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4 years ago

How does Bohr’s model of the atom compare with Thomson’s model?

2 answers:

5 0

Bohr's model of the atom states that electrons orbit an atom's nucleus in electron shells of increasing energy. The electrons are free of physical contact with the nucleus, but are held in orbit due to the electrons' attraction to the oppositely charged protons in the nucleus.
Thomson's model, or the plum pudding model, it describes atoms as spheres of positively charged matter, in which electrons are embedded in.
The key difference is the locations of, and the motions of the electrons.

maria [59]4 years ago

4 0

J.J.Thompson discovered that cathode rays are negatively charged particles. According to J.J.Thompson’s plum pudding model, electrons are negative charges that embedded in a sea of positive charge like a pudding. According to Bohr’s atomic model electrons are found in discrete orbits around the nucleus. Bohr proposed that the electrons have quantized energy hence they do not spin and fall into nucleus thus making an atom highly stable.

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Which arrows represents the phase transitions in which he's energy is gained

prisoha [69]

Answer:

Do you have a picture of this?

6 0

3 years ago

A solution of 0.0470 M HCl is used to titrate 26.0 mL of an ammonia solution of unknown concentration. The equivalence point is

DanielleElmas [232]

The pH at equivalence point is 12.46

At equivalence point, number of moles of acid, n equals number of moles of base, n'

So, n = n'

CV = C'V' where

C = concentration of acid (HCl) = 0.0470 M,
V = volume of acid = 16.0 mL,
C' = concentration of base (ammonia solution) and
V' = volume of base = 26.0 mL.

<h3>Concentration of ammonia solution</h3>

Making C' subject of the formula, we have

C' = CV/V'

Substituting the values of the variables into the equation, we have

C' = CV/V'

C' = 0.0470 M × 16.0 mL/26.0 mL

C' = 0.752 MmL/26.0 mL

C' = 0.0289 M

<h3>The concentration of acid at equivalence point</h3>

We know that the ion-product of water Kw is

Kw = [H⁺][OH⁻] = where

[H⁺] = concentration of HCl at equivalence point,
[OH⁻] = C' = concentration of ammonia solution = 0.0289 M and
Kw = 1.01 × 10⁻¹⁴

Making [H⁺] subject of the formula, we have

[H⁺} = Kw/[OH⁻]

[H⁺] = 1.01 × 10⁻¹⁴/0.0289

[H⁺] = 34.95 × 10⁻¹⁴

[H⁺] = 3.495 × 10⁻¹³

<h3>pH at equivalence point</h3>

Since pH = -㏒[H⁺]

pH = -㏒[3.495 × 10⁻¹³]

pH = -㏒[3.495] + (-㏒10⁻¹³)

pH = -㏒[3.495] + [-13(-㏒10)]

pH = 13 - 0.5434

pH = 12.4566

pH ≅ 12.46

So, the pH at equivalence point is 12.46

Learn more about pH at equivalence point here:

brainly.com/question/25487920

3 0

2 years ago

Help me out please!!

malfutka [58]

Answer: $6.48*10^{-2}$

Explanation:

alright, dawg, lets get this bread. CHEMISTRY? OH YEAH I LOVE CHEMISTRY.

what is a mol? do you know who avogadro is? sounds like avocado. free shavocado. ok so you MUST REMEMBER THIS NUMBER PLEASE.

please remember this number and commit it to your memory: avogadros number

$6.02 * 10^{23}$

this is how much a mole is. you know how a pair is 2 and a dozen is 12? ok so a mole is $6.02 * 10^{23}$ it is confusing at first but hopefully this helps you to understand.

now that we understand this..... lets perform this calculation with a calculator

$(3.90 * 10^{22}) / (6.02 * 10^{23})$

notice i divide the question by the avogadros number to find out how many moles are in the number. ok but listen... it gets into a tough area here... because HOW ARE WE TO DIVIDE SUCH A HUMONGOUS NUMBER BY ANOTHER HUMONGOUS NUMBER?!?!?

its easy, its cake, just listen this is how you do it. only focus on the numbers NOT the 10 exponential ones. so just 3.90 and 6.02 ok? lets divide these two numbers 3.90 / 6.02 and we get 0.6478... how interesting... ok now lets deal with the exponents of 10. notice that we are DIVIDING these numbers so think of it as subtracting the exponents of ten..... 22 minus 23 equals -1

so we have $0.6478 * 10^{-1}$

now this negative 1 thing is annoying so lets just make it to the power of 0

$0.06478 * 10^{0}$

and anything to the power of 0 just becomes 1.

0.06478

so this is our answer but keep in mind we need 3 sig figs. if we round then we get 0.0648

put this into scientific notation we get $6.48*10^{-2}$

5 0

3 years ago

At 35.0°c and 3.00 atm pressure, a gas has a volume of 1.40 l. what pressure does the gas have at 0.00°c and a volume of 0.950 l

Leona [35]

Answer : The pressure of gas will be, 3.918 atm and the combined gas law is used for this problem.

Solution :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 3 atm

$P_2$ = final pressure of gas = ?

$V_1$ = initial volume of gas = 1.40 L

$V_2$ = final volume of gas = 0.950 L

$T_1$ = initial temperature of gas = $35^oC=273+35=308K$

$T_2$ = final temperature of gas = $0^oC=273+0=273K$

Now put all the given values in the above equation, we get the final pressure of gas.

$\frac{3atm\times 1.40L}{308K}=\frac{P_2\times 0.950L}{273K}$

$P_2=3.918atm$

Therefore, the pressure of gas will be, 3.918 atm and the combined gas law is used for this problem.

4 0

3 years ago

Read 2 more answers

A 25.00 ml sample of h3po4 solution is neutralized by exactly 54.93 ml of 0.04345 m ca(oh)2. what is the molarity of the h3po4 s

son4ous [18]

The neutralization equation is:
3 Ca(OH)₂ + 2 H₃PO₄ → Ca₃(PO₄)₂ + 6 H₂O

From this equation we can see that 3 moles of Ca(OH)₂ react with 2 moles of H₃PO₄

Numbers of mmol of Ca(OH)₂ = M x V = 0.04345 x 54.93 = 2.387 mmol

Number of mmol of H₃PO₄ = 2.387 x (2/3) = 1.591 mmol

Molarity of solution = n (mmol) / V(ml) = 1.591 / 25 = 0.0636 M

8 0

3 years ago

Read 2 more answers