Answer:
Explanation:
Hello,
In this case, given that a typical aspirin tablet contains 5.00 grains of pure aspirin, the first step here is to compute the mass of those grans per tablet given that 1.00 g = 15.4 grains:
In such a way, the number of aspirin tablets are computed considering the total mass of aspirin and the mass per tablet:
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Answer:
A = 0.75 ×10² KJ.
B = 3.9 ×10³ dg
C = 0.22 × 10² μl.
Explanation:
A = 7.5 ×10⁴ j to kilo joules
7.5 ×10⁴ / 1000 = 0.75 ×10² KJ.
Joule is the smaller unit while kilo joule is the larger unit. One kilo joule equals to the thousand joule that's why we will divide the given value by 1000 in order to convert into KJ.
B = 3.9 ×10⁵ mg to decigrams.
3.9 ×10⁵ / 100 = 3.9 ×10³ dg
Decigram is larger unit while milligram is smaller unit. One decigram is equal to the 100 milligram. In order to convert the given value into decigram we have to divide the value by 100.
C = 2.21 ×10⁻⁴ dL to micorliters
2.21 ×10⁻⁴ ×10⁵ = 0.22 × 10² μl.
Deciliter is bigger unit then micro liter . One deciliter equals to the 100000 micro liters. In order to convert the dL into micro liter we have to multiply the given value with 100000.
Metal rusts when it oxidized around moisture.
0.000001ppm
Explanation:
Mass of fluoride = 500g
Volume of water = 500000liters
Unknown:
Parts per million of fluoride = ?
Solution:
The parts per million is the amount of solute in milligram dissolved in a liter of water or milligram per kilogram of solvent
It is a unit used to express very small concentration.
we need to convert g - mg
500g = 500 x 10⁻³mg = 0.5mg
Concentration in parts per million = 
Concentration in parts per million = 
= 0.000001ppm
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Answer:
2Cu2^+ + 2I^- ----> 2Cu^+ + I2
Explanation:
The reaction performed in the experiment is;
2 Cu(NO3)2 + 4 KI → 2 CuI (s) + 4 KNO3 + I2
The iodide ions reduces Cu^2+ to Cu^+ which is insoluble in water hence the precipitate. This is so because iodine is a good oxidizing agent seeing that it requires one electron to fill its outermost shell. Potassium on the other hand is a good reducing agent since it easily looses its one electron.
The oxidation - reduction equation is as follows;
2Cu2^+ + 2e ----> 2Cu^+ reduction half equation
2I^- ----> I2 + 2e. Oxidation half equation
Balanced redox reaction equation;
2Cu2^+ + 2I^- ----> 2Cu^+ + I2
