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3 years ago

"what is the concentration of positive charge and mass in rutherford's atomic model called?"

1 answer:

3 0

The concentration of positive charge and mass in Rutherford's atomic model is called the nucleus. Rutherford's experiments involving the use of alpha particle beams directed onto thin metal foils demonstrated the existence of the nucleus. The nucleus of an atom contains positively charge particles called protons and other uncharged particles called neutrons. According to this model most volume of an atom is made up of an empty space.

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What is the total E associated with one mole of photons (a unit known as the Einstein) of 3.91x1019 Hz?

sashaice [31]

Answer:

$E=1.56\times 10^{10}J$

Explanation:

The<em> energy of a photon</em>, E, can be calculated with the Planck-Einstein equation:

$E=hf$

Where:

h is Planck's constant 6.626×10⁻³⁴ J.s, and

f is the frequency of the photon or electromagnetic radiation.

Substituting with your data:

$E=6.626\times 10^{-34}J.s\times 3.91\times 10^{19}s^{-1}=2.5908\times 10^{-14}J$

Now multiply by Avogadro's number to obtain the energy of one mole of photons:

$E=2.5908\times 10^{-14}J\times 6.022\times 10^{23}=1.56\times 10^{10}J$

8 0

3 years ago

The practical limit to ages that can be determined by radiocarbon dating is about 41000-yr-old sample, what percentage of the or

Aloiza [94]

Answer:

In percentage, the sample of C-4 remains = 0.7015 %

Explanation:

The Half life Carbon 14 = 5730 year

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac {ln\ 2}{t_{1/2}}$

$k=\frac {ln\ 2}{5730}\ hour^{-1}$

The rate constant, k = 0.000120968 year⁻¹

Time = 41000 years

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

So,

$\frac {[A_t]}{[A_0]}=e^{-0.000120968\times 41000}$

$\frac {[A_t]}{[A_0]}=0.007015$

<u>In percentage, the sample of C-4 remains = 0.7015 %</u>

3 0

3 years ago

Write a balanced half-reaction describing the reduction of gaseous dichlorine to aqueous chloride anions.

Annette [7]

Answer:- $Cl_2(g)+2e^-\rightarrow 2Cl^-(aq)$

Explanations:- In reduction the electrons are accepted and so they are written on the reactant side. When an atom accepts electrons then it forms anion. Chlorine has 7 valence electrons and it needs one more electron to complete it's octet. Since, dichlorine has two Cl atoms and each Cl atom needs one more electron to complete it's octet, two electrons are accepted by dichorine to make aqueous chloride ion. For balancing the equation, there would be two chloride ions as the reactant side has two chlorine atoms.

$Cl_2(g)+2e^-\rightarrow 2Cl^-(aq)$

4 0

3 years ago

Please help me do this Qustion

Helen [10]

I think the answer is D

7 0

3 years ago

When 2−bromo−3,3−dimethylbutane is treated with K⁺ −OC(CH₃)₃, a single product T having molecular formula C₆H₁₂. When 3,3−dimeth

madam [21]

Answer:

T = 3,3-dimethylbut-1-ene

U = 2,3-dimethylbut-2-ene

Explanation:

The chemical reactions for the two reactions and the mechanisms of reaction are shown in the two attachments to this answer.

Hope it helps!

7 0

4 years ago