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3 years ago

Some metals corrode very readily. Metals from which groups might be used to provide a protective coating? Justify the answer

Chemistry

1 answer:

Andreyy893 years ago

8 0

Explanation:

Cadmium, nickel, chromium, and silver are sometimes used as protective platings. Metals have a wide range of corrosion resistance. The most active metals (i.e., those that tend to lose electrons easily)--such as magnesium and aluminum--corrode easily and are listed at the top of Table 2-1.

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In the reaction mg 2hcl right arrow. mgcl2 h2 which element’s oxidation number does not change? cl mg h

netineya [11]

The correct answer is Cl.

Chlorine is a substance that's employed in industry and is present in a number of household goods. There are times when chlorine takes the form of toxic gas. Chlorine gas can be converted into a liquid by applying pressure and cooling so that it can be transported and stored. The term "oxidation number" refers to the number of electrons that an element's atom either loses or gains during the production of a compound. The charge that an atom seems to have when forming ionic connections with other heteroatoms is used to define an atom's oxidation number. Even if it develops a covalent bond, an atom with a higher electronegativity is given a negative oxidation state.

Learn more about oxidation numbers here:-

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5 0

1 year ago

Which of the following an ionic bond like what holds Na+ and Cl- together in NaCl?

galina1969 [7]

Answer:

Explanation:

Rb look closely

8 0

3 years ago

Read 2 more answers

Catalyst

Hoochie [10]

a. volume of NO : 41.785 L

b. mass of H2O : 18 g

c. volume of O2 : 9.52 L

<h3>Further explanation</h3>

Given

Reaction

4 NH₃ (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l)

Required

a. volume of NO

b. mass of H2O

c. volume of O2

Solution

Assume reactants at STP(0 C, 1 atm)

Products at 1000 C (1273 K)and 1 atm

a. mol ratio NO : O2 from equation : 4 : 5, so mo NO :

$\tt \dfrac{4}{5}\times 0.5=0.4$

volume NO at 1273 K and 1 atm

$\tt V=\dfrac{nRT}{P}=\dfrac{0.4\times 0.08206\times 1273}{1}=41.785~L$

b. 15 L NH3 at STP ( 1mol = 22.4 L)

$\tt \dfrac{15}{22.4}=0.67~mol$

mol ratio NH3 : H2O from equation : 4 : 6, so mol H2O :

$\tt \dfrac{6}{4}\times 0.67=1$

mass H2O(MW = 18 g/mol) :

$\tt mass=mol\times MW=1\times 18=18~g$

c. mol NO at 1273 K and 1 atm :

$\tt n=\dfrac{PV}{RT}=\dfrac{1\times 35.5}{0.08206\times 1273}=0.34$

mol ratio of NO : O2 = 4 : 5, so mol O2 :

$\tt \dfrac{5}{4}\times 0.34=0.425$

Volume O2 at STP :

$\tt 0.425\times 22.4=9.52~L$

5 0

3 years ago

A student placed 18.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then

mamaluj [8]

Answer:

1.30464 grams of glucose was present in 100.0 mL of final solution.

Explanation:

$Molarity=\frac{moles}{\text{Volume of solution(L)}}$

Moles of glucose = $\frac{18.5 g}{180 g/mol}=0.1028 mol$

Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)

Molarity of the solution = $\frac{0.1028 mol}{0.1 L}=1.028 mol/L$

A 30.0 mL sample of above glucose solution was diluted to 0.500 L:

Molarity of the solution before dilution = $M_1=1.208 mol$

Volume of the solution taken = $V_1=30.0 mL$

Molarity of the solution after dilution = $M_2$

Volume of the solution after dilution= $V_2=0.500L = 500 mL$

$M_1V_1=M_2V_2$

$M_2=\frac{M_1V_1}{V_2}=\frac{1.208 mol/L\times 30.0 mL}{500 mL}$

$M_2=0.07248 mol/L$

Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:

Volume of solution = 100.0 mL = 0.1 L

$0.07248 mol/L=\frac{\text{moles of glucose}}{0.1 L}$

Moles of glucose = $0.07248 mol/L\times 0.1 L=0.007248 mol$

Mass of 0.007248 moles of glucose :

0.007248 mol × 180 g/mol = 1.30464 grams

1.30464 grams of glucose was present in 100.0 mL of final solution.

4 0

3 years ago

Calculate the ph of a 0.20 m solution of iodic acid (hio3, ka = 0.17).

saveliy_v [14]

Iodic acid partially dissociates into H+ and IO3-
Assuming that x is the concentration of H+ at equilibrium, and sine the equation says the same amount of IO3- will be released as that of H+, its concentration is also X. The formation of H+ and IO3- results from the loss of HIO3 so its concentration at equilibrium is 0.20 M - x
Ka = [H+] [IO3-] / [HIO3];
Initially, [H+] ≈ [IO3-] = 0 and [HIO3] = 0.20; 
At equilibrium [H+] ≈ [IO3-] = x and [HIO3] = 0.20 - x; 
so 0.17 = x² / (0.20 - x); 
Solving for x using the quadratic formula: 
x = [H+] = 0.063 M or pH = - log [H+] = 1.2.

7 0

3 years ago