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3 years ago

87 g of oxygen gas would occupy how many liters of volume at stp

1 answer:

8 0

Number of mole : mass/ molar mass

87g/ (2 x 16) = 2.719 mol

volume = number of mole x molar volume
volume = 2.719 x 22.4
volume = 60.906 L

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Using the balanced equation below, how many moles of water can be produced when 4 grams of Hydrogen react? 2H2(g)+O2(g) —> 2H

gavmur [86]

Answer:

4 × 10 g

Explanation:

Step 1: Write the balanced equation

2 H₂(g) + O₂(g) ⇒ 2 H₂O(I)

Step 2: Calculate the moles corresponding to 4 g of H₂

The molar mass of H₂ is 2.02 g/mol.

4 g × 1 mol/2.02 g = 2 mol

Step 3: Calculate the moles of H₂O produced from 2 moles of H₂

The molar ratio of H₂ to H₂O is 2:2. The moles of H₂O produced are 2/2 × 2 mol = 2 mol.

Step 4: Calculate the mass corresponding to 2 moles of H₂O

The molar mass of H₂O is 18.02 g/mol.

2 mol × 18.02 g/mol = 4 × 10 g

7 0

3 years ago

How many molecules are there in 5H20?

Svetach [21]

Answer:

The number before any molecular formula applies to the entire formula. So here you have five molecules of water with two hydrogen atoms and one oxygen atom per molecule. Thus you have ten hydrogen atoms and five oxygen atoms in total.

3 0

3 years ago

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Calculate the mass, in grams, of Avogadro's number of fluorine atoms. Enter your answer in the provided box. I g/mol

juin [17]

Answer:

Every substance, whether an element or a compound that contains Avogadro's number, is said to form a mole of it. Therefore for the fluorine atoms that form the Avogadro number we say that we have one mole of fluorine atoms and this weighs 19 g, so 19 g/mol

Explanation:

You can search the F in the Periodic Table where you have the molar mass, 19 g/mol but if you don't have a Periodic Table you can calculate like this:

1 atom of F weighs 19 Da (unified atomic mass unit) and 1 Da is 1,661 x10*-24 g, so try to apply a rule of three, twice.

1 Da ........... 1,661x10*-24 g

19 Da .......... x

x= (19 Da . 1,661x10*-24 g ) / 1 Da = 3,1559x10*-23 g

1 atom F .............. 3,1559x10*-23 g

6,022x10*23 atoms F ............. x

X= (6,022x10*23 atoms F . 3,1559x10*-23 g) / 1 atom F = 19.0048 g

3 0

3 years ago

How many moles of sulfur are needed to produce 15.0 mol of sulfur dioxide?

Brut [27]

<span>S + O2 = SO2

1 mole S ------------- 1 mole SO2
( moles S ) --------- 15.0 moles SO2

moles S = 15.0 x 1 / 1

moles S = 15.0 / 1

= 15.0 moles S</span>

4 0

3 years ago

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If a single gold atom has a diameter of 2.9×x10−8 cm, how many atoms thick was rutherford's foil

ale4655 [162]

27,586

<h3>Further explanation</h3>

<u>Given:</u>

A single gold atom has a diameter of $\boxed{ \ 2.9 \times 10^{-8} \ cm. \ }$

From a reference, the Rutherford gold foil used in his scattering experiment had a thickness of approximately $\boxed{ \ 8 \times 10^{-3} \ mm. \ }$

<u>Question:</u>

How many atoms thick were Rutherford's foil?

<u>The Process:</u>

Convert thickness from mm to cm.

$\boxed{ \ 8 \times 10^{-3} \ mm = 8 \times 10^{-3} \times 10{-1} \ cm \ } \rightarrow \boxed{ \ 8 \times 10^{-4} \ cm \}$

The number of atoms is calculated from gold foil thickness divided by the atomic diameter.

$\boxed{ \ = \frac{8 \times 10^{-4} \ cm}{2.9 \times 10^{-8} \ cm} \ }$

$\boxed{ \ =2.7586 \times 10^4 \ atoms \ }$

Therefore, we get an atomic thickness of 27,586 atoms.

<u>Notes:</u>

In 1909-1910, Ernest Rutherford with two of his assistants, namely Hans Geiger and Ernest Marsden, conducted a series of experiments to find out more about the arrangement of atoms. They fired at a very thin gold plate with high-energy alpha particles.
One of their observations is that a small portion of alpha particles are reflected. This greatly surprised Rutherford. The reflected alpha particle must have hit something very dense in the atom. This fact is incompatible with the atomic model proposed by J.J. Thomson where the atoms are described as homogeneous in all parts with electrons and protons evenly distributed.
In 1911, Rutherford was able to explain the scattering of alpha rays by proposing ideas about atomic nuclei. According to him, most of the mass and positive charge of atoms are concentrated at the center of the atom, hereinafter referred to as the nucleus.

<h3>Learn more</h3>

The energy density of the stored energy brainly.com/question/9617400
The theoretical density of platinum which has the FCC crystal structure. brainly.com/question/5048216
Compound microscope brainly.com/question/4000241

Keywords: if a single gold atom, has a diameter of 2.9 x 10⁻⁸ cm, how many, atoms thick, Rutherford's foil, his scattering experiment

7 0

3 years ago

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