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natka813 [3]
2 years ago
11

List six forms of electromagnetic radiation from the shortest waves(highest energy) to the longest waves (lowest energy)

Physics
1 answer:
romanna [79]2 years ago
8 0
That's the reverse of RIVUXG
so your answer is
gamma rays
x rays
ultraviolet light
visible light
infrared
radio waves
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According to the ideal gas law, a 1.074 mol sample of oxygen gas in a 1.746 L container at 267.6 K should exert a pressure of 13
luda_lava [24]

Answer:

% differ  1.72%

Explanation:

given data:

P_ideal = 13.51 atm

n = 1.074 mol

V = 1.746 L

T = 267.6 K

According to ideal gas law we have

(P+ \frac{n^2 *a}{v^2}) (V - nb) = nRT

(P+ (\frac{1.074^2 *1.360}{1.746^2})) (1.746 - 1.074*3.183*10^{-2}) = 1.074*0.0821*267.6

(P+0.514)(1.711) = 23.59

P_v = 13.276 atm

% differ = \frac{ P_I - P_v}{P_I} *100

             = \frac{13.51 - 13.27}{13.51} *100

             = 1.72%

3 0
3 years ago
Will each of the following actions increase the distance between the diffraction spots? a) Increase the distance between the scr
diamong [38]

Answer:

A) no

B) no

C) yes

D) no

E) yes

5 0
2 years ago
Two common ions of manganese are Mn2+ and Mn3+. Which binary ionic compound is likely to form? A) Mn2N3 B) MnBr4 C) MnCl D) MnO
meriva
Given that Oxygen has an oxidation state of  2 -, you can combine Mn 3+ with O 2- ions to form Mn2O3, and you can combine Mn 2+ with O 2- to form Mn2O2 which is MnO.

The other compounds imply oxidation states of N, Br and Cl that does not exist.

Therefore, the answer is the option D. MnO
4 0
3 years ago
Read 2 more answers
Prior to determining the experimental design, a scientist typically? A. makes observations. B. forms a hypothesis. C. performs a
Umnica [9.8K]

Prior to determining the experimental design, a scientist typically forms a hypothesis. The answer is letter B. this is to prepare the scientist, the possible outcome of their research before the experimental design whether they are wrong or not.

7 0
2 years ago
Two large non-conducting plates of surface area A = 0.25 m 2 carry equal but opposite charges What is the energy density of the
Stells [14]

Answer:

5.1*10^3 J/m^3

Explanation:

Using E = q/A*eo

And

q =75*10^-6 C

A = 0.25

eo = 8.85*10^-12

Energy density = 1/2*eo*(E^2) = 1/2*eo*(q/A*eo)^2 = [q^2] / [2*(A^2)*eo]

= [(75*10^-6)^2] / [2*(0.25)^2*8.85*10^-12]

= 5.1*10^3 J/m^3

8 0
3 years ago
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