A lot of molecules will be in 2.570 moles of H2
<u>Answer:</u> The mass percent of lead in lead (IV) carbonate is 63.32 %
<u>Explanation:</u>
The given chemical formula of lead (IV) carbonate is 
To calculate the mass percentage of lead in lead (IV) carbonate, we use the equation:

Mass of lead = (1 × 207.2) = 207.2 g
Mass of lead (IV) carbonate = [(1 × 207.2) + (2 × 12) + (6 × 16)] = 327.2 g
Putting values in above equation, we get:

Hence, the mass percent of lead in lead (IV) carbonate is 63.32 %
Isotope 1: 89.905 * 51.45 = 4625.61225 / 100 = 46.2561225
Isotope 2: 90.906 * 11.22 = 1019.96532 / 100 = 10.1996532
Isotope 3: 91.905 * 17.15 = 1576.17175 / 100 = 15.7617075
Isotope 4: 93.906 * 17.38 = 1632.08628 / 100 = 16.3208628
Isotope 5: 95.908 * 2.08 = 268.5424 / 100 = 2.685424
46.2561225 + 10.1996532 + 15.7617075 + 16.3208628 + 2.685424 = 91.22377
actual mass Zr = about 91.22
Answer: 2.4 ml
Solution :
Molar mass of
= 17 g/mole
Given,: 28% w/w of
solution means 28 g of ammonia in 100 g of solution.
Mass of solution = 100 g
Now we have to calculate the volume of solution.
Molarity : It is defined as the number of moles of solute present in one liter of solution.

where,
n = moles of solute 
= volume of solution in liter = 0.11 L
Now put all the given values in the formula of molarity, we get

Using molarity equation:



Answer:
8.547 x 10⁴disintegrations per second
Explanation:
To calculate the disintegrations per second as -
Given ,
2.31 μCi of sulfur -35 .
Since ,
1 Ci = 3.7 * 10 ¹⁰ Bq
1 μCi = 10 ⁻⁶ Ci
Hence ,
conversation is done as follows -
2.31 ( 1 * 10⁻⁶) * ( 3.7 * 10¹⁰)
= 8.547 x 10⁴
Hence ,
8.547 x 10⁴disintegrations per second , the sample undergo for it to be brand new .