A population of rabbits triples every 2 months If there are 2 rabbits initially, how long will it take for the population to inc
1 answer:
Answer:
The rabbit population will reach 500 after 10 months.
Explanation:
According to the given data:
The initial number of rabbit's equals 2.
Number of rabbit's after 2 months =2x3= 6
Number of rabbit's after 4 months = 6x3=18
Number of rabbit's after 6 months = 18x3=54
Number of rabbit's after 8 months = 54x3=162
Thus we can see that the number of rabbit's form a Geometric series with common ratio =3 and initial term = 2
Now the general term of a geometric series with first term 'a' and common ratio 'r' is given by
Thus we need to find when the term becomes 500 thus using the given data we have
Thus the fifth term (excluding the start term) will have a rabbit count of 500 now since each term has a time difference of 2 months thus sixth term will occur after
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Answer : The value of at this temperature is 66.7
Explanation : Given,
Pressure of at equilibrium = 0.348 atm
Pressure of at equilibrium = 0.441 atm
Pressure of at equilibrium = 10.24 atm
The balanced equilibrium reaction is,
The expression of equilibrium constant for the reaction will be:
Now put all the values in this expression, we get :
Therefore, the value of at this temperature is 66.7
First of all, I need to know what these five salts are. Luckily, I found a similar problem from another website which is shown in the attached picture. The Ksp is the solubility product constant. It follows the formula:
Ksp = [cation concentration]ᵃ[anion concentration]ᵇ
where a and b are the subscripts of the metal and nonmetal, respectively.
For the solutions ahead, let x be the concentration of the cation.
A. The formula is PbCr₂O₄.
2.8×10⁻¹³ = [x][x]
Solving for x, x = 5.29×10⁻⁷ M
B. The formula is Co(OH)₂.
1.3×10⁻¹⁵ = [x][x]²
Solving for x, x = 1.09×10⁻⁵ M
C. The formula is CoS.
5×10⁻²² = [x][x]
Solving for x, x = 2.24×10⁻¹¹ M
D. The formula is Cr(OH)₃.
1.6×10⁻³⁰ = [x][x]³
Solving for x, x = 3.56×10⁻⁶ M
E. The formula is Ag₂S.
6×10⁻⁵¹ = [x]²[x]
Solving for x, x = 1.82×10⁻¹⁷ M
<em>Thus,the highest concentration is letter B, Cobalt (II) Hydroxide.</em>
Ksp = [cation concentration]ᵃ[anion concentration]ᵇ
where a and b are the subscripts of the metal and nonmetal, respectively.
For the solutions ahead, let x be the concentration of the cation.
A. The formula is PbCr₂O₄.
2.8×10⁻¹³ = [x][x]
Solving for x, x = 5.29×10⁻⁷ M
B. The formula is Co(OH)₂.
1.3×10⁻¹⁵ = [x][x]²
Solving for x, x = 1.09×10⁻⁵ M
C. The formula is CoS.
5×10⁻²² = [x][x]
Solving for x, x = 2.24×10⁻¹¹ M
D. The formula is Cr(OH)₃.
1.6×10⁻³⁰ = [x][x]³
Solving for x, x = 3.56×10⁻⁶ M
E. The formula is Ag₂S.
6×10⁻⁵¹ = [x]²[x]
Solving for x, x = 1.82×10⁻¹⁷ M
<em>Thus,the highest concentration is letter B, Cobalt (II) Hydroxide.</em>