Tuesday if your hypnosis is correct and has been examined than it will be come a theory
Wednesday because without it your experiment would fail
Thursday volume
Friday I don't know
Following is the balanced <span>radioactive decay series:
</span><span>
Particle/radiations generated during the reaction are labeled in bold at end of reaction.
Care must be taken that, atomic number and atomic mass number should be balanced in each of these reactions.
1) 92 238U </span>→ <span> 90 234Th + 2 4He(</span>α particle<span>)
A = </span>90 234Th because alpha particle is emitted along with it. So atomic number of daughter element has to be 92 - 2 = 90. This corresponds to Th. <span>
2) 90 234Th </span>→<span> 91 234Pa + -1 0e (electron)
B = -1 0e i.e electron because after radioactive disintegration atomic number of daughter element (Pa) is +1 as compared to parent element (Th)
3) 91 234Pa </span>→<span> 92 234U + –1 0e (electron)
</span>C = 92 234U because electron is emitted along with it. So atomic number of daughter element has to be 91 - (-1) = 92. This corresponds to U. <span>
4) 92 234U </span>→ 90 230Th + 2 4He (α particle<span>)
</span><span>In this case, 92 234U undergoes nuclear disintegration to generate 90 230Th and alpha particle
5) 90 230Th </span>→<span> 88 226Ra + 2 4He </span>(α particle)
D = 88 226Ra because alpha particle is emitted along with it. So atomic number of daughter element has to be 90 - 2 = 88. This corresponds to Ra.
<span>6) 88 266Ra </span>→ 86 222Rn + 2 4He (α particle)
E = alpha particle because during nuclear disintegration, 88 266Ra is converted into 86 222Rn. Hence, for mass balance we have 88 - 86 = 2. It corresponds to alpha particles.
<span>
7) 86 222Rn </span>→<span> 84 218Po + 2 4He </span>(α particle)
Again, F = alpha particle because during nuclear disintegration, 86 222Rn is converted into 84 218Rn. Hence, for mass balance we have 86 - 84 = 2. It corresponds to alpha particles.
<span>
8) 84 218Po </span>→<span> 82 214Pb + 2 4He </span>(α particle)
G = 82 214Pb because alpha particle is emitted along with it. So atomic number of daughter element has to be 84 - 2 = 82. This corresponds to Pb.
<span>
9) 82 214Pb </span>→<span> 83 214Bi + -1 0e (electron)
H = </span>-1 0e because after radioactive disintegration atomic number of daughter element (Bi) is +1 as compared to parent element (Pb)<span>
10) 83 214Bi </span>→<span> 84 214Po + –1 0e (electron)
I = </span>84 214Po because electron is emitted along with it. So atomic number of daughter element has to be 83 - (-1) = 84. This corresponds to Po.<span>
11) 84 214Po </span>→<span> 82 210Pb + 2 4He </span>(α particle)
J = 82 210Pb because alpha particle is emitted along with it. So atomic number of daughter element has to be 84 - 2 = 82. This corresponds to Pb.
Answer:
5.06atm
Explanation:
Using the combined gas law equation;
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (atm)
P2 = final pressure (atm)
V1 = initial volume (Litres)
V2 = final volume (Litres)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the information provided in this question;
P1 = 1.34 atm
P2 = ?
V1 = 5.48 L
V2 = 1.32 L
T1 = 61 °C = 61 + 273 = 334K
T2 = 31 °C = 31 + 273 = 304K
Using P1V1/T1 = P2V2/T2
1.34 × 5.48/334 = P2 × 1.32/304
7.34/334 = 1.32P2/304
Cross multiply
334 × 1.32P2 = 304 × 7.34
440.88P2 = 2231.36
P2 = 2231.36/440.88
P2 = 5.06
The final pressure is 5.06atm
Answer:
Water's high heat of vaporization allows dogs to cool themselves by panting.
Explanation:
Vaporization or Evaporation is the process in which the substance is converted into gas (vapors). This is an endothermic reaction and requires energy to be carried out. In dogs this phenomenon is being carried out in order to cool the body.
As the water molecules are evaporated it results in the removal of highly energetic molecules leaving a low energy molecules behind. Hence, cooling is achieved.
Answer:
91.33 L
Explanation:
For an ideal gas, the product between the gas pressure and the gas volume is proportional to the absolute temperature of the gas. Mathematically:
where:
p is the pressure of the gas
V is the volume of the gas
T is the absolute temperature of the gas
For a transformation of the gas, we can write:
For the sample of krypton in this problem, we have:
(initial pressure)
(initial volume)
(initial temperature)
(final pressure, at stp)
(temperature at stp)
Therefore, the volume at stp is: