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2 years ago

Which piece of technology allowed scientists to look for life on Mars? a telescope a rover a microscope probeware

Chemistry

2 answers:

Rzqust [24]2 years ago

8 0

Answer:

<h2>A Rover</h2><h2 />

I hope this helps

Nana76 [90]2 years ago

6 0

Answer:

A rover

I hope this helps!

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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

2 years ago

Read 2 more answers

Can someone please help me on #2

Colt1911 [192]

To much letters and numbers

8 0

2 years ago

Using the Rydberg formula, calculate the initial energy level when an electron in a hydrogen atom transitions into n= 2 and emit

EleoNora [17]

Answer:

Explanation:

Using the Rydberg formula as:

$\frac {1}{\lambda}=R_H\times Z^2\times (\frac {1}{n_{1}^2}-\frac {1}{n_{2}^2})$

where,

λ is wavelength of photon

R = Rydberg's constant (1.097 × 10⁷ m⁻¹)

Z = atomic number of atom

n₁ is the initial final level and n₂ is the final energy level

For Hydrogen atom, Z= 1

n₂ = 2

Wavelength = 410.1 nm

Also,

1 nm = 10⁻⁹ m

So,

Wavelength = 410.1 × 10⁻⁹ m

Applying in the formula as:

$\frac {1}{410.1\times 10^{-9}}=1.097\times 10^7\times 1^2\times (\frac {1}{n_{1}^2}-\frac {1}{2^2})$

Solving for n₁ , we get

n₁ ≅ 1

3 0

3 years ago

Read 2 more answers

If electromagnetic radiation has a frequency of 2.0×1015s−1, what is the radiation’s energy?(1 point)

Alika [10]

Frequency=v=2×10^15Hz
Energy=E

Using planks quantum theory

$\\ \sf\longmapsto E=hv$

$\\ \sf\longmapsto E=6.626\times 10^{-34}Js\times 2\times 10^{15}s^{-1}$

$\\ \sf\longmapsto E=13.252\times 10^{-19}J$

$\\ \sf\longmapsto E=1.3\times 10^{-18}J$

7 0

2 years ago

Describe, on a molecular level, how you would expect these lipids to behave in water.

Gelneren [198K]

Lipids are hydrophobic; They would be insoluble, group together, and float to the top

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2 years ago