The answer is: A
C-14 is not stable and this is the reason why it goes through radioactive decay.
Answer:
0.73L
Explanation:
The following data were obtained from the question :
V1 = 0.65 L
P1 = 3.4 atm
T1 = 19°C = 19 + 273 = 292K
V2 =?
P2 = 3.2 atm
T2 = 36°C = 36 + 273 = 309K
The bubble's volume near the top can be obtain as follows:
P1V1 /T1 = P2V2 /T2
3.4 x 0.65/292 = 3.2 x V2 /309
Cross multiply to express in linear form as shown below:
292 x 3.2 x V2 = 3.4 x 0.65 x 309
Divide both side by 292 x 3.2
V2 = (3.4 x 0.65 x 309) /(292 x 3.2)
V2 = 0.73L
Therefore, the bubble's volume near the top is 0.73L
We know,
AgNO3 + NaCl ⇒ NaNO3 + AgCl(s)
The moles of Na+ present:
0.5 L * 0.001 mol/L
= 5 x 10⁻⁴ mol
Moles of Ag+ present:
0.5 * 0.02
= 0.01 mol
The limiting reactant is Na
Therefore, the moles of Ag reacted:
5 x 10⁻⁴
AgCl is insoluble in water; therefore, the AgCl formed will precipitate
The answer is leucine would be in the interior, and serine would be on the exterior of the globular protein.
The side chain (R group) of the amino acid serine is CH₂OH. The side chain of the amino acid leucine is CH₂CH(CH₃)₂. In globular protein, leucine found in the interior, and serine found on the exterior. The nature of side chain decides the amino acid position in the globular protein , as CH₂CH(CH₃)₂ this is hydrophobic and CH₂OH is hydrophlic.
Answer:
1935100 Bq
Explanation:
Let us recall that:
If 1 μCi can be equivalent to 37000 Bq
Then; the activity of 52.3 μCi will be:
