Answer:
Option A
Explanation:
Number of millimoles of Na3PO4 = 1 × 100 = 100
Number of millimoles of AgNO3 = 1 × 100 = 100
When 1 mole of Na3PO4 is dissociated we get 3 moles of sodium ions and 1 mole of phosphate ion
When 1 mole of AgNO3 is dissociated, we get 1 mole of Ag+ and 1 mole of NO3-
As Ag+ concentration is negligible, the dissociated Ag+ ion must have form the precipitate with phosphate ion and as number of moles of Ag+ and phosphate ion are same, therefore the concentration of phosphate ion must be negligible
Here as 100 millimoles of Na3PO4 is there, we get 300 millimoles of Na+ and 100 millimoles of PO43-
And as 100 millimoles of AgNO3 is there, we get 100 millimoles of Ag+ and 100 millimoles of NO3-
∴ Increasing order of concentration will be PO43- < NO3- < Na+
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Answer: 
Explanation:- Volume percentage is the ratio of volume of solute to the volume of solution defined in terms of percentage.
Given: volume of solute = 29.4 ml
Volume of solution= 359 ml
<h3>The average atomic mass of Iodine : 126.86 amu</h3><h3>Further explanation</h3>
Given
80% 127I, 17% 126I, and 3% 128I.
Required
The average atomic mass
Solution
The elements in nature have several types of isotopes
Atomic mass is the average atomic mass of all its isotopes
Mass atom X = mass isotope 1 . % + mass isotope 2.% + ... mass isotope n.%
Atomic mass of Iodine = 0.8 x 127 + 0.17 x 126 + 0.03 x 128
Atomic mass of Iodine = 101.6 + 21.42 + 3.84
Atomic mass of Iodine = 126.86 amu
Answer:
The number on the lag label should be 15.
Explanation:
It seems your question is incomplete, as it is lacking the working values. An internet search showed me the full question, you can see it in the attached picture.
Let's say we have 100 g of the fertilizer.
- <em>45 g are of ammonium phosphate</em> ( (NH₄)₃PO₄ ), of which:
- 45 g (NH₄)₃PO₄ *
= 12.7 g are of Nitrogen.
(We used the molar mass of ammonium phosphate in the denominator and three times the molar mass of nitrogen in the numerator)
- <em>18 g are of calcium nitrate</em> (Ca(NO₃)₂), of which:
- 16 g Ca(NO₃)₂ *
= 2.73 g are of Nitrogen.
So in total there are (12.7+2.73) 15.43 g of Nitrogen in 100 g of the fertilizer. So the percent by mass of nitrogen is 15.43%.
Rounding to the nearest percent the answer is 15.
