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3 years ago

9

What is the ratio of blue-tailed to orange-talied critter pup?

1 answer:

exis [7]3 years ago

4 0

Blue to orange
blue:orange
$\frac{blue}{orange}$

You might be interested in

the spectral lines observed for hydrogen arise from transitions from excited states back to the n=2 principle quantum level. Cal

Sunny_sXe [5.5K]

Rydberg formula is given by:

$\frac{1}{\lambda } = R_{H}\times (\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} )$

where, $R_{H}$ = Rydberg constant = $1.0973731568508 \times 10^{7} per metre$

$\lambda$ = wavelength

$n_{1}$ and $n_{2}$ are the level of transitions.

Now, for $n_{1}$ = 2 and $n_{2}$ = 6

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{6^{2}} )$

= $1.0973731568508 \times 10^{7} \times (\frac{1}{4}-\frac{1}{36} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.0278 )$

= $1.0973731568508 \times 10^{7} \times 0.23$

= $0.2523958\times 10^{7}$

$\lambda = \frac{1}{0.2523958\times 10^{7}}$

= $3.9620\times 10^{-7} m$

= $396.20\times 10^{-9} m$

= $396.20 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 5

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{5^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.04 )$

= $1.0973731568508 \times 10^{7} \times (0.21 )$

= $0.230 \times 10^{7}$

$\lambda= \frac{1}{0.230 \times 10^{7}}$

= $4.3478 \times 10^{-7} m$

= $434.78\times 10^{-9} m$

= $434.78 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 4

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{4^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.0625 )$

= $1.0973731568508 \times 10^{7} \times (0.1875 )$

= $0.20575 \times 10^{7}$

$\lambda= \frac{1}{0.20575 \times 10^{7}}$

= $4.8602 \times 10^{-7} m$

= $486.02 \times 10^{-9} m$

= $486.02 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 3

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{3^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.12 )$

= $1.0973731568508 \times 10^{7} \times (0.13 )$

= $0.1426585\times 10^{7}$

$\lambda= \frac{1}{0.1426585\times 10^{7}}$

= $7.0097 \times 10^{-7} m$

= $700.97 \times 10^{-9} m$

= $700.97 nm$

5 0

3 years ago

Read 2 more answers

How many atoms are present in a sample of Potassium (K) weighing 33.49g?

madam [21]

Answer:

5.158 × 10²³ atoms K

General Formulas and Concepts:

<u>Chemistry - Atomic Structure</u>

Reading a Periodic Table
Using Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

<u>Step 1: Define</u>

33.49 g K

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

Avogadro's Number

Molar Mass of K - 39.10 g/mol

<u>Step 3: Convert</u>

<u /> $33.49 \ g \ K(\frac{1 \ mol \ K}{39.10 \ g \ K} )(\frac{6.022 \cdot 10^{23} \ atoms \ K}{1 \ mol \ K} )$ = 5.15797 × 10²³ atoms K

<u>Step 4: Check</u>

<em>We are given 4 sig figs. Follow sig figs and round.</em>

5.15797 × 10²³ atoms K ≈ 5.158 × 10²³ atoms K

3 0

3 years ago

Light elements with an atomic number less than 20 generally have a neutron to proton ratio equal to

Natalija [7]

Atomic elements consist of a nucleus that contains protons and neutrons. Protons carry a positive charge whilst neutrons are electrically neutral. In light elements with an atomic number less than 20, the neutron to proton ratio is generally equal to 1:1. This changes for heavier elements since the Coulomb interaction between many protons gets stronger and demands more neutrons for the nucleus to remain stable.

6 0

3 years ago

The equilibrium constant Kc for the reaction below is 0.00584 at a certain temperature. Br2(g) ⇌ 2Br(g) If the initial concentra

Vinil7 [7]

Answer:

Explanation:

Given that:

The chemical equation for the reaction is:

Br2(g) ⇌ 2Br(g)

Initially 0.0345M 0.0416M

$Q_C = \dfrac{[Br]^2}{[Br_2]} = \dfrac{(0.0416)^2}{(0.0345)}= 0.05016$

$Q_C =0.05016 >>> K_c(0.00584)$

Thus, the given reaction will proceed in the backward direction

The I.C.E table is as follows:

Br2(g) ⇌ 2Br(g)

I 0.0345 0.0416

C +x -2x

E (0.0345+x) (0.0416 -2x)

$K_c = \dfrac{[Br]^2}{[Br_2]} = \dfrac{(0.0416-2x)^2}{(0.0345+x)} = 0.00584$

= 0.00173056 - 0.0832x - 0.0832x + 4x² = 0.00584 (0.0345 +x)

= 0.00173056 - 0.166x + 4x² = 2.0148× 10⁻⁴ + 0.00584x

= 0.00173056 - 2.0148× 10⁻⁴ - 0.166x - 0.00584x + 4x²

= 0.00152908 - 0.17184x + 4x²

Solving by using Quadratic formula

x = 0.03038 or 0.0126

For x = 0.03038

At equilibrium

[Br₂] = (0.0345 + 0.03038) = 0.06488 M

[Br] = (0.0416 -2(0.03038)) = - 0.01916 M

Since we have a negative value for [Br], we discard the value for x

For x = 0.0126

At equilibrium

[Br₂] = (0.0345 + 0.0126) = 0.0471 M

[Br] = (0.0416 -2(0.0126)) = 0.0164 M

4 0

3 years ago

When light waves pass through solids, the speed of the wave ___________.

bearhunter [10]

Answer:

i think that the answer is that it would decrease

Explanation:

hope this helps

sorry if the answer is wrong

8 0

3 years ago