<h3>Answer </h3>
After another 5730 years ( three half lives or 17190 years) 17.5 /2 = 8.75mg decays and 8.75g remains left. after three half lives or 17190 years, 8.75 g of C-14 will be
Explanation:
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Answer is: concentration of hydrogenium ions is 9,54·10⁻⁵ M.
c(HNO₂) = 0,075 M.
c(NaNO₂) = 0,035 M.
Ka(HNO₂) = 4,5·10⁻⁵.
This is buffer solution, so use <span>Henderson–Hasselbalch equation:
pH = pKa + log(c(</span>NaNO₂) ÷ c(HNO₂)).
pH = -log(4,5·10⁻⁵) + log(0,035 M ÷ 0,075 M).
pH = 4,35 - 0,33.
pH = 4,02.
<span>[H</span>₃O⁺] = 10∧(-4,02).
<span>[H</span>₃O⁺] = 0,0000954 M = 9,54·10⁻⁵ M.
Answer:
9.72 grams.
Explanation:
From the equation, 4 moles of NH₃ produce 6 moles of water.
Therefore the reaction to product ratio of NH₃ to H₂O is 4:6
and 2:3 into its simplest form.
The number of moles of NH₃ in 6.12 g is:
Number of moles=mass/ RMM
=6.12 g/17 G/mol
=0.36 moles.
Therefore the number of moles of H₂O produced is calculated as follows.
(0.36 Moles×3)2 = 0.54 moles
Mass= Number of moles × RMM
=0.54 moles×18g/mol
=9.72 grams.
Answer:
balanced equation mole ratio 5 2 mol NO/1 mol O2
10.00 g O2 3 1 mol O2/32.00 g O2 5 0.3125 mol O2
20.00 g NO 3 1 mol NO/30.01 g NO 5 0.6664 mol NO
actual mole ratio 5 0.6664 mol NO/0.3125 mol O2 5 2.132 mol NO/1.000 mol O2
Because the actual mole ratio of NO:O2 is larger than the balanced equation mole
ratio of NO:O2, there is an excess of NO; O2 is the limiting reactant.
Mass of NO used 5 0.3125 mol O2 3 2 mol NO/1 mol O2 5 0.6250 mol NO
0.6250 mol NO 3 30.01 g NO/1 mol NO 5 18.76 g NO
Mass of NO2 produced 5 0.6250 mol NO2 3 46.01 g NO2/1 mol NO2 5 28.76 g NO2
Excess NO 5 20.00 g NO 2 18.76 g NO 5 1.24 g N
Explanation:
Answer:
Explanation:
In a nuclear reaction, the total mass and total atomic number remains the same.
Am has an atomic number of 95. So correct reaction is:-
To calculate A:
Total mass on reactant side = total mass on product side
343 + 4 = A + 2
A = 345
To calculate Z:
Total atomic number on reactant side = total atomic number on product side
95 + 2 = Z + 0
Z = 97
Hence, the isotopic symbol of unknown element is