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3 years ago

9

MARK ALL of the objects that can be found in our Solar System

2 answers:

lapo4ka [179]3 years ago

7 0

Answer A E F C

Explanation: star, planets, moons, dwarf planets, comets, asteroids, gas, and dust.

Kipish [7]3 years ago

7 0

Answer:A,E,F,C- I hope this could help

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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

2 years ago

Read 2 more answers

Is zinc carbide a metallic bond?

Feliz [49]

Metallic bonds are found in metals like zinc.

3 0

2 years ago

0.50 g of hydrogen chloride (HCl) is dissolved in water to make 4.0 L of solution. What is the pH of the resulting hydrochloric

svlad2 [7]

Explanation:

Given the mass of HCl is ---- 0.50 g

The volume of solution is --- 4.0 L

To determine the pH of the resulting solution, follow the below-shown procedure:

1. Calculate the number of moles of HCl given by using the formula:

$number of moles of a substance=\frac{given mass of the substance}{its molecular mass}$

2. Calculate the molarity of HCl.

3. Calculate pH of the solution using the formula:

$pH=-log[H^+]$

Since HCl is a strong acid, it undergoes complete ionization when dissolved in water.

$HCl(aq)->H^+(aq)+Cl^-(aq)$

Thus, $[HCl]=[H^+]$

Calculation:

1. Number of moles of HCl given:

$number of moles of a substance=\frac{given mass of the substance}{its molecular mass}\\=0.50g/36.5g/mol\\=0.0137mol$

2. Concentration of HCl:

$Molarity of HCl=\frac{number of moles of HCl}{its molar mass}\\=\frac{0.0137 mol}{4.0 L} \\= 0.003425 M$

3. pH of the solution:

$pH=-log[H^+]\\=-log(0.003425)\\=2.47$

Hence, pH of the given solution is 2.47.

3 0

3 years ago

Periods On the periodic table represent elements

Natalka [10]

Answer:When you look at the periodic table, each row is called a period (Get it? Like PERIODic table.). All of the elements in a period have the same number of atomic orbitals. For example, every element in the top row (the first period) has one orbital for its electrons.

Explanation:

7 0

2 years ago

The ratio of the lengths of two sides of a right triangle forms a

Delicious77 [7]

Answer:

perimeter

Explanation:

6 0

3 years ago