T is amount after time t
<span>Ao is initial amount </span>
<span>t is time </span>
<span>HL is half life </span>
<span>log (At) = log [ Ao x (1/2)^(t/HL) ] </span>
<span>log (At) = log Ao + log (1/2)^(t/HL) </span>
<span>log (At) = log Ao + (t/HL) x log (1/2) </span>
<span>( log At - log Ao) / log (1/2) = t / HL </span>
<span>log (At/Ao) / log (1/2) = t / HL </span>
<span>HL = t / [( log (At / Ao)) / log (1/2) ] </span>
<span>HL = 14.4 s / [ ( log (12.5 / 50) / log (1/2) ] </span>
<span>HL = 14.4 s / 2 = 7.2 seconds </span>
Answer:
a) molarity of CCl3F = 1.12 × 10^-11 mol/dm³
Molarity of CCl2F2 = 2.20 × 10^-11 mol/dm³
B) molarity of CCL3F = 7.96 × 10 ^-13 mol/dm³
Molarity of CCl2F2 = 1.55 × 10^-12 mol/dm³
Explanation:
Using the ideal gas equation:
PV = nRT
Further explanations are found in the attachment below.
Answer:
86.2 g/mol
Explanation:
Before you can find the molar mass, you first need to calculate the number of moles of the gas. To find this value, you need to use the Ideal Gas Law:
PV = nRT
In this equation,
-----> P = pressure (mmHg)
-----> V = volume (L)
-----> n = moles
-----> R = Ideal Gas constant (62.36 L*mmHg/mol*K)
-----> T = temperature (K)
After you convert the volume from mL to L and the temperature from Celsius to Kelvin, you can use the equation to find the moles.
P = 760 mmHg R = 62.36 L*mmHg/mol*K
V = 250 mL / 1,000 = 0.250 L T = 20 °C + 273.15 = 293.15 K
n = ? moles
PV = nRT
(760 mmHg)(0.250 L) = n(62.36 L*mmHg/mol*K)(293.15 K)
190 = n(18280.834)
0.0104 = n
The molar mass represents the mass (g) of the gas per every 1 mole. Since you have been given a mass and mole value, you can set up a proportion to determine the molar mass.
<----- Proportion
<----- Cross-multiply
<----- Divide both sides by 0.0104
Best Answer: 4.52 X 10²³ atoms. Use Avogadro's number to multiply 0.750 moles times 6.022 X 10²³
Answer: 2.223
Explanation:
HClO4, also known as perchloric acid, is a very strong. This means we can assume that complete dissociation occurs.
This means that:
[H+] = [HClO4] = <span>0.00598
pH = -log[H+]
pH = -log[</span><span>0.00598] = 2.223
Note: The base of the log is 10 (conventionally we don't write the base if it is 10)</span>