The element produced by this nuclear reaction, I believe would be a metal.
Explanation:
The given balanced reaction is as follows.

It is given that mass of ammonium nitrate is 86.0 kg.
As 1 kg = 1000 g. So, 86.0 kg = 86000 g.
Hence, moles of
present will be as follows.
Moles of
= 
= 
= 1074.42 mol
Therefore, moles of
,
and
produced by 1074.42 mole of
will be as follows.
Moles of
= 
= 537.21 mol
Moles of
= 
= 1074.42 mol
Moles of
= 
= 2148.84 mol
Therefore, total number of moles will be as follows.
537.21 mol + 1074.42 mol + 2148.84 mol
= 3760.47 mol
According to ideal gas equation, PV = nRT. Hence, calculate the volume as follows.
PV = nRT
1 atm \times V = 3760.47 mol \times 0.0821 L atm/mol K \times 580 K[/tex] (as
= 307 + 273 = 580 K)
V = 179066.06 L
Thus, we can conclude that total volume of the gas is 179066.06 L.
<span>Using proportions,
if x is the unknown volume we have:
1.84 / 1 = 42.5 / x </span>
<span>
Multiply both sides above by x:
1.84*x / 1 = 42.5
1.84*x = 42.5
Divide both sides by 1.84:
x = 42.5 / 1.84</span>
<span>X = 23.10 mL</span>
Answer:

Explanation:
Data:
p₁ = 1.00 atm; V₁ = 350. L
p₂ = ?; V₂ = 2.00 L
Calculation:

I mole of water has an Avogadro number of molecules.
1 mole = 6.02 * 10^ 23 molecules.
6.02 * 10^ 23 molecules = 1 mole of water
1 molecule = 1/(6.02 * 10^23) mole of water
2.0 * 10^22 molecules would have = (2*10^22) * 1/(6.02*10^23)
= 0.033
2* 10 ^22 molecules of water would have 0.033 moles of water.