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Assoli18 [71]
3 years ago
11

The scientists involved in this research are specialists in

Chemistry
1 answer:
Rudik [331]3 years ago
5 0

Answer:

Sep 23, 2019 — It's in everyone's benefit if physicians participate in research. By Mukesh K. Jain, Tadataka Yamada and Robert Lefkowitz. Drs. Jain, Yamada ...

Explanation:

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One type of electromagnetic radiation has a frequency of 107.1MHz,another type has wavelength of of 2.12x10^-10 m,and another ty
Andreas93 [3]

Each type of electromagnetic radiation and placed in order of increasing photon energy and increasing frequency

  • FM Radio waves < Visible light < X - rays
  • Order of increasing frequency.
  • FM Radio waves < Visible light < X - rays

This is further explained below.

<h3>What is electromagnetic radiation?</h3>

Parameters

Frequency =107 100 000 hertz

Wavelength, λ = c / ν => 2.8 m => 2.12 * 10-10 m

Energy = 3.97 * 10 -19 J/ photon.

Generally, the equation for Wavelength is  mathematically given as

λ = h c / E

Therefore

λ =6.626 * 10-34  * 3 * 108  / (3.97* 10 -19 J)

λ =5 * 10-7 m

In conclusion, This is considered to be part of the viewable area (green Increasing photon energy from lowest to highest.

  • FM Radio waves < Visible light < X - rays
  • Order of increasing frequency.
  • FM Radio waves < Visible light < X - rays

Read more about electromagnetic radiation

brainly.com/question/11895806

#SPJ1

5 0
1 year ago
For this exercise, you can simulate the described conditions by changing the values in the run experiment tool of the simulation
Furkat [3]
1) ideal gas law: p·V = n·R·T.
p - pressure of gas.
V -volume of gas.
n - amount of substance.
R - universal gas constant.
T - temperature of gas.
n₁ = 0,04 mol, V₁ = 0,06 l.
n₂ = 0,07 mol, V₂ = 0,06 · 0,07 ÷ 0,04 = 0,105 l.
2) V₁ = 0,06 l, T₁ = 240,00 K.
T₂ = 340,00 K, V₂ = 340 · 0,06 ÷ 240 = 0,05 l.
3 0
2 years ago
Read 2 more answers
If 1.08 g of sodium sulfate reacts with an excess of phosphoric acid, how much sulfuric acid is produced?
vfiekz [6]

Answer:  0.745 g of H_2SO_4 will be produced from  1.08 g of sodium sulfate

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} Na_2SO_4=\frac{1.08g}{142.04g/mol}=0.0076moles  

3Na_2SO_4+2H_3PO_4\rightarrow 2Na_3PO_4+3H_2SO_4

Na_2SO_4 is the limiting reagent as it limits the formation of product and H_3PO_4 is the excess reagent.

According to stoichiometry :

3 moles of Na_2SO_4 produce = 3 moles of H_2SO_4

Thus 0.0076 moles of Na_2SO_4 will require=\frac{3}{3}\times 0.0076=0.0076moles  of H_2SO_4

Mass of H_2SO_4=moles\times {\text {Molar mass}}=0.0076moles\times 98.1g/mol=0.745g

Thus 0.745 g of H_2SO_4 will be produced from  1.08 g of sodium sulfate

3 0
2 years ago
An aqueous solution containing 17.5 g of an unknown molecular (nonelectrolyte) compound in 100.0 g of water has a freezing point
Sonbull [250]

Answer:

molar mass = 180.833 g/mol

Explanation:

  • mass sln = mass solute + mass solvent

∴  solute: unknown molecular (nonelectrolyte)

∴ solvent: water

∴ mass solute = 17.5 g

∴ mass solvent =  100.0 g = 0.1 Kg

⇒ mass sln = 117.5 g

freezing point:

  • ΔTc = - Kc×m

∴ ΔTc = -1.8 °C

∴ Kc H2O = 1.86 °C.Kg/mol

∴ m: molality (mol solute/Kg solvent)

⇒ m = ( - 1.8 °C)/( - 1.86 °C.Kg/mol)

⇒ m = 0.9677 mol solute/Kg solvent

  • molar mass (Mw) [=] g/mol

∴ mol solute = ( m )×(Kg solvent)

⇒ mol solute = ( 0.9677 mol/Kg) × ( 0.100 Kg H2O )

⇒ mol solute = 0.09677 mol

⇒ Mw solute = ( 17.5 g ) / ( 0.09677 mol )

⇒ Mw solute = 180.833 g/mol

6 0
2 years ago
What is the daughter nucleus (nuclide) produced when 64 Cu Cu64 undergoes beta decay by emitting an electron? Replace each quest
kenny6666 [7]

<u>Answer:</u> The daughter nuclide formed by the beta decay of given isotope is _{30}^{64}\textrm{Zn}

<u>Explanation:</u>

Beta decay is defined as the process in which beta particle is emitted. In this process, a neutron gets converted to a proton and an electron.

The released beta particle is also known as electron.

_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta

We are given:

Parent isotope = _{29}^{64}\textrm{Cu}

The chemical equation for the beta decay process of _{29}^{64}\textrm{Cu} follows:

_{29}^{64}\textrm{Cu}\rightarrow _{30}^{64}\textrm{Zn}+_{-1}^0\beta

Hence, the daughter nuclide formed by the beta decay of given isotope is _{30}^{64}\textrm{Zn}

4 0
3 years ago
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