Hi there!
We can use the following conversions to solve:
Total mass --> amount of mols --> amount of atoms (Avogadro's number)
Begin by calculating the amount of boron trifluoride in 3.61 grams:
3.61 g * (1 mol BF₃ / 67.8 g) ≈ 0.0532 mol BF₃
Use avogadro's number to convert:
0.0532 mol * 6.02× 10²³atoms / 1 mol = 3.203 × 10²² atoms
Answer:
-490.7 K
Explanation:
Given:
[Ni^2+]= 0.4 M
[Pb^2+]=0.002 M
∆V= -0.012 V
VNi= -0.250V
VPb= -0.126V
F= 96500 C
R= 8.314 JK-1 mol-1
n= 2
From
T= -nF/R [∆V-(VNi-VPb)/ln [Pb2+]/[Ni2+]]
T= 2(96500)/8.314[ (-0.012) -(-0.250) - (-0.126))/ln[0.002]/[0.4]
T= 23213.856(0.112/(-5.298))
T= -490.7 K
3.124mg of I-131 is present after 32.4 days.
The 131 I isotope emits radiation and particles and has an 8-day half-life. Orally administered, it concentrates in the thyroid, where the thyroid gland is destroyed by the particles.
What is Half life?
The time required for half of something to undergo a process: such as. a : the time required for half of the atoms of a radioactive substance to become disintegrated.
Half of the iodine-131 will still be present after 8.1 days.
The amount of iodine-131 will again be halved after 8.1 additional days, for a total of 8.1+8.1=16.2 days, reaching (1/2)(1/2)=1/4 of the initial amount.
The quantity of iodine-131 will again be halved after 8.1 more days, for a total of 16.2+8.1+8.1=32.4 days, to (1/4)(1/2)(1/2)=1/16 of the initial quantity.
If the original dose of iodine-131 was 50mg, the residual dose will be (50mg)*(1/16)=3.124mg after 32.4 days.
Learn more about the Half life of radioactie element with the help of the given link:
brainly.com/question/27891343
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Answer:
\large \boxed{\textbf{609 kJ}}
Explanation:
The formula for the heat absorbed is
q = mCΔT
Data:
m = 2.07 kg
T₁ = 23 °C
T₂ = 191 °C
C = 1.75 J·°C⁻¹g⁻¹
Calculations:
1. Convert kilograms to grams
2.07 kg = 2070 g
2. Calculate ΔT
ΔT = T₂ - T₁ = 191 - 23 = 168 °C
3. Calculate q
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