Answer:
The situation given here is imaginary such that the life of Rock has to be found using the half-life of the element lokium that has been found inside the rock.
Half-life of any material is the amount of time taken by that particular material to decay. Now the amount of lokium found in rock can show after how many half-lives this amount has been left out.
The time elapsed will be log (L) atoms X half-life.
Explanation:
Answer is: concentration ammonia is higher than concentration of ammonium ion.
Chemical reaction of ammonia in water: NH₃ + H₂O → NH₄⁺ + OH⁻.
Kb(NH₃) = 1,8·10⁻⁵.
c₀(NH₃) = 0,8 mol/L.
c(NH₄⁺) = c(OH⁻) = x.
c(NH₃) = 0,8 mol/L - x.
Kb = c(NH₄⁺) · c(OH⁻) / c(NH₃).
0,000018 = x² / 0,8 mol/L - x.
solve quadratic equation: x = c(NH₄⁺) = 3,79·10⁻³ mol/L.
Considering the definition of molarity, the molar concentration is 0.294 .
Molarity reflects the concentration of a solution indicating the number of moles of solute that are dissolved in a given volume.
The molarity of a solution is calculated by dividing the moles of the solute by the volume of the solution:
Molarity is expressed in units .
In this case, you know:
- amount of moles of solute= 0.250 moles
- volume= 0.850 L
Replacing in the definition of molarity:
Solving:
molarity= 0.294
Finally, the molar concentration is 0.294 .
Learn more about molarity with this example: brainly.com/question/15406534?referrer=searchResults
Polaris the north star is very bright and it doesn’t move
<u>Answer:</u>
<u>For 2:</u> The % yield of the product is 92.34 %
<u>For 3:</u> 12.208 L of carbon dioxide will be formed.
<u>Explanation:</u>
The percent yield of a reaction is calculated by using an equation:
......(1)
Given values:
Actual value of the product = 78.4 g
Theoretical value of the product = 84.9 g
Plugging values in equation 1:
Hence, the % yield of the product is 92.34 %
The number of moles is defined as the ratio of the mass of a substance to its molar mass.
The equation used is:
......(2)
Given mass of carbon dioxide = 24 g
Molar mass of carbon dioxide = 44 g/mol
Plugging values in equation 1:
<u>At STP conditions:</u>
1 mole of a gas occupies 22.4 L of volume
So, 0.545 moles of carbon dioxide will occupy = of volume
Hence, 12.208 L of carbon dioxide will be formed.