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Yuri [45]
3 years ago
8

Plz help (it’s a picture)

Chemistry
1 answer:
Feliz [49]3 years ago
7 0
Umm I’ll figure it out rn! Will come back in 1 min
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based on the cause and effect relationship between the magnets, which phenomenon is the teacher modeling for her students.
faltersainse [42]

Answer:

Someone breaking up.

Explanation:

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2 years ago
Standard reduction half-cell potentials at 25∘c half-reaction e∘ (v) half-reaction e∘ (v) au3 (aq) 3e−→au(s) 1.50 fe2 (aq) 2e−→f
Zanzabum

<em>K</em> = 5.0 × 10^25

<h2>Part (a). Calculate <em>E</em>° for the reaction </h2>

<em>Step 1.</em> Write the equations for the two half-reactions

2H^(+)(aq) + 2e^(-) → H2(g); _0.00 V

Zn^(2+)(aq) + 2e^(-) → Zn(s); -0.76 V

<em>Step 2.</em> Identify the cathode and the anode

The half-cell with the more negative <em>E</em>° (Zn) is the anode.

<em>Step 3.</em> Calculate <em>E</em>°

Zn(s) → Zn^(2+)(aq) + 2e^(-); _________+0.76 V

2H^(+)(aq) + 2e^(-) → H2(g); __________0.00 V

Zn(s) + 2H^(+)(aq) → Zn^(2+)(aq) + H2(g); +0.76 V

<em>E</em>° = +0.76 V

<h2>Part (b). Calculate <em>K</em> for the reaction </h2>

The relation between <em>E</em>° and <em>K</em> is

<em>E</em>° = (<em>RT</em>)/(<em>nF</em>)ln<em>K </em>

where

<em>R</em> = the universal gas constant: 8.314 J·K^(-1)mol^(-1)

<em>T</em> = the Kelvin temperature

<em>n</em> = the moles of electrons transferred

<em>F</em> = the Faraday constant: 96 485 J·V^(-1)mol^(-1)

Then

0.76 V = [8.314 J·K^(-1)mol^(-1) × 298.15 K]/[2 × 96 485 J·V^(-1)mol^(-1)]ln<em>K</em>

0.76 = 0.012 85 ln<em>K</em>

ln<em>K</em> = 0.76/0.012 85 = 59.16

<em>K</em> =e^59.16 = 5.0 × 10^25

4 0
3 years ago
Calculate the radius of a tantalum (ta) atom, given that ta has a bcc crystal structure, a d ensity of 16.6 g/cm3, and an atomic
Sidana [21]

solution:

Ta is BCC, a=\frac{4}{\sqrt{3}r},n=2atoms/cell\\p=16.6gram/cm^3,A=180.9gram/mol\\P=\frac{A\times n}{N_{a}\times v}=\frac{180.9\frac{gram}{mol}\times 2 \frac{atom}{cell}}{6.023\times10^23\frac{atom}{mol}\times(\frac{4}{\sqrt{3}\times r})^3\frac{cm^3}{cell}}\\=16.6\frac{gram}{cm^3}\\r=(\frac{180.9\times2}{16.6\times6.023\times10^23\times(\frac{4}{\sqrt{3}})^3})^\frac{1}{3}\times10^8


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3 years ago
A researcher claims that an ancient scroll originated from greek scholars in about 500 bce. a measure of its carbon-14 decay rat
Naddik [55]
The amount of substance present in a certain object with a given half-life in terms of h can be expressed through the equation,

     A(t) = (A(o))(0.5)^(t/h)

where A(t) is the amount of substance after t years and A(o) is the original amount. In this item we are given that A(t)/A(o) is equal to 0.89. Substituting the known values,

     0.89 = (0.5)(t / 5730 years)

The value of t from the equation is 963.34 years.

<em>Answer: 963 years</em>
8 0
3 years ago
Draw a schematic of the hydrogen atom with the single proton in the nucleus, and the n=1, n=2, n=3, and n=4 energy level options
max2010maxim [7]

Explanation:

According to Bohr's postulates, the electron in the present in the lower energy level can absorb energy and exits to higher energy level. Also, when this electron returns back to its orbit, it emits some energy.

Since the hydrogen consists of 1 electron and 1 proton. The lowest energy configuration of the hydrogen is when n =1 or, when the electron is present in the K-shell or the ground state.

The possible transition for the electron given in the question is :

n = 2, 3 and 4

The schematic diagram of the hydrogen atom consisting of these four quantum levels in which the electron can jump (Absorption) and comeback to from these energy levels (emission) .

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3 years ago
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