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DiKsa [7]
3 years ago
12

The discovery of the atomic nucleus, or perhaps better stated the development of theory of the nucleus, is credited to Ernest Ru

therford in 1911. He was a kiwi physicist who worked in the laboratory of JJ. Tomson, who developed the earlier Plum Pudding model of the atom. A few years earlier, in 1904, Japanese physicist Hantaro Nagaoka postulated the saturnian model of the atom where he envisioned a very heavy positively charged center of the atom with electrons orbiting much like the rings of Saturn. What are the similarities and differences between the Nagaoka and Rutherford models of the atom? If he developed his theory about 7 years earlier than Rutherford, why is Nagaoka not credited with the discovery of the nucleus? Hint: what's a nucleus anyways
Chemistry
1 answer:
Greeley [361]3 years ago
3 0

Answer:

Both of the studies said that the mass of the atom is centered in the nucleus, which is positive, and there are electrons (negative particles) orbiting it. So, Rutheford and Nagaoka discovered that the atom can be divisible and it has an empty space.

But, in the model of Nagaoka, the nucleus was huge, and for Rutherford, the nucleus was really small, and the mass was concentrated. By his experiment with the gold sheets, the theory was appropriated. That's why Rutherford is credited with the discovery of the nucleus. Nagaoka was incorrect in his suppositions.

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irina1246 [14]

Answer:

The answer is -consumer to consumer interaction

7 0
3 years ago
In a metabolic pathway, succinate dehydrogenase catalyzes the conversion of succinate to fumarate. the reaction is inhibited by
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Competitive inhibitor

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3 years ago
In the laboratory you dissolve 13.9 g of potassium phosphatein a volumetric flask and add water to a total volume of 250mL.
tensa zangetsu [6.8K]

Answer:

Molarity of the solution? 0,262 M.

Concentration of the potassium cation? 0,786 M.

Concentration of the phosphate anion? 0,262 M.

Explanation:

Potassium phosphate (K₃PO₄; 212,27 g/mol) dissolves in water thus:

K₃PO₄ → 3 K⁺ + PO₄³⁻ <em>(1)</em>

Molarity is an unit of chemical concentration given in moles of solute (K₃PO₄) per liters of solution.

There are 250 mL of solution≡0,25 L

The moles of K₃PO₄ are:

13,9 g of K₃PO₄ ×\frac{1mol}{212,27 g} = 0,0655 moles of K₃PO₄

The molarity of the solution is:

\frac{0,0655 moles}{0,25L} = 0,262 M

In (1) you can see that 1 mole of K₃PO₄ produces 3 moles of potassium cation. The moles of potassium cation are:

0,0655 moles×3 = 0,1965 moles

The concentration is:

\frac{0,1965 moles}{0,25L} = 0,786 M

The moles of K₃PO₄ are the same than moles of PO₄³⁻, thus, concentration of phosphate anion is the same than concentration of K₃PO₄. 0,262 M

I hope it helps!

4 0
3 years ago
You mix 200. mL of 0.400M HCl with 200. mL of 0.400M NaOH in a coffee cup calorimeter. The temperature of the solution goes from
GuDViN [60]

Answer : The enthalpy of neutralization is, 56.012 kJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol

\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol

The balanced chemical reaction will be,

HCl+NaOH\rightarrow NaCl+H_2O

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.08 mole of HCl neutralizes by 0.08 mole of NaOH

Thus, the number of neutralized moles = 0.08 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = 200mL+200L=400mL

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 400mL=400g

Now we have to calculate the heat absorbed during the reaction.

q=m\times c\times (T_{final}-T_{initial})

where,

q = heat absorbed = ?

c = specific heat of water = 4.18J/g^oC

m = mass of water = 400 g

T_{final} = final temperature of water = 27.78^oC=273+25.10=300.78K

T_{initial} = initial temperature of metal = 25.10^oC=273+27.78=298.1K

Now put all the given values in the above formula, we get:

q=400g\times 4.18J/g^oC\times (300.78-298.1)K

q=4480.96J

Thus, the heat released during the neutralization = -4480.96 J

Now we have to calculate the enthalpy of neutralization.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of neutralization = ?

q = heat released = -4480.96 J

n = number of moles used in neutralization = 0.08 mole

\Delta H=\frac{-4480.96J}{0.08mole}=-56012J/mole=-56.012kJ/mol

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 56.012 kJ/mole

8 0
3 years ago
How many liters are in 2.75 ounces? Use the conversion factor: 1 liter = 33.814 ounces Rounded to the result to 3 significant fi
Serggg [28]
1L = 33.814 oz
xL = 2.75 oz

so it's a proportion

1L / 33.814 oz = xL / 2.75

solve for x

(1/33.814) * 2.75 = 0.0813272609 on your calculator, but it's not the answer.

the number in your problem, 2.75 oz, has 3 significant figures. so you can only round this number to 3 significant figures too.

your equipment isn't accurate enough to give a reading to 10 significant figures if that makes sense. you have to give the answer in terms of the term you use with the lowest significant figures.

so with 3 significant figures,
0.0813272609 rounds to
0.0813 L
4 0
3 years ago
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