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Kazeer [188]
3 years ago
5

Match the scientist with their scientific idea. Question 2 options: Substances combine or break apart to create new, different s

ubstances. Gases are formed from tiny particles so small you can’t see them. The particles are spread out into a certain area and move when people walk through them. Matter was made of different kinds of things. 1. Democritus 2. Bernoulli 3. Priestley
Chemistry
1 answer:
aev [14]3 years ago
4 0

Answer:

1. Democritus - Matter was made of different kinds of things.

2. Bernoulli- Gases are formed from tiny particles so small you can’t see them. The particles are spread out into a certain area and move when people walk through them.

3. Priestley- Substances combine or break apart to create new, different substances.

Explanation:

1. Democritus was a philosopher who because of his immense contribution to atomic theory was regarded as the "FATHER OF SCIENCE". His idea was that matter was made of different kinds of things called which he called ATOMOS meaning ATOM.

2. Daniel Bernoulli was another outstanding mathematician and scientist who played a major part in the kinetic molecular theory of gases via his "HYDRODYNAMICA". His idea, which states that "gases are formed from tiny particles so small you can’t see them. The particles are spread out into a certain area and move when people walk through them"

was based on fluid dynamics.

3. Joseph Priestley is remarkably known for his 1774 discovery of oxygen. His findings that air is a mixture of gases and not just a substance made of one element paved the way for his idea that " Substances combine or break apart to create new, different substances".

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Read 2 more answers
combustion analysis of a hydrocarbon produced 33.01g CO2 and 13.51g H2O. Calculate the empirical formula for the hydrocarbon
masya89 [10]

Answer:

\rm CH_2.

Explanation:

Carbon and hydrogen are the only two elements in a hydrocarbon. When a hydrocarbon combusts completely in excess oxygen, the products would be \rm CO_2 and \rm H_2O. The \rm C and \rm H would come from the hydrocarbon, while the \rm O atoms would come from oxygen.

Look up the relative atomic mass of these three elements on a modern periodic table:

  • \rm C: 12.011.
  • \rm H: 1.008.
  • \rm O: \rm 15.999.

Calculate the molar mass of \rm CO_2 and \rm H_2O:

M(\mathrm{CO_2}) = 12.011 + 2 \times 15.999 = 44.009\; \rm g \cdot mol^{-1}.

M(\mathrm{H_2O}) = 2 \times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}

Calculate the number of moles of \rm CO_2 molecules in 33.01\; \rm g of \rm CO_2\!:

\displaystyle n(\mathrm{CO_2}) = \frac{m(\mathrm{CO_2})}{M(\mathrm{CO2})} = \frac{33.01\; \rm g}{44.009\; \rm g\cdot mol^{-1}} \approx 0.7501\; \rm mol.

Similarly, calculate the number of moles of \rm H_2O molecules in 13.51\; \rm g of \rm H_2O\!:

\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{13.51\; \rm g}{18.015\; \rm g\cdot mol^{-1}} \approx 0.7499\; \rm mol.

Note that there is one carbon atom in every \rm CO_2 molecule. Approximately0.7501\; \rm mol of \rm CO_2\! molecules would correspond to the same number of \rm C atoms. That is: n(\mathrm{C}) \approx 0.7501\; \rm mol.

On the other hand, there are two hydrogen atoms in every \rm H_2O molecule. approximately 0.7499\; \rm mol of \rm H_2O molecules would correspond to twice as many \rm H\! atoms. That is: n(\mathrm{H}) \approx 2 \times 0.7499 \; \rm mol\approx 1.500\; \rm mol.

The ratio between the two is: n(\mathrm{C}): n(\mathrm{H}) \approx 1:2.

The empirical formula of a compound gives the smallest whole-number ratio between the elements. For this hydrocarbon, the empirical formula would be \rm CH_2.

6 0
3 years ago
The half-life of cesium-137 is 30 years. Suppose we have a 200-mg sample. (a) Find the mass that remains after t years.
scZoUnD [109]

Given:

Half life(t^ 1/2) :30 years

A0( initial mass of the substance): 200 mg.

Now we know that

A= A0/ [2 ^ (t/√t)]

Where A is the mass that remains after t years.

A0 is the initial mass

t is the time

t^1/2 is the half life

Substituting the given values in the above equation we get

A= [200/ 2^(t/30) ] mg


Thus the mass remaining after t years is [200/ 2^(t/30) ] mg

5 0
3 years ago
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