Answer:
Half-life = 3 minutes
Explanation:
Using the radioactive decay equation we can solve for reaction constant, k. And by using:
K = ln2 / Half-life
We can find half-life of polonium-218
Radioactive decay:
Ln[A] = -kt + ln [A]₀
Where:
[A] could be taken as mass of polonium after t time: 1.0mg
k is Reaction constant, our incognite
t are 12 min
[A]₀ initial amount of polonium-218: 16mg
Ln[A] = -kt + ln [A]₀
Ln[1.0mg] = -k*12min + ln [16mg]
-2.7726 = - k*12min
k = 0.231min⁻¹
Half-life = ln 2 / 0.231min⁻¹
<h3>Half-life = 3 minutes</h3>
Answer:
Cu(s) + 2AgNO3(aq)→Cu(NO3)2(aq)+2Ag(s)
This chemical equation means:
One mole of solid copper plus two moles of aqueous silver nitrate produce one mole of copper(II) nitrate plus two moles of solid silver.
This is a single replacement reaction in which the metal copper replaces the metal silver.
Let's eliminate these one by one.
The first pair would not be the same, as X would most likely be in group IA, and Y would be in group VIIA, because of their tendency to gain and lose electrons.
The second pair would also violate the same rule, but X would most likely be in group IIA, and Y would most likely be in group VIA.
The third pair would not be the same, as X is most likely in group VIIA, and since Y has eight valence electrons, it is most likely a noble gas.
The final pair has X with atomic number 15, making it phosphorous. Phosphorous wants to gain 3 electrons to have a full octet of 8 outer "valence" electrons, and Y would also like to gain 3 electrons. This means it is possible that the final pair would be in the same group.
Answer:
Acceleration = (change in speed) / (time for the change)
Change in speed= (0 - 26 km/hr) = -26 km/hr
(-26 km/hr) x (1,000 m/km) x (1 hr / 3,600 sec) = -7.222 m/sec
Average acceleration = (-7.222 m/s) / (22 min x 60sec/min) = -0.00547 m/sec²
Average speed during the stopping maneuver =
(1/2) (start speed + end speed) = 13 km/hr = 3.6111 m/sec
Explanation:
