Can someone help me please

Put the following names in the correct alphabetic indexing order:(1) Topper & Casey Plumbing(2) KST Enterprises(3) Leland an

dangina [55]

Answer: the correct option is 2, 3, 4, 1.

Explanation:alphabetic indexing order is the order in which files or names are being arranged according to the alphabet. In the following names:

2) KST Enterprises

3)Leland and Son Graphics

4)Lucinda Topper

1) Topper & Casey Plumbing.

While arranging alphabetically, the first letters are usually considered but in a scenario where alphabet occurs twice( 3 And 4) the second letter is considered. I hope this helps, thanks

5 0

3 years ago

Read 2 more answers

How much energy does an X-ray with an 8 nm (8 x 10-9 m) wavelength have?

Vlada [557]

Answer: 2.48×10^-17 J

Explanation:

Given the following :

Wavelength = 8nm (8 x 10^-9 m)

Energy(e) of X-ray =?

Energy=[speed of light(c) × planck's constant (h)] ÷ wavelength

Speed of light = 3×10^8m/s

Planck's constant = 6.626×10^-34 Js

Wavelength = 8 x 10^-9 m

Energy = [(3×10^8) * (6.626×10^-34)] / 8 x 10^-9

Energy = [19.878×10^(8-34)] / 8 x 10^-9

Energy = 2.48475 × 10^(-26+9)

Energy = 2.48×10^-17 J

8 0

3 years ago

Pls can someone help me with this pls

GenaCL600 [577]

Answer:

not sure if this is the right way to answer this question but PbS is Phosphate-buffered saline (i think)

Explanation:

5 0

3 years ago

Arrange the following H atom electron transitions in order of decreasing wavelength of the photon absorbed or emitted:

Katyanochek1 [597]

The decreasing order of wavelengths of the photons emitted or absorbed by the H atom is : b → c → a → d

Rydberg's formula :

$\frac{1}{\lambda} = R_h (\frac{1}{n_1^2}-\frac{1}{n_2^2} )$ ,

where λ is the wavelength of the photon emitted or absorbed from an H atom electron transition from $n_1$ to $n_2$ and $R_h$ = 109677 is the Rydberg Constant. Here $n_1$ and $n_2$ represents the transitions.

(a) $n_1$ =2 to $n_2$ = infinity

$\frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{\infty^2} )$ = 109677/4 [since 1/infinity = 0] Therefore, $\lambda$ = 4 / 109677 = 0.00003647 m

(b) $n_1$ =4 to $n_2$ = 20

$\frac{1}{\lambda} = 109677\times (\frac{1}{4^2}-\frac{1}{20^2} )$ = 6580.62

Therefore, $\lambda$ = 1 / 6580.62 = 0.000152 m

(c) $n_1$ =3 to $n_2$ = 10

$\frac{1}{\lambda} = 109677\times (\frac{1}{3^2}-\frac{1}{10^2} )$ = 11089.56

Therefore, $\lambda$ = 1 / 11089.56 = 0.00009 m

(d) $n_1$ =2 to $n_2$ = 1

$\frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{1^2} )$ = - 82257.75

Therefore, $\lambda$ = 1 /82257.75 = - 0.0000121 m

[Even though there is a negative sign, the magnitude is only considered because the sign denotes that energy is emitted.]

So the decreasing order of wavelength of the photon absorbed or emitted is b → c → a → d.

Learn more about the Rydberg's formula athttps://brainly.com/question/14649374

#SPJ4

8 0

1 year ago

Name: _________________ Temperature o fwater_25_degreecent.YOU MUST SHOW ALL CALCULATIONS TO RECEIVE CREDIT FOR THEM! DATA ANALY

Anton [14]

Answer:

1. 0.02 M

2. 0.01 M

3. 4×10⁻⁶

Explanation:

We know that V₁S₁ = V₂S₂

1.

Concentration of HCl = 0.05 M

end point comes at = 10 ml

So, concentration of OH⁻(aq) = [OH⁻(aq)] ⇒ (0.05 × 10) ÷ 25 ⇒ 0.02 M

2.

2mol of OH⁻(aq) ≡ 1 mole of Ca²⁺(aq)

[Ca²⁺] = 0.02 ÷ 2 = 0.01 M

3.

$K_{sp}$ = [Ca²⁺(aq)] [OH⁻(aq)]²

Ca(OH)₂ (aq) ⇄ Ca²⁺ (aq) + 2OH⁻ (aq)

$K_{sp}$ = [0.01 × (0.02)²] = 4×10⁻⁶

4.

If reaction is exothermic which means heat energy will get evolved as a result temperature of the reaction media will get increased during the course of the reaction. If temperature is externally increased, the reaction will go backward to accumulate extra heat energy.

5.

$K_{sp}$ value describes the solubility of a particular ionic compound. The higher the $K_{sp}$ value, the higher the Solubility will be.

6.

This may be due to uncommon ion effect. The process of other ions (K⁺ or Na⁺) may increase the solubility

4 0

3 years ago