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3 years ago

The deflection of alpha particles in Rutherford’s gold foil experiments resulted in what change to the atomic model?

2 answers:

5 0

Hello.

This experiment proved that atoms are not massive, they have large blanks inside the structure. Therefore, models like Thomsom's and Dalton's weren't good enough to describe the comportment of these units. A model of an atom like a <em>planetary system</em> was elaborated after this.

STALIN [3.7K]3 years ago

4 0

Rutherford performed the gold foil experiment and the observation was as follows:

Most of the alpha particles passed undeflected through the atom proving that most of the atom is empty space.
A certain fraction of alpha particles deflected by 90 degrees that indicated the center of the atom is positive.
Another small fraction of alpha particles deflected by more than 90 degrees that indicated that most of the atom's mass lies in a small volume at the center of the atom.

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A crow flies 100 meters east in 5 minutes. What is the crow’s velocity?

torisob [31]

<em>Answer:</em> 20 m/min east

<em>Explanation:</em> The speed is distance divided by time

distance between start-point and end-point = 100 meters

<em>time to cover the distan</em>ce = 5 minutes

100 m ÷ 5 min = 20 m/min east

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3 0

3 years ago

What does the electron sea model for metals suggest?

fgiga [73]

Answer:

electron sea model for metals suggest that valence electrons drift freely around the metal cations.

Explanation:

Explanation: In electron sea model, the valence electrons in metals are delocalized instead of orbiting around the nucleus. ... These electrons are free to move within the metal atoms. Thus, we can conclude that the electron sea model for metals suggest that valence electrons drift freely around the metal cations.

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4 years ago

1. Two boys balanced steady in a sea-saw game

anygoal [31]

Mutual

They are balanced steadily which means they’re at the same point

4 0

3 years ago

Read 2 more answers

A 32.5 g iron rod, initially at 22.4 ∘C, is submerged into an unknown mass of water at 63.0 ∘C, in an insulated container. The f

a_sh-v [17]

Answer:

$m_{H_2O}=39.0g$

Explanation:

Hello,

In this case, is possible to infer that the thermal equilibrium is governed by the following relationship:

$\Delta H_{iron}=-\Delta H_{H_2O}\\m_{iron}Cp_{iron}(T_{eq}-T_{iron})=-m_{H_2O}Cp_{H_2O}(T_{eq}-T_{H_2O})$

Thus, both iron's and water's heat capacities are: 0.444 and 4.18 J/g°C respectively, so one solves for the mass of water as shown below:

$m_{H_2O}=\frac{m_{iron}Cp_{iron}(T_{eq}-T_{iron})}{-Cp_{H_2O}(T_{eq}-T_{H_2O}} \\\\m_{H_2O}=\frac{32.5g*0.444\frac{J}{g^0C}*(59.7-22.4)^0C}{-4.18\frac{J}{g^0C}*(59.7-63.0)^0C} \\\\m_{H_2O}=39.0g$

Best regards.

8 0

3 years ago

How Many Moles Of HCl Need To Be Added To 150.0 ML Of 0.50 M NaZ To Have A Solution With A PH Of 6.50

Aleks04 [339]

The number of mole of HCl needed for the solution is 1.035×10¯³ mole

<h3>How to determine the pKa</h3>

We'll begin by calculating the pKa of the solution. This can be obtained as follow:

Equilibrium constant (Ka) = 2.3×10¯⁵
pKa =?

pKa = –Log Ka

pKa = –Log 2.3×10¯⁵

pKa = 4.64

<h3>How to determine the molarity of HCl </h3>

pKa = 4.64
pH = 6.5
Molarity of salt [NaZ] = 0.5 M
Molarity of HCl [HCl] =?

pH = pKa + Log[salt]/[acid]

6.5 = 4.64 + Log[0.5]/[HCl]

Collect like terms

6.5 – 4.64 = Log[0.5]/[HCl]

1.86 = Log[0.5]/[HCl]

Take the anti-log

0.5 / [HCl] = anti-log 1.86

0.5 / [HCl] = 72.44

Cross multiply

0.5 = [HCl] × 72.44

Divide both side by 72.44

[HCl] = 0.5 / 72.4

[HCl] = 0.0069 M

<h3>How to determine the mole of HCl </h3>

Molarity of HCl = 0.0069 M
Volume = 150 mL = 150 / 1000 = 0.15 L

Mole of HCl =?

Mole = Molarity x Volume

Mole of HCl = 0.0069 × 0.15

Mole of HCl = 1.035×10¯³ mole

<h3>Complete question</h3>

How many moles of HCl need to be added to 150.0 mL of 0.50 M NaZ to have a solution with a pH of 6.50? (Ka of HZ is 2.3 x 10 -5 .) Assume negligible volume of the HCl

Learn more about pH of buffer:

brainly.com/question/21881762

7 0

2 years ago