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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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A sample of phosphonitrilic bromide, PNBr2, contains 2.01 mol of the compound. Determine the amount (in mol) of each element pre

nordsb [41]

Answer:

2.01 moles of P → 1.21×10²⁴ atoms

2.01 moles of N → 1.21×10²⁴ atoms

4.02 moles of Br → 2.42×10²⁴ atoms

Explanation:

We begin from this relation:

1 mol of PNBr₂ has 1 mol of P, 1 mol of N and 2 moles of Br

Then 2.01 moles of PNBr₂ will have:

2.01 moles of P

2.01 moles of N

4.02 moles of Br

To determine the number of atoms, we use the relation:

1 mol has NA (6.02×10²³) atoms

Then: 2.01 moles of P will have (2.01 . NA) = 1.21×10²⁴ atoms

2.01 moles of N (2.01 . NA) = 1.21×10²⁴ atoms

4.02 moles of Br (4.02 . NA) = 2.42×10²⁴ atoms

6 0

3 years ago

Read 2 more answers

For the following example, list the given and unknown information (including gratis or moles)

Setler79 [48]

Answer:

9.6 moles O2

Explanation:

I'll assume it is 345 grams, not gratis, of water. Hydrogen's molar mass is 1.01, not 101.

The molar mass of water is 18.0 grams/mole.

Therefore: (345g)/(18.0 g/mole) = 19.17 or 19.2 moles water (3 sig figs).

The balanced equation states that: 2H20 ⇒ 2H2 +02

It promises that we'll get 1 mole of oxygen for every 2 moles of H2O, a molar ratio of 1/2.

get (1 mole O2/2 moles H2O)*(19.2 moles H2O) or 9.6 moles O2

6 0

2 years ago

Write ionic equations for any three of the following:

worty [1.4K]

The ionic equation is:

H⁺(aq) + NO₃⁻(aq) + Na⁺(aq) + OH⁻(aq) → Na⁺(aq) + NO₃⁻(aq) + H₂O(l)

HNO₃(aq) + NaOH(aq) → NaNO₃(aq) + H2O(l)

The required dissociations are

HNO₃(aq) → H⁺(aq) + NO₃⁻(aq)

NaOH(aq) → Na⁺(aq) + OH⁻(aq)

NaNO₃(aq) → Na⁺(aq) + NO₃⁻(aq)

So, the ionic equation is

H⁺(aq) + NO₃⁻(aq) + Na⁺(aq) + OH⁻(aq) → Na⁺(aq) + NO₃⁻(aq) + H₂O(l)

The ionic equation is

H⁺(aq) + Cl⁻(aq) + K⁺(aq) + OH⁻(aq) → K⁺(aq) + Cl⁻(aq) + H₂O(l)

HCl(aq) + KOH(aq) → KCl(aq) + H2O(l)

The required dissociations are

HCl(aq) → H⁺(aq) + Cl⁻(aq)

KOH(aq) → K⁺(aq) + OH⁻(aq)

KCl(aq) → K⁺(aq) + Cl⁻(aq)

So, the ionic equation is

H⁺(aq) + Cl⁻(aq) + K⁺(aq) + OH⁻(aq) → K⁺(aq) + Cl⁻(aq) + H₂O(l)

The ionic equation is

2H⁺(aq) + SO₄²⁻(aq) + Mg²⁺(aq) + 2OH⁻(aq) → Mg²⁺(aq) + SO₄²⁻(aq)

+ 2H₂O(l)

H₂SO₄(aq) + Mg(OH)₂(aq) → MgSO₄(aq) + 2H2O(l)

The required dissociations are

H₂SO₄(aq) → 2H⁺(aq) + SO₄²⁻(aq)

Mg(OH)₂(aq) → Mg²⁺(aq) + 2OH⁻(aq)

MgSO₄(aq) → Mg²⁺(aq) + SO₄²⁻(aq)

So, the ionic equation is

2H⁺(aq) + SO₄²⁻(aq) + Mg²⁺(aq) + 2OH⁻(aq) → Mg²⁺(aq) + SO₄²⁻(aq)

+ 2H₂O(l)

Learn more about ionic equations here:

brainly.com/question/11628165

5 0

2 years ago

What is the longest wavelength in the Balmer series? (Hint: the Rydberg constant for Hydrogen is 1.096776×107 1/m, and the Balme

boyakko [2]

<u>Answer:</u> The longest wavelength of light is 656.5 nm

<u>Explanation:</u>

For the longest wavelength, the transition should be from n to n+1, where: n = lower energy level

To calculate the wavelength of light, we use Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant = $1.096776\times 10^7m^{-1}$

$n_f$ = Higher energy level = $n_i+1=(2+1)=3$

$n_i$ = Lower energy level = 2 (Balmer series)

Putting the values in above equation, we get:

$\frac{1}{\lambda }=1.096776\times 10^7m^{-1}\left(\frac{1}{2^2}-\frac{1}{3^2} \right )\\\\\lambda =\frac{1}{1.5233\times 10^6m^{-1}}=6.565\times 10^{-7}m$

Converting this into nanometers, we use the conversion factor:

$1m=10^9nm$

So, $6.565\times 10^{-7}m\times (\frac{10^9nm}{1m})=656.5nm$

Hence, the longest wavelength of light is 656.5 nm

4 0

3 years ago

A 0.02 molar solution of sodium chloride contains 0.1 mole of solute. What is the volume of the solution

Deffense [45]

Answer:

Volume of solution = 5 L

Explanation:

Given data:

Molarity of solution = 0.02 M

Moles of solute = 0.1 mol

Volume of solution = ?

Solution:

Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.

Formula:

Molarity = number of moles of solute / L of solution

by putting values,

0.02 M = 0.1 mol / volume of solution

Volume of solution = 0.1 mol / 0.02 M

Volume of solution = 5 L

3 0

2 years ago