What happens to elemental copper when it loses two electrons?

How did Ernest Ruth ford's discovery affect the model of the atom?

Wewaii [24]

By 1911 the components of the atom had been discovered. The atom consisted of subatomic particles called protons and electrons. However, it was not clear how these protons and electrons were arranged within the atom. J.J. Thomson suggested the"plum pudding" model. In this model the electrons and protons are uniformly mixed throughout the atom: Rutherford tested Thomson's hypothesis by devising his "gold foil" experiment. Rutherford reasoned that if Thomson's model was correct then the mass of the atom was spread out throughout the atom. Then, if he shot high velocity alpha particles (helium nuclei) at an atom then there would be very little to deflect the alpha particles. He decided to test this with a thin film of gold atoms. As expected, most alpha particles went right through the gold foil but to his amazement a few alpha particles rebounded almost directly backwards. These deflections were not consistent with Thomson's model. Rutherford was forced to discard the Plum Pudding model and reasoned that the only way the alpha particles could be deflected backwards was if most of the mass in an atom was concentrated in a nucleus. He thus developed the planetary model of the atom which put all the protons in the nucleus and the electrons orbited around the nucleus like planets around the sun.

5 0

3 years ago

Is the follow trait inherited or acquired? “speaking English”

AnnZ [28]

Acquired, because everyone has to learn how to speak English you can’t just be born and know how to speak.

5 0

4 years ago

5. List four examples of neutralization reactions and four for acid-base reactions. You can revisit last week's lesson #

garik1379 [7]

Answer:

Reacciones de neutralización

Mg(OH)2

NaHCO3

Bicarbonato de sodio

Cloruro de sodio

Reacciones acido base

HCl + NaOH → NaCl + H2O.

H2SO4 + 2 NaOH → Na2SO4 + H2O.

HCl + NH3 → NH4Cl + H2O.

HCN + NaOH → NaCN + H2O.

Explanation:

7 0

3 years ago

What volume of nitrogen dioxide is formed at 735 torr and 28.2 °C by reacting 3.56 cm3 of copper (d = 8.95 g/cm3) with 200 mL of

weqwewe [10]

Answer:

25.76 L

Explanation:

Given, Volume of Copper = 3.56 cm³

Density = 8.95 g/cm³

Considering the expression for density as:

$Density=\frac {Mass}{Volume}$

So,

So, Mass= Density * Volume = 8.95 g/cm³ * 3.56 cm³ = 31.862 g

Mass of copper = 31.862 g

Molar mass of copper = 63.546 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{31.862\ g}{63.546\ g/mol}$

Moles of copper = 0.5014 moles

Given, Volume of nitric acid solution = 200 mL = 200 cm³

Density = 1.42 g/cm³

Considering the expression for density as:

$Density=\frac {Mass}{Volume}$

So,

So, Mass= Density * Volume = 1.42 g/cm³ * 200 cm³ = 284 g

Also, Nitric acid is 68.0 % by mass. So,

Mass of nitric acid = $\frac {68}{100}\times 284\ g$ = 193.12 g

Molar mass of nitric acid = 63.01 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{193.12\ g}{63.01\ g/mol}$

Moles of nitric acid = 3.0649 moles

According to the reaction,

$Cu_{(s)}+4HNO_3_{(aq)}\rightarrow Cu(NO_3)_2_{(aq)} + 2NO_2_{(g)} + 2H_2O_{(l)}$

1 mole of copper react with 4 moles of nitric acid

Thus,

0.5014 moles of copper react with 4*0.5014 moles of nitric acid

Moles of nitric acid required = 2.0056 moles

Available moles of nitric acid = 3.0649 moles

Limiting reagent is the one which is present in small amount. Thus, nitric acid is present in large amount, copper is the limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of copper on reaction forms 2 moles of nitrogen dioxide

So,

0.5014 mole of copper on reaction forms 2*0.5014 moles of nitrogen dioxide

Moles of nitrogen dioxide = 1.0028 moles

Given:

Pressure = 735 torr

The conversion of P(torr) to P(atm) is shown below:

$P(torr)=\frac {1}{760}\times P(atm)$

So,

Pressure = 735 / 760 atm = 0.9632 atm

Temperature = 28.2 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (28.2 + 273.15) K = 301.35 K

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.9632 atm × V = 1.0028 mol × 0.0821 L.atm/K.mol × 301.35 K

⇒V = 25.76 L

4 0

3 years ago

Why was Newland’s periodic table rejected?

Rina8888 [55]

He put iron with sulfur and oxygen and it had a few more errors and iron is a metal the other two are nonmetals

5 0

3 years ago