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madam [21]
3 years ago
15

What was Ernest Rutherford's atom model name

Chemistry
2 answers:
Mashcka [7]3 years ago
8 0
It was called the "Rutherford model".
gregori [183]3 years ago
7 0
The model was called The plum pudding atom.
You might be interested in
A series of lines involving a common level in the spectrum of atomic hydrogen lies at 656.46 nm, 486.27 nm, 434.17 nm, and 410.2
alexandr402 [8]

Answer:

1) the wavelength of the next line in the series is 397.2 nm

2) The ionization energy is  3.3996 eV

Explanation:

Step 1: Data given

A series of lines involving a common level in the spectrum of atomic hydrogen lies at 656.46 nm, 486.27 nm, 434.17 nm, and 410.29 nm

Step 2: Calculate n₂

The visible spectrum of light from hydrogen displays four wavelengths, 410 nm, 434 nm, 486 nm, and 656 nm,

1/λ = Rh *(1/n₁² - 1/n₂² )

⇒with λ = the wavelength

⇒with Rh = Rydberg constant for hydrogen, 1.09677583 * 10^7 m

⇒ with n₁ = the principal quantum number of an energy level

⇒with  n₂ = the principal quantum number of an energy level for the atomic electron transition

λ * Rh = n₁²* (n₁+1)² / (2n₁² + 1)

656.46 nm * 109677 cm = n₁²* (n₁+1)² / (2n₁² + 1)

7.20 = n₁²* (n₁+1)² / (2n₁² + 1)

n1 = 2

All those are in the visible spectrum and are called Balmer series, or Balmer lines.

n1 (the principal quantum number of an energy level) for Balmer series is: n1 = 2

Step 3: calculate he wavelength of the next line in the series?

1/λ = Rh *(1/n₁² - 1/n₂² )

 ⇒with  n₂ = the principal quantum number of an energy level for the atomic electron transition = 7

1/λ = 109677.6 / cm * (1/2² - 1/7²)

1/λ = 109677.6 / cm * (1/4 - 1/51.84)

λ = 397.2 nm

the wavelength of the next line in the series is 397.2 nm

Step 4: What is the ionization energy of the atom when it is in the lower state of the transitions?

The energy required to ionize the atom is:

n₂ → ∞

V∞ = 1/λ = 109677.6 / cm * (1/4 - 0)

V∞ = 109677.6 * 1 eV/ 8065.5 cm-1

V∞ = 27419.25 * 1 eV / 8065.5 cm-1

V∞ = 3.3996 eV

The ionization energy is  3.3996 eV

8 0
3 years ago
7.24 miles of Ag(PO4)2
Vladimir [108]
7.42 moles of Ag(PO4)2 x 297.8109g / 1 moles = 2210g Ag(PO4)2
7 0
3 years ago
Construct a three-step synthesis of 3-bromo-3-methyl-2-butanol from 2-methyl-2-butene by dragging the appropriate formulas into
juin [17]

Answer:

See explanation

Explanation:

In order to do this, we need to use 3 reagents to get the final product.

The first one, and logic is the halogenation of the alkene. Doing this, with Br2/CCl4, we'll get an alkane with two bromines, one in carbon 2 and the other in carbon 3.

Then, the next step is to eliminate one bromine of the reactant. The best way to do this, is using sodium ethoxide in ethanol. This is because sodium ethoxide is a relatively strong base, and it will promove the product of elimination in major proportions rather than the sustitution product. If we use NaOH is a really strong base, and it will form another product.

When the sodium ethoxide react, it will form a double bond between carbon 1 and 2 (The carbon where one bromine was with the methyl, changes priority and it's now carbon 3).

The final step, is now use acid medium, such H3O+/H2O or H2SO4/H2O. You can use any of them. This will form an carbocation in carbon 2 (it's a secondary carbocation, so it's more stable that in carbon 1), and then, the water molecule will add to this carbon to form the alcohol.

