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2 years ago

Why does the atomic radius decrease across a period

1 answer:

4 0

Because the nuclear charge increases across a period and so it has a stronger pull on the outer electrons and will pull in the radius

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What three compounds are produced by the combustion of hydrocarbon fuels?

MakcuM [25]

The most common hydrogen carbon fuels are ethanol and diesel and their product of combustion is carbon dioxide, water and heat .

8 0

2 years ago

In order to become stable, Potassium, K, will _______ and have a resulting charge of _____.

svlad2 [7]

Answer:

Explanation:

it doesn't matter how many protons k has all the answers have electron with the positive charge when is negative the question is trying to distract you D is the only one that would use the electrons sign correctly

4 0

2 years ago

What is the Ka of a 0.0796 M solution of nitrous acid (HNO2) with a pH of 2.95?

Vadim26 [7]

Answer:

Coefficient = 1.58

Exponent = - 5

Explanation:

pH = 2.95

Molar concentration = 0.0796M

Ka = [H+]^2 / [HA]

Ka = [H+]^2 / 0.0796

Therefore ;

[H+] = 10^-2.95

[H+] = 0.0011220 = 1.122 × 10^-3

Ka = [H+] / molar concentration

Ka = [1.122 × 10^-3]^2 / 0.0796

Ka = (1.258884 × 10^-6) / 0.0796

Ka = 15.815 × 10^-6

Ka = 1.58 × 10^-5

Coefficient = 1.58

Exponent = - 5

4 0

3 years ago

What is the mass of 3.77mol of K3N?

AleksAgata [21]

<h3>Answer:</h3>

495 g K₃N

<h3>General Formulas and Concepts:</h3>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.77 mol K₃N

<u>Step 2: Identify Conversions</u>

Molar Mass of K - 39.10 g/mol

Molar Mass of N - 14.01 g/mol

Molar Mass of K₃N - 3(39.10) + 14.01 = 131.31 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 3.77 \ mol \ K_3N(\frac{131.31 \ g \ K_3N}{1 \ mol \ K_3N})$
Multiply/Divide: $\displaystyle 495.039 \ g \ K_3N$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

495.039 g K₃N ≈ 495 g K₃N

5 0

2 years ago

Calculate the wavelength, in nanometers, of the spectral line produced when an electron in a hydrogen atom undergoes the transit

Lapatulllka [165]

<u>Answer:</u> The wavelength of spectral line is 656 nm

<u>Explanation:</u>

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

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

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant = $1.097\times 10^7m^{-1}$

$n_f$ = Final energy level = 2

$n_i$ = Initial energy level = 3

Putting the values in above equation, we get:

$\frac{1}{\lambda }=1.097\times 10^7m^{-1}\left(\frac{1}{2^2}-\frac{1}{3^2} \right )\\\\\lambda =\frac{1}{1.524\times 10^6m^{-1}}=6.56\times 10^{-7}m$

Converting this into nanometers, we use the conversion factor:

$1m=10^9nm$

So, $6.56\times 10^{-7}m\times (\frac{10^9nm}{1m})=656nm$

Hence, the wavelength of spectral line is 656 nm

6 0

2 years ago