Answer:
The products are SnPO4 and LiC2H3O2
Explanation:
The reactants are LiPO4 + Sn(C2H3O2)2
This is a double replacement reaction
So what you do is switch the elements the other way around.
To do that, all you have to do switch Sn with PO4 since Sn is a cation and PO4 is an anion.
Then you switch Li with C2H3O2 because Li is a cation and C2H3O2 is an anion.
After that, check the charges. PO4 has -3 charge
So just leave Sn the way it is without a subscript.
In word form, the product would be Tin(III) Phosphate
C2H3O2 has a -1 charge Li has a +1 charge
So leave both of them the way it is without any subscripts.
In word form, the product would be Lithium Acetate
Answer:
226.8 mg of mupirocin powder are required
Explanation:
Given that;
weight of standard pack = 22 g
mupirocin by weight = 2%
so weight of mupirocin = 2% × 22 = 2/100 × 22 = 0.44 g
so by adding the needed quantity of mupirocin powder to prepare a 3% w/w mupirocin ointment
mg of mupirocin powder are required = ?, lets rep this with x
Total weight of ointment = 22 + x g
Amount of mupirocin = 0.44 + x g
percentage of mupirocin in ointment is 3?
so
3/100 = 0.44 + x g / 22 + x g
3( 22 + x g ) = 100( 0.44 + x g )
66 + 3x g = 44 + 100x g
66 - 44 = 100x g - 3x g
97 x g = 22
x g = 22 / 97
x g = 0.2268 g
we know that; 1 gram = 1000 Milligram
so 0.2268 g = x mg
x mg = 0.2268 × 1000
x mg = 226.8 mg
Therefore, 226.8 mg of mupirocin powder are required
PH and pOH are opposites, so pH can be determined by subtracting pOH from 14
pH = 14 - 7.9 = 6.1
Answer:
Empirical Formula = NH₄NO₃ (Ammonium Nitrate)
Solution:
Step 1: Calculate Moles of each Element;
Moles of N = %N ÷ At.Mass of N
Moles of N = 35.0 ÷ 14
Moles of N = 2.5 mol
Moles of O = %O ÷ At.Mass of O
Moles of O = 59.96 ÷ 16
Moles of O = 3.7475 mol
Moles of H = [100% - (%N + %O)] ÷ At.Mass of H
Moles of H = [100% - (35.0 + 59.96)] ÷ 1.008
Moles of H = [100% - 94.96] ÷ 1.008
Moles of H = 5.04 ÷ 1.008
Moles of H = 5 mol
Step 2: Find out mole ratio and simplify it;
N H O
2.5 5 3.7475
2.5/2.5 5/2.5 3.7475/2.5
1 2 1.5
Multiply Mole Ratio by 2,
2 4 3
Result:
Empirical Formula = N₂H₄O₃
Or,
Empirical Formula = NH₄NO₃
This empirical formula is also a Molecular Formula for Ammonium Nitrate a well known Fertilizer and often misused in the formation of Explosives.
