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olga nikolaevna [1]
3 years ago
14

Ways of preventing rusting​

Chemistry
2 answers:
Flura [38]3 years ago
5 0

Answer:

Here is your answer......

lesya [120]3 years ago
4 0
Use an Alloy. Many outdoor structures, like this bridge, are made from COR-TEN steel to reduce the effects of rust. ...
Apply Oil. ...
Apply a Dry Coating. ...
Paint the Metal. ...
Store Properly. ...
Galvanize. ...
Blueing. ...
Powder Coating.
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Which of the following is a physical change?
Wewaii [24]

Answer:

O ice melts at 0°C that is the answer

5 0
3 years ago
Which of the following compounds would contain an electrostatic attraction as bonds between ions?
4vir4ik [10]

Answer:

iron sulfate

Explanation:

Electrostatic attraction as bonds between ions is characteristic of the electrovalent bond or the ionic bond

In this type of bond, there is electron transfer from one atom to another. The atom that looses electrons become positively charged while the atom that gains electrons becomes negatively charged.

In iron sulfate, there is electrostatic attraction between Fe II ions and sulphate ions, making iron sulfate an ionic compound.

6 0
3 years ago
It is recommended that drinking water contain 1.6 ppm fluo- ride (F) to prevent tooth decay. Consider a cylindrical reservoir wi
Ne4ueva [31]

Answer:

32,127.02 grams of hydrogen hexafluorosilicate will contain this 25,434  grams of F.

Explanation:

Volume of cylindrical reservoir = V

Radius of the cylindrical reservoir = r = d/2

d = diameter of the cylindrical reservoir = d =4.50\times 10^1 m=45 m

r = d/2 = 22.5 m

Depth of the reservoir = h =  10.0 m

V=\pi r^2 h

=3.14\times (22.5 m)^2\times 10.0 m=15,896.25 m^3=15,896,250 L

1 m^3=1000 l

Volume of water cylindrical reservoir : V

Density of water,d = 1 kg/L

Mass of water cylindrical reservoir =  m

m=d\times V=1 kg/L\times 15,896,250 L=15,896,250 kg

1.6 kilogram of fluorine per million kilograms of water. (Given)

Concentration of fluorine in water = 1.6 kg/ 1000,000 kg of water

In 1000,000 kg of water = 1.6 kg of fluorine

Then 15,896,250 kg of water have x mass of fluorine:

\frac{x}{15,896,250 kg\text{kg of water}}=\frac{1.6 kg}{1000,000 \text{kg of water}}

x=\frac{1.6 kg}{1000,000}\times 15,896,250 kg=25.434 kg

15,896,250 kg water of contains mass 25.434 kg of fluorine.

25.434 kg = 25434 g

25,434  grams of fluorine  should be added to give 1.6 ppm.

Percentage of fluorine in hydrogen hexafluorosilicate :

Molar mass hydrogen hexafluorosilicate = 144 g/mol

F\%=\frac{6\times 19 g/mol}{144 g/mol}\times 100=79.16\%

Total mass of hydrogen hexafluorosilicate = m'

79.16\%=\frac{25,434 g}{m'}\times 100

m' = 32,127.02 g

32,127.02 grams of hydrogen hexafluorosilicate will contain this 25,434  grams of F.

5 0
3 years ago
4. Which elements would have similar properties to potassium?
iren [92.7K]

Answer:

The answer is A. Lithium and Sodium

5 0
3 years ago
In terms of the number of valence electrons, describe one cyclic property that can be observed in the model.
Crazy boy [7]
Will you please tell about which model are you talking about? So, that I can help you because your seems to be incomplete.
6 0
3 years ago
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