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romanna [79]
2 years ago
7

In chemistry class, you are given a density column with 3 liquids (shown in the

Chemistry
1 answer:
Zielflug [23.3K]2 years ago
3 0

Answer:

A)

Explanation:

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angiosperms

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dangina [55]
Democritus was the first to propose the idea of the atom. He said the atom was just this tiny, solid sphere. However, he used no scientific evidence to support his claim, so a guy named John Dalton did some experimenting and basically backed up Democritus' claim with evidence. Then, a guy named J.J. Thompson came along and said the atom was not solid and that is consisted of tiny negatively charged particles(electrons) and he came up with the Plum Pudding model which is just a tiny sphere with a punch of random scattered dots in it. After that, Ernest Rutherford did experiments and found that the tiny sphere is made up of mostly empty space with a tiny, dense, positively charged sphere inside of it, and the negatively charged particles just randomly float around it. Neils Bohr then said that the electrons take specific, circular, evenly spaced paths. Then, finally, we come to the Quantum Mechanical Model which is the one accepted today. This model basically vetos Bohr's idea and has a nucleus inside of an electron cloud, which is where the electrons are found.
6 0
3 years ago
Limiting Reactants—————-
denis-greek [22]

Answer:

21.8 grams.

Explanation:

Molar mass data from a modern periodic table:

  • Mg: 24.301;
  • O: 15.999.

How many moles of MgO will be produced if Mg is the limiting reactant?

Number of moles of Mg:

\displaystyle n = \frac{m}{M} = \frac{16.3}{24.301} = 0.670644\;\text{mol}.

The ratio between the coefficient of Mg and that of MgO is 2:2. Two moles of Mg will make two moles of MgO. 0.670644 moles of MgO will be produced if Mg is the limiting reactant.

How many moles of MgO will be produced if O₂ is the limiting reactant?

Number of moles of O₂:

\displaystyle n = \frac{m}{M} = \frac{4.33}{15.999} = 0.270642\;\text{mol}.

The ratio between the coefficient of O₂ and that of MgO is 1:2. One mole of O₂ will make two moles of MgO. 2\times 0.270642 = 0.541284\;\text{mol} of MgO will be produced if O₂ is in excess.

How many moles of MgO will be produced?

0.541284 is smaller than 0.670644. Only 0.541284 moles of MgO will be produced since O₂ will run out before all 16.3 grams of Mg is consumed.

What's the mass of 0.541284 moles of MgO?

Formula mass of MgO:

24.301 + 15.999 = 40.300\;\text{g}\cdot\text{mol}^{-1}.

Mass of 0.541284 moles of MgO:

m = n \cdot M = 0.541284\times 40.300 = 21.8\;\text{g}.

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3 years ago
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barxatty [35]
I think the answer is three
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