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2 years ago

12

What is the lightest particle in the following 1)electron 2)proton 3)neutron 4)photon

2 answers:

goldfiish [28.3K]2 years ago

6 0

The answer would be electron

Anuta_ua [19.1K]2 years ago

5 0

Answer:

1) Electron

Explanation:

It carries a negative charge of 1.602176634 × 10−19 coulomb, which is considered the basic unit of electric charge. The rest mass of the electron is 9.1093837015 × 10−31 kg, which is only 1/1,836the mass of a proton.

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The ratio would be 12.6 of carbon acid

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3 years ago

A molecule of composition is replicated in a solution containing unlabeled (not radioactive) GTP, CTP, and TTP plus adenine nucl

Lapatulllka [165]

Answer:

Please find the complete question in the attached file.

Explanation:

It would only be radioactive if the DNA molecule that employed the poly-T rand as templates. Its other molecule of the daughter would not have been radioactive as it did not need dATP for its replication. While each strand of the second molecule includes t, simultaneous reproduction dATP from both daughter molecules is needed so that each of those is radioactive.

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2 years ago

What is the (a) frequency and (b) wavelength (in nm) of a source of purple light at the edge of the visible light spectrum?

stiv31 [10]

Answer: a) $Frequency=666\times10^{12}Hz-789\times10^{12}Hz$

b) $Wavelength=380nm-450nm$

Explanation: Visible light is a form of electromagnetic wave.

Wavelength is defined as the distance between two successive crests of a wave. It is represented as symbol $\lambda$ .

Frequency is defined as the number of complete cycles happening per second. It is represented by the symbol $\nu$ .

Wavelength and Frequency follow inverse relation,

$\nu=\frac{c}{\lambda}$

where, c = speed of light = $3\times10^8m/s$

Wavelength and frequency of a wave is usually expressed in a range.

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7 0

3 years ago

Diatomic iodine [I2] decomposes at high temperature to form I atoms according to the reaction I2(g)⇌2I(g), Kc = 0.011 at 1200∘C

umka2103 [35]

Answer:

The concentration of I at equilibrium = 3.3166×10⁻² M

Explanation:

For the equilibrium reaction,

I₂ (g) ⇄ 2I (g)

The expression for Kc for the reaction is:

$K_c=\frac {\left[I_{Equilibrium} \right]^2}{\left[I_2_{Equilibrium} \right]}$

Given:

$\left[I_2_{Equilibrium} \right]$ = 0.10 M

Kc = 0.011

Applying in the above formula to find the equilibrium concentration of I as:

$0.011=\frac {\left[I_{Equilibrium} \right]^2}{0.10}$

So,

$\left[I_{Equilibrium} \right]^2=0.011\times 0.10$

$\left[I_{Equilibrium} \right]^2=0.0011$

$\left[I_{Equilibrium} \right]=3.3166\times 10^{-2}\ M$

<u>Thus, The concentration of I at equilibrium = 3.3166×10⁻² M</u>

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3 years ago

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Alex17521 [72]

Answer:

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3 years ago

Read 2 more answers