Answer:
There is 54.29 % sample left after 12.6 days
Explanation:
Step 1: Data given
Half life time = 14.3 days
Time left = 12.6 days
Suppose the original amount is 100.00 grams
Step 2: Calculate the percentage left
X = 100 / 2^n
⇒ with X = The amount of sample after 12.6 days
⇒ with n = (time passed / half-life time) = (12.6/14.3)
X = 100 / 2^(12.6/14.3)
X = 54.29
There is 54.29 % sample left after 12.6 days
Converting temperature of 68°F to °C gives 20 °C.
Converting temperature of 68°F to K gives 293 K.
<h3>What is temperature conversion?</h3>
Temperature conversion is the process of converting the measurement units of the temperature recorded in a particular unit to another unit.
The various units of Temperature include;
- degree Celsius
- degree Fahrenheit
- degree Kelvin
Temperature is measured with thermometer and it records the hotness or coldness of a body.
<h3>Converting 68°F to °C</h3>
F = 1.8C + 32
(F - 32/1.8) = C
(68 - 32) / 1.8 = C
20 ⁰C = 68 ⁰F
<h3>Converting 20°C to K</h3>
0 °C = 273 K
20 °C = 273 + 20 = 293 K
Learn more about temperature conversion here: brainly.com/question/23419049
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We can use two equations for this problem.<span>
t1/2 = ln
2 / λ = 0.693 / λ
Where t1/2 is the half-life of the element and λ is
decay constant.
20 days = 0.693 / λ
λ = 0.693 / 20 days
(1)
Nt = Nο eΛ(-λt) (2)
Where Nt is atoms at t time, No is the initial amount of substance, λ is decay constant and t is the time
taken.
t = 40 days</span>
<span>No = 200 g
From (1) and (2),
Nt = 200 g eΛ(-(0.693 / 20 days) 40 days)
<span>Nt = 50.01 g</span></span><span>
</span>Hence, 50.01 grams of isotope will remain after 40 days.
<span>
</span>
Answer:
Q= 245 =2.5 * 10^2
Explanation:
ΔG = ΔGº + RTLnQ, so also ΔGº= - RTLnK
R= 8,314 J/molK, T=298K
ΔGº= - RTLnK = - 6659.3 J/mol = - 6.7 KJ/mol
ΔG = ΔGº + RTLnQ → -20.5KJ/mol = - 6.7 KJ/mol + 2.5KJ/mol* LnQ
→ 5.5 = LnQ → Q= 245 =2.5 * 10^2
1) Calculate the volume from d = m/V => V = m/d = 2.0*10^-23 g / 1.0*10^14 g/cm^3 = 2.0*10^-9 cm^3
2) Now use the formula of volume for a sphere: V = (4/3)π(r^3) =>
r =∛[3V/(4π)] = ∛[(3*2.0*10^-9 cm^3) / (4π)] = 0.48*10^-3 cm = 4.8*10^-4 cm = 0.00048cm
