Answer:
116 years
Explanation:
To solve this, we will use the half life equation;
A(t) = A_o(½)^(t/t_½)
Where;
A(t) is the amount of strontium left after t years;
A_o is the initial quantity of strontium that will undergo decay;
t_½ is the half-life of strontium
t is the time it will take to decay
We are given;
A(t) = 7.5 g
A_o = 120 g
From online values, half life of strontium-90 is 29 years. Thus, t_½ = 29
Thus;
7.5 = 120 × ½^(t/29)
Divide both sides by 120 to get;
7.5/120 = ½^(t/29)
0.0625 = ½^(t/29)
In 0.0625 = (t/29) In ½
-2.772589 = (t/29) × (-0.693147)
(t/29) = -2.772589/(-0.693147)
t/29 = 4
t = 29 × 4
t = 116 years
The answer is (3) pH 3 to pH 1. The pH is related to the concentration of H3O+ with the relationship: pH = -lg c(H+). So when concentration of H3O+ increase, the pH will decrease. And decrease 2 when when increase hundredfold. Because 100=10^2.
Answer : The mass of Allura Red present in an 10 fluid ounce glass of the beverage is, 7.634 mg.
Explanation : Given,
Volume of solution = 10 fl oz = 295.735 mL
Molar mass of Allura Red = 496.42 g/mole
Concentration of Allura Red (Molarity) = 0.000052 M
Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.
Formula used :
Now put all the given values in this formula, we get:
Therefore, the mass of Allura Red present in an 10 fluid ounce glass of the beverage is, 7.634 mg.
The volume of chlorine required is 7.71 L.
The reaction between phosphorus and chlorine is:
2P + 5Cl₂→ 5PCl₅
Therefore, 2 moles of P requires 5 moles of chlorine to react with it.
Given mass of P =3.39 g
Molar mass of P=30.97 g/mol
No. of moles of P = given mass/ molar mass = 3.39 / 30.97 = 0.109 moles
2 moles of P requires 5 moles of chlorine
0.109 moles of P will require 0.109 x 5/2 = 0.2725 moles of chlorine
According to ideal gas equation
PV=nRT
2.04 x V = 0.2725 x 0.0821 x 703
V = 0.2725 x 0.0821 x 703 / 2.04
V = 7.71L
Learn more about ideal gas equation here:
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Answer:
120 g of NaCl in 300 g H20 at 90 C
Explanation:
At x = 90 go vertical to the line for NaCl...then go left to the y-axis to find the solubility in 100 g H20 = 40
we want 300 g H20 so multiply this by 3 to get 120 gm of NaCl in 300 g
