# chemistry - mathematical lesson (PPT

```“When the well is
dry, we know the
worth of
water”
Benjamin Franklin
Mineral water
(integrated mathematic-chemistry lesson)
The aim of the lesson:

to get acquainted with variety
of mineral waters of Lithuania.
The goals of the lesson:

To explore general mineralization of mineral
water.
 To measure qualitative components of mineral
water “Vytautas”.
 To learn to solve integrated mathematicschemistry tasks.
The classification of mineral water

Mineral waters
wich have very few
minerals: the
concentration of
minerals ≤ 50mg/l.

Examples:
“Neptūnas”
“Rytas”
“Perrier”
“Vittel”

Mineral waters
wich have few
minerals: the
concentration of
minerals ≤ 500mg/l.
Mineral waters
wich have a lot of
minerals: the
concentration of
minerals > 1500mg/l.
Examples:
“Rasa”
“Vytautas”
“Naleczovanka”
“Augustovianka”
0
605
877
Naleczovianka
403
Augustovianka
2745
Fuldataler
349
Perrier
500
Vittel
4000
Vichy Celestines
Neptūnas
2000
Tiche
3000
ruskininkų Rasa
1000
Aqville
Birutė
8000
Vytautas
Total Dissolved Solids (TDS)
mg/l
7200
7000
6000
5000
3378
2515
2017
1378
713
Group 1 : Ag+(aq) + Cl-(aq) → AgCl(s)
Group 2 : Ca2+(aq) + CO32-(aq) → CaCO3(s)
Group 3 : Ba2+(aq) + SO42-(aq) → BaSO4(s)
Group 4 : H+(aq) + HCO3-(aq) → CO2(g) + H2O(l)
n
1. C =
V
534 mg/l · 1l = 534 mg
534
= 0,534 g
1000
n(Ca2+ )=
0,534g
= 0,013mol
40g / m ol
0,013 mol
 2  0,013 mol / l
C=
2l
2. Ca2+(aq) + Na2CO3(aq) → CaCO3(s) + 2Na+(aq)
n(Ca2+) = 0,013 mol
n(Na2CO3) = 0,013 mol
n
C=
V
0,013m ol
V=
0,5m ol/ l
= 0,026l
3. “Vytautas”
3437mg/l · 1l = 3437mg
3437 mg
=
= 3,437g
1000
3,437g
w(Cl ) =
· 100% = 47,98%
7,165g
m(Cl-)
“Perrier”
21,5mg/l · 1l = 21,5mg
21,5mg
m(Cl ) =
1000
0,0215g
w(Cl ) =
· 100% = 3,56%
0,6046g
1.
Let us assume, that:
x is the number of small bottles (0,5l),
y is the number of large bottles (1,5l).
Make an equation system:
x + y = 36
0,5x + 1,5y = 42
x = 36 – y
0,5x + 1,5y = 42
0,5 · (36 – y) + 1,5y = 42
18 – 0,5y + 1,5y = 42
y = 42 – 18
y = 24
x = 36 – 24 = 12
The answer: there are 12 small bottles (0,5l) and 24 large
bottles (1,5l)
2.
Let us assume, that:
Let us assume, that:
x is the number of
small bottles (0,5l) of
mineral water “Vytautas”
y is the number of
small bottles (0,5l) of
mineral water “Rasa”
t is the number of large
bottles (1,5l) of mineral
water “Tiche”
z is the number of large
bottles (1,5l) of mineral
water “Vytautas”
Make an equation
system:
Make an equation
system:
x + y = 12
x=y+2
y + 2 + y = 12
2y = 12 – 2
2y = 10
/ ÷2
y=5
x=5+2=7
t + z = 24
t = 3z
3z + z = 24
4z = 24
/ ÷4
z=6
t = 3 · 6 = 18
2.
Count the praise:
5 · 1,39 + 7 · 1,49 + 6 · 2,29 + 18 · 2,09 = 68,74
3.
Convert to Euros:
68,74 ÷ 3,4528 ≈ 19,9
Convert to Zlotys:
68,74 ÷ 0,81765 ≈ 84,07
Conclusion

Mineral water is underwater which contains
minerals and has curative features;

1 litre of mineral water should contain not less
than 1gr of minerals;

The composition of salts in mineral water should
be constant (stable);

Ionic composition of mineral water and the
concentration of ions are indicated on the bottle
sticker (label).
```
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