Answer:
The rate of decay of atoms in container A is greater than the rate of decay of atoms in container B.
Explanation:
From the question,
Container A contains 1000 atoms
Container B contains 500 atoms
<u>The rate of decay of atoms in container A is greater than the rate of decay of atoms in container B.</u>
The reason for such is due to the difference in the concentration of the isotopes. Container A which contains higher number of atoms will have the more changes of the release of the neutron as the changes of the hitting and splitting increases as the density of the atoms increases.
<u>Thus, the atoms in the container A will therefore decay faster than the atoms in the container B. </u>
Answer:
4.8 g/mL is the density of chloroform vapor at 1.00 atm and 298 K.
Lunch of a patient has 3 oz skinless chicken, 3 oz of broccoli, 1 medium apple, and 1 cup of nonfat milk
Energy content of 3 oz skinless chicken is = 110 kcal
Energy content of 3 oz broccoli = 30 kcal
Energy content of 1 medium apple = 60 kcal
Energy content of 1 cup non-fat milk = 90 kcal
So the kilocalories of energy patient obtained from lunch
= 110 kcal+ 30 kcal + 60 kcal + 90 kcal = 290 kcal
The total number of digits that are counted while performing a measurement, are called significant figures or significant numbers. In a measured number, all the digits are certain except the last digit. The last digit which is uncertain in a measured number is called estimated digit. In the measurement 543.1267 inches, the last digit 7 is estimated digit. While taking into account the total number of significant figures, we count all the certain digits plus the estimated digit. Therefore, the number 543.1267 has 7 significant figures.
To get the value of ΔG we need to get first the value of ΔG°:
when ΔG° = - R*T*㏑K
when R is constant in KJ = 0.00831 KJ
T is the temperature in Kelvin = 25+273 = 298 K
and K is the equilibrium constant = 4.5 x 10^-4
so by substitution:
∴ ΔG° = - 0.00831 * 298 K * ㏑4.5 x 10^-4
= -19 KJ
then, we can now get the value of ΔG when:
ΔG = ΔG° - RT*㏑[HNO2]/[H+][NO2]
when ΔG° = -19 KJ
and R is constant in KJ = 0.00831
and T is the temperature in Kelvin = 298 K
and [HNO2] = 0.21 m & [H+] = 5.9 x 10^-2 & [NO2-] = 6.3 x 10^-4 m
so, by substitution:
ΔG = -19 KJ - 0.00831 * 298K* ㏑(0.21/5.9x10^-2*6.3 x10^-4 )
= -40
