<h3>
Answer:</h3>
28.52 seconds
<h3>
Explanation:</h3>
Initial number of atoms of Nitrogen 12,000 atoms
Half-life = 7.13
Number of atoms after decay = 750 atoms
We are required to determine the time taken for the decay.
Note that half life is the time taken for a radioactive isotope to decay to a half of its original amount.
Using the formula;
Remaining amount = Initial amount × (1/2)^n , where n is the number of half lives
In our case;
750 atoms = 12,000 atoms × (1/2)^n
0.0625 = 0.5^n
n = log 0.0625 ÷ log 0.5
n = 4
But, 1 half life =7.13 seconds
Therefore;
Time taken = 7.13 seconds × 4
= 28.52 seconds
Therefore, the time taken for 12,000 atoms of nitrogen to decay to 750 atoms is 28.52 seconds
Answer is: excess of hydrazine is 16 grams.
Chemical reaction: N₂O₄(l) + 2N₂H₄(l) → 3N₂(g) + 4H₂<span>O(g).
</span>m(N₂H₄) = 80,1 g.
m(N₂O₄) = 92,0 g.
n(N₂H₄) = m(N₂H₄) ÷ M(N₂H₄).
n(N₂H₄) = 80,1 g ÷ 32 g/mol.
n(N₂H₄) = 2,5 mol.
n(N₂O₄) = 92 g ÷ 92 g/mol.
n(N₂O₄) = 1 mol; limiting reactant.
From chemical reaction: n(N₂H₄) : n(N₂O₄) = 2 : 1.
n(N₂H₄) = 2 mol reacts.
Δn(N₂H₄) = 2,5 mol - 2 mol = 0,5 mol.
Δm(N₂H₄) = 0,5 mol · 32 g/mol = 16 g.
Answer:
[KOH] = 0.10M in KOH
Explanation:
Molar Concentration [M] = moles solute/volume solution in liters
moles KOH = 0.56g/56g/mole = 0.01mole
Volume of solution = 100cm³ = 100ml = 0.10 liter
[KOH] = 0.01 mole KOH / 0.10 liter solution = 0.10M in KOH
Answer:
6.66 mol
Explanation:
(atm x L) ÷ (0.0821 x K)
(0.875 x 250) ÷ (0.0821 x 400)
=6.66108
