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3 years ago

Who were the scientists who won the nobel prize for the atomic model?

Chemistry

1 answer:

AleksAgata [21]3 years ago

4 0

Answer:

Here's what I find.

Explanation:

Many scientists contributed to our model of the atom.

Among those who received the Nobel Prize for their work are:

1906 — J.J. Thomson — discovery of the electron

1908 — Ernest Rutherford — nuclear model of the atom

1922 — Niels Bohr — planetary model of the atom

1922 — Albert Einstein — quantum mechanical model of the atom

1935 — James Chadwick — discovery of the neutron

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Find the mass of a 60 ML volume of water if the density of water is 1 g/mL

valkas [14]

Answer: 60 grams

Explanation: (60 ml)*(1g/ml) = 60g

3 0

3 years ago

The following reactions have the indicated equilibrium constants at a particular temperature: N2(g) + O2(g) ⇌ 2NO(g) Kc = 4.3 ×

Anuta_ua [19.1K]

Answer:

$Kc=~1.49x10^3^4}$

Explanation:

We have the reactions:

A: $N_2_(_g_) + O_2_(_g_) 2NO_(_g_)~~~~~~Kc = 4.3x10^-^2^5$

B: $2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~~~Kc = 6.4x10^9$

Our target reaction is:

$4NO_(_g_) N_2_(_g_) + 2NO_2_(_g_)$

We have $NO_(_g_)$ as a reactive in the target reaction and $NO_(_g_)$ is present in A reaction but in the products side. So we have to flip reaction A.

A: $2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}$

Then if we add reactions A and B we can obtain the target reaction, so:

A: $2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}$

B: $2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~Kc=6.4x10^9$

For the final Kc value, we have to keep in mind that when we have to add chemical reactions the total Kc value would be the multiplication of the Kc values in the previous reactions.

$4NO_(_g_) N_2_(_g_) + 2NO_2_(_g_)~~~Kc=\frac{6.4x10^9}{4.3x10^-^2^5}$

$Kc=~1.49x10^+^3^4}$

3 0

3 years ago

Write the name for: Mg3O2

sveticcg [70]

Answer:

you sure it is the right sign your using

4 0

2 years ago

Use the ideal gas law to calculate the concentrations of nitrogen and oxygen present in air at a pressure of 1.0 atm and a tempe

notka56 [123]

Answer:

[N₂] = 0.032 M

[O₂] = 0.0086 M

Explanation:

Ideal Gas Law → P . V = n . R . T

We assume that the mixture of air occupies a volume of 1 L

78% N₂ → Mole fraction of N₂ = 0.78

21% O₂ → Mole fraction of O₂ = 0.21

1% another gases → Mole fraction of another gases = 0.01

In a mixture, the total pressure of the system refers to total moles of the mixture

1 atm . 1L = n . 0.082L.atm/mol.K . 298K

n = 1 L.atm / 0.082L.atm/mol.K . 298K → 0.0409 moles

We apply the mole fraction to determine the moles

N₂ moles / Total moles = 0.78 → 0.78 . 0.0409 mol = 0.032 moles N₂

O₂ moles / Total moles = 0.21 → 0.21 . 0.0409 mol = 0.0086 moles O₂

4 0

3 years ago

Two moles of magnesium (Mg) and five moles of oxygen (O2) are placed in a reaction vessel. When magnesium is ignited, it reacts

SOVA2 [1]

Answer:

$\boxed{\text{Mg is the limiting reactant}}$

Explanation:

We are given the amounts of two reactants, so this is a limiting reactant problem.

We know that we will need moles, so, lets assemble the data in one place.

2Mg + O₂ ⟶ 2MgO

n/mol: 2 5

Calculate the moles of MgO we can obtain from each reactant.

From Mg:

The molar ratio of MgO:Mg is 2:2

$\text{Moles of MgO} = \text{2 mol Mg} \times \dfrac{\text{2 mol MgO}}{\text{2 mol Mg}} = \text{2 mol MgO}$

From O₂:

The molar ratio of MgO:O₂ is 2:1.

$\text{Moles of MgO} = \text{5 mol O}_{2} \times \dfrac{\text{2 mol MgO}}{\text{1 mol O}_{2}} = \text{10 mol MgO}\\\\\boxed{\textbf{Mg is the limiting reactant}} \text{ because it gives the smaller amount of MgO}$

6 0

3 years ago