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1 year ago

Which hypothesis of thomson's atomic model was later found to be not true by niels bohr and why?

1 answer:

3 0

The hypothesis of Thomson's atomic model was that an atom is posively charged sphere and in which negatively charged electron are embedded in it but later it was found to be not true by niels Bohr.

The above hypothesis of Thomson's atomic model was false because according to niels Bohr 'The mass of any atoms is localize at the center of an atom which is positively charged while negatively charged electrons revolved around the positive charged'.

The J.J Thomson give the world Thomson model of the atom in 1904 . Thomson model is also known as the plum pudding model .

To learn more about Thomson's atomic model

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A crate holds 20 boxes of sewing supplies. Each box contains 8 spools of thread. Each spool has 250 meters of thread wrapped aro

exis [7]

Answer: 40000

Explanation:

250*8=2,000

2000*20=40000

5 0

1 year ago

If an element has 10 protons and 11 neutrons and 10 electrons, which expression correctly identifies the element? Atomic numbers

Bezzdna [24]

Answer:

Explanation:

Firstly, it should be noted that atomic number (number of protons) determines element. And the element with the atomic number 10 (10 protons) is Neon. Hence, Neon-10 (₁₀Ne) is the answer.

Note that sodium has an atomic number of 11. Also, number of protons is usually equal to the number of electrons in neutral atoms, this is because the total number of positive particles (protons) must be equal to the total number of negative particles (electrons) to give a neutral atom.

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2 years ago

If the cylindrical pistons are 25.000 cm in diameter at 20.0 ∘c, what should be the minimum diameter of the cylinders at that te

attashe74 [19]

Refer to the diagram shown below.

The piston supports the same load W at both temperatures.
The ideal gas law is
$pV=nRT$
where
p = pressure
V = volume
n = moles
T = temperature
R = gas constant

State 1:
T₁ = 20 C = 20+273 = 293 K
d₁ = 25 cm piston diameter

State 2:
T₂ = 150 C = 423 K
d₂ = piston diameter

Because V, n, and R remain the same between the two temperatures, therefore
$\frac{p_{1}}{T_{1}} = \frac{p_{2}}{T_{2}}$

If the supported load is W kg, then
$p_{1} = \frac{W \, N}{ \frac{\pi}{4} d_{1}^{2}} = \frac{4W \, N}{\pi (0.25 \, m)^{2}} = 20.3718W \, Pa$
Similarly,
$p_{2} = \frac{4W}{\pi d_{2}^{2}} \, Pa$

$\frac{p_{1}}{p_{2}} = \frac{20.3718 \pi d_{2}^{2}}{4} = 16 d_{2}^{2}$

Because p₁/p₂ = T₁/T₂, therefore
$16d_{2}^{2} = \frac{293}{423} \\\\ d_{2}^{2} = \frac{0.6927}{16} \\\\ d_{2} = 0.2081 \, m$

The minimum piston diameter at 150 C is 20.8 cm.

Answer: 20.8 cm diameter

8 0

3 years ago

If I have 251.14 grams of citric acid and calcium bicarbonate mixture how much citrus acid do I have??

Snezhnost [94]

About 255.3 grams is the answer I believe. I just had a school work packet and had that question on it, and that is the answer that I had put on it I believe.

6 0

3 years ago

Calculate the solubilities of the following compounds in a 0.02 M solution of barium nitrate using molar concentrations, first i

Law Incorporation [45]

Answer:

a. 1.7 × 10⁻⁴ mol·L⁻¹; b. 5.5 × 10⁻⁹ mol·L⁻¹

c. 2.3 × 10⁻⁴ mol·L⁻¹; 5.5 × 10⁻⁸ mol·L⁻¹

Explanation:

a. Silver iodate

Let s = the molar solubility.

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq); Ksp = 3.0 × 10⁻⁸

E/mol·L⁻¹: s s

$K_{sp} =\text{[Ag$^{+}$][IO$_{3}$$^{-}$]} = s\times s = s^{2} = 3.0\times 10^{-8}\\s = \sqrt{3.0\times 10^{-8}} \text{ mol/L} = 1.7 \times 10^{-4} \text{ mol/L}$

b. Barium sulfate

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq); Ksp = 1.1 × 10⁻¹⁰

I/mol·L⁻¹: 0.02 0

C/mol·L⁻¹: +s +s

E/mol·L⁻¹: 0.02 + s s

$K_{sp} =\text{[Ba$^{2+}$][SO$_{4}$$^{2-}$]} = (0.02 + s) \times s \approx 0.02s = 1.1\times 10^{-10}\\s = \dfrac{1.1\times 10^{-10}}{0.02} \text{ mol/L} = 5.5 \times 10^{-9} \text{ mol/L}$

c. Using ionic strength and activities

(i) Calculate the ionic strength of 0.02 mol·L⁻¹ Ba(NO₃)₂

The formula for ionic strength is

$\mu = \dfrac{1}{2} \sum_{i} {c_{i}z_{i}^{2}}\\\\\mu = \dfrac{1}{2} (\text{[Ba$^{2+}$]}\cdot (2+)^{2} + \text{[NO$_{3}$$^{-}$]}\times(-1)^{2}) = \dfrac{1}{2} (\text{0.02}\times 4 + \text{0.04}\times1)= \dfrac{1}{2} (0.08 + 0.04)\\\\= \dfrac{1}{2} \times0.12 = 0.06$

(ii) Silver iodate

a. Calculate the activity coefficients of the ions

$\log \gamma = -0.51z^{2}\sqrt{I} = -0.051(1)^{2}\sqrt{0.06} = -0.51\times 0.24 = -0.12\\\gamma = 10^{-0.12} = 0.75$

b. Calculate the solubility

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq)

$K_{sp} =\text{[Ag$^{+}$]$\gamma_{Ag^{+}}$[IO$_{3}$$^{-}$]$\gamma_{IO_{3}^{-}}$} = s\times0.75\times s \times 0.75 =0.56s^{2}= 3.0 \times 10^{-8}\\s^{2} = \dfrac{3.0 \times 10^{-8}}{0.56} = 5.3 \times 10^{-8}\\\\s =2.3 \times 10^{-4}\text{ mol/L}$

(iii) Barium sulfate

a. Calculate the activity coefficients of the ions

$\log \gamma = -0.51z^{2}\sqrt{I} = -0.051(2)^{2}\sqrt{0.06} = -0.51\times16\times 0.24 = -0.50\\\gamma = 10^{-0.50} = 0.32$

b. Calculate the solubility

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq

$K_{sp} =\text{[Ba$^{2+}$]$\gamma_{ Ba^{2+}}$[SO$_{4}$$^{2-}$]$\gamma_{ SO_{4}^{2-}}$} = (0.02 + s) \times 0.32\times s\times 0.32 \approx 0.02\times0.10s\\2.0\times 10^{-3}s = 1.1 \times 10^{-10}\\s = \dfrac{1.1\times 10^{-10}}{2.0 \times 10^{-3}} \text{ mol/L} = 5.5 \times 10^{-8} \text{ mol/L}$

7 0

3 years ago