during the winter.
Hello, Ginamuhs2!
The oxidizing agent in 2H2S + 3O2 → 2SO2 + 2H2O is the oxygen.
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Answer is: the ratio of the effusion rate is 1.59 : 1.
1) rate of effusion of carbon monoxide gas = 1/√M(CO).
rate of effusion of carbon monoxide gas = 1/√28.
rate of effusion of carbon monoxide gas = 0.189.
2) rate of effusion of chlorine = 1/√M(Cl₂).
rate of effusion of chlorine = 1/√70.9.
rate of effusion of chlorine = 0.119.
rate of effusion of carbon monoxide : rate of effusion of chlorine =
= 0.189 : 0.119 / ÷0.119.
rate of effusion of carbon monoxide : rate of effusion of chlorine = 1.59 : 1.
5.367 ml of the concentrated acid must be added to obtain a total volume of 100 ml of the dilute solution.
Dilution is defined as the process in which the concentration of a sample is decreased by adding more solvent. The dilution formula is given below.
C₁V₁ = C₂V₂
where C₁ = initial concentration of sample = 3.00 m
V₁ = initial volume of sample
C₂ = final concentration after dilution = 0.161 m
V₂ = total final volume after dilution = 100 ml
Plug in the values to the formula and solve for the volume of the concentrated acid that must be added.
C₁V₁ = C₂V₂
3.00 m (V₁) = 0.161 m (100 ml)
V₁ = 5.367 ml
Learn more about dilution here: brainly.com/question/1615979
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Answer:
Approximately 75%.
Explanation:
Look up the relative atomic mass of Ca on a modern periodic table:
There are one mole of Ca atoms in each mole of CaCO₃ formula unit.
- The mass of one mole of CaCO₃ is the same as the molar mass of this compound:
. - The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element:
.
Calculate the mass ratio of Ca in a pure sample of CaCO₃:
.
Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be 
of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio 
:
.
In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:
.