See the attached picture for the mechanism of this.

4 0
3 years ago
Why is the Hubble telescope located in space? *
dolphi86 [110]

Answer: I’m not one hundred percent sure, but based off of what I know, I believe it is most likely “So it’s images aren’t distorted by the Earth’s atmosphere.”

Explanation:

7 0
2 years ago
(i) Based on the graph, determine the order of the decomposition reaction of cyclobutane at 1270 K. Justify your answer.
Leni [432]

Answer:

(c)(i) The order of the reaction based on the graph provided is first order.

(ii) 99% of the cyclobutane would have decomposed in 53.15 milliseconds.

d) Rate = K [Cl₂]

K = rate constant

The justification is presented in the Explanation provided below.

e) A catalyst is a substance that alters the rate of a reaction without participating or being used up in the reaction.

Cl₂ is one of the reactants in the reaction, hence, it participates actively and is used up in the process of the reaction, hence, it cannot be termed as a catalyst for the reaction.

So, this shows why the student's claim is false.

Explanation:

To investigate the order of a reaction, a method of trial and error is usually employed as the general equations for the amount of reactant left for various orders are known.

So, the behaviour of the plot of maybe the concentration of reactant with time, or the plot of the natural logarithm of the concentration of reactant with time.

The graph given is evidently an exponential function. It is a graph of the concentration of cyclobutane declining exponentially with time. This aligns with the gemeral expression of the concentration of reactants for a first order reaction.

C(t) = C₀ e⁻ᵏᵗ

where C(t) = concentration of the reactant at any time

C₀ = Initial concentration of cyclobutane = 1.60 mol/L

k = rate constant

The rate constant for a first order reaction is given

k = (In 2)/T

where T = half life of the reaction. It is the time taken for the concentration of the reactant to fall to half of its initial concentration.

From the graph, when the concentration of reactant reaches half of its initial concentration, that is, when C(t) = 0.80 mol/L, time = 8.0 milliseconds = 0.008 s

k = (In 2)/0.008 = (0.693/0.008) = 86.64 /s

(ii) Calculate the time, in milliseconds, that it would take for 99 percent of the original cyclobutane at 1270 K to decompose

C(t) = C₀ e⁻ᵏᵗ

when 99% of the cyclobutane has decomposed, there's only 1% left

C(t) = 0.01C₀

k = 86.64 /s

t = ?

0.01C₀ = C₀ e⁻ᵏᵗ

e⁻ᵏᵗ = 0.01

In e⁻ᵏᵗ = In 0.01 = -4.605

-kt = -4.605

t = (4.605/k) = (4.605/86.64) = 0.05315 s = 53.15 milliseconds.

d) The reaction mechanism for the reaction of cyclopentane and chlorine gas is given as

Cl₂ → 2Cl (slow)

Cl + C₅H₁₀ → HCl + C₅H₉ (fast)

C₅H₉ + Cl → C₅H₉Cl (fast)

The rate law for a reaction is obtained from the slow step amongst the the elementary reactions or reaction mechanism for the reaction. After writing the rate law from the slow step, any intermediates that appear in the rate law is then substituted for, using the other reaction steps.

For This reaction, the slow step is the first elementary reaction where Chlorine gas dissociates into 2 Chlorine atoms. Hence, the rate law is

Rate = K [Cl₂]

K = rate constant

Since, no intermediates appear in this rate law, no further simplification is necessary.

The obtained rate law indicates that the reaction is first order with respect to the concentration of the Chlorine gas and zero order with respect to cyclopentane.

e) A catalyst is a substance that alters the rate of a reaction without participating or being used up in the reaction.

Cl₂ is one of the reactants in the reaction, hence, it participates actively and is used up in the process of the reaction, hence, it cannot be termed as a catalyst for the reaction.

So, this shows why the student's claim is false.

Hope this Helps!!!

6 0
3 years ago
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