Which of the following is one way to prevent the corrosion of iron? A. Paint exposed iron p21 with protective paint.
B. Protect the iron from polluted air.
C. Add carbon to the iron.
D. Let the iron develop a natural coat of carbonate.

Answers

Answer 1
Answer: The best way to prevent the corrosion of iron is to paint the exposed iron with protective paint, because that prevents water from rusting the iron.

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How much energy (in Joules) would it take to warm 3.11 grams of gold by 7.7 oC(specific heat capacity of gold = 0.130 J/g °C)

Answers

Answer:

It would take 3.11 J to warm 3.11 grams of gold

Explanation:

Step 1: Data given

Mass of gold = 3.11 grams

Temperature rise = 7.7 °C

Specific heat capacity of gold = 0.130 J/g°C

Step 2: Calculate the amount of energy

Q = m*c*ΔT

⇒ Q = the energy required (in Joules) = TO BE DETERMINED

⇒ m = the mass of gold = 3.11 grams

⇒ c = the specific heat of gold = 0.130 J/g°C

⇒ ΔT = The temperature rise = 7.7 °C

Q = 3.11 g * 0.130 J/g°C * 7.7 °C

Q = 3.11 J

It would take 3.11 J to warm 3.11 grams of gold

In which compound does chlorine have the highest oxidation number?(1) NaClO (3) NaClO3(2) NaClO2 (4) NaClO4

Answers

Answer:

              NaClO₄

Solution:

Let us calculate the oxidation state of Chlorine in each compound one by one;

1) NaClO

               O.N of Na = +1

               O.N of O = -2

So,

               (+1) + (Cl) + (-2) = 0

               1 + Cl - 2 = 0

               Cl = -1 + 2

               Cl = +1

2) NaClO₃

               (+1) + (Cl) + (-2)3 = 0

               (+1) + (Cl) - 6 = 0

               1 + Cl - 6 = 0

               Cl = -1 + 6

               Cl = +5

3) NaClO₂

               (+1) + (Cl) + (-2)2 = 0

               (+1) + (Cl) + (-4) = 0

               1 + Cl - 4 = 0

               Cl = -1 + 4

               Cl = +3

4) NaClO₄

               (+1) + (Cl) + (-2)4 = 0

               (+1) + (Cl) + (-8) = 0

               1 + Cl - 8 = 0

               Cl = -1 + 8

               Cl = +7

Elements from which two families would most likely combine with each other to create salts?

Answers

Also could be an alkali metal (such as Na) and a halogen (such as Cl)
Metals and non metals

How many milliliters of 0.183 m hcl would be required to titrate 5.93 g koh?

Answers

  Molar mass:
KOH = 56.0 g/mol

Number of moles of KOH :

5.93  / 36.5 => 0.1624 moles

KOH + HCl = KCl + H₂O 

1 mole KOH --------------> 1 mole  HCl
0.1624 moles KOH ----> ?

moles HCl = 0.1624 x 1 / 1

= 0.1624 moles of HCl

V ( HCl ) = moles / molarity

V(HCl) = 0.1624 / 0.183

V (HCl) = 0.887 L x 1000 = 887 mL

hope this helps!



The following three solutions are mixed: 100.0mL of 0.100M Na₂SO₄, 50.0mL of 0.300M ZnCl₂, and 100.0mL of 0.200M Ba(CN)₂. Cyanide (CN-) solubilities are not in most tables, but Ba(CN)₂ is soluble, Zn(CN)₂ is not. a. What ionic compounds will precipitate out of solution? b. What is the molarity of each ion remaining in the solution, assuming complete precipitation of all insoluble compounds, and assuming that volumes are additive?

Answers

a. To determine the ionic compounds that will precipitate out of solution, we need to consider the solubility rules. According to the solubility rules:

1. All sodium (Na+) salts are soluble, so Na₂SO₄ will remain in solution.
2. Zinc (Zn2+) salts are generally soluble, except for zinc sulfide (ZnS) and zinc hydroxide (Zn(OH)₂). However, in this case, we are adding ZnCl₂ to the solution, which contains chloride (Cl-) ions. Chloride ions form soluble salts with most cations, including Zn2+. Therefore, ZnCl₂ will remain in solution.
3. Barium (Ba2+) salts are generally soluble, except for barium sulfate (BaSO₄) and barium carbonate (BaCO₃). However, in this case, we are adding Ba(CN)₂ to the solution, which contains cyanide (CN-) ions. Cyanide ions form insoluble salts with most cations, including Ba2+. Therefore, Ba(CN)₂ will precipitate out of solution as Ba(CN)₂ is not soluble.

b. Assuming complete precipitation of all insoluble compounds and that volumes are additive, we can calculate the molarity of each ion remaining in the solution.

For Na₂SO₄:
- Sodium (Na+) ion concentration: 2 * 0.100 M = 0.200 M
- Sulfate (SO₄2-) ion concentration: 0.100 M

For ZnCl₂:
- Zinc (Zn2+) ion concentration: 0.300 M
- Chloride (Cl-) ion concentration: 2 * 0.300 M = 0.600 M

For Ba(CN)₂:
- Barium (Ba2+) ion concentration: 0.200 M
- Cyanide (CN-) ion concentration: 2 * 0.200 M = 0.400 M

Therefore, the molarity of each ion remaining in the solution, assuming complete precipitation of all insoluble compounds and that volumes are additive, are as follows:
- Sodium (Na+) ion: 0.200 M
- Sulfate (SO₄2-) ion: 0.100 M
- Zinc (Zn2+) ion: 0.300 M
- Chloride (Cl-) ion: 0.600 M
- Barium (Ba2+) ion: 0.200 M
- Cyanide (CN-) ion: 0.400 M

Please! Which statement correctly describes glucose (C6H12O6)?
A)It is made of matter and contains twenty-four molecules.
B)It is an element made of three types of atoms.
C)It is an atom made up of carbon, hydrogen, and oxygen.
D)It is a compound made of twenty-four total atoms of three different elements.

Answers

Answer:

The correct answer is D)It is a compound made of twenty-four total atoms of three different elements

Explanation:

Glucose (C₆H₁₂O₆) is a molecule (not an atom neither an element, so the options B and C are not completely correct). It is composed by three different elements (oxygen: O, carbon: C and hydrogen: H), and it has twenty-four atoms in its molecule: (6 x C) + (12 x H) + (6 x O)= 24. So, the option D is the one that best describes glucose.


C) it is made up of carbon, hydrogen, and oxygen

D) it is a compound made of twenty-four total atoms of three different elements

they are both right but I would go with D