A 23.9g object was submerged into 50.5ml of water. If the water rose to 89.4ml then what is the object's density?

In this diagram, medium A represents water and medium B represents air. The phenomenon in this diagram, moving from A to B, is c

Anarel [89]

Answer:

In this diagram, the phenomenon moving from medium A which represents water to medium B that represents air is called refraction.

Explanation:

Refraction takes place when light passes through a water surface due to its refractive index which is 1.33. Air possesses a refractive index of about 1. Looking at the phenomenon moving between medium A and B, it seems to bend at the water's surface. This happens because of the bending of the light rays as they move from the water to the air. The time when the rays reach the eyes, the eyes trace them back as straight lines or lines of sight. These lines divide at a higher position than where the actual rays originated. This causes the phenomenon to appear higher and the water to appear shallower than what it is in reality.

5 0

3 years ago

In the equation below

yulyashka [42]

Answer:

6.25 moles of N₂ is produced, and 18.8 moles of Cu and H₂O is produced.

Explanation:

We are given the chemical equation:

$\displaystyle 2\text{NH$_3$}_\text{(g)} + 3\text{CuO}_\text{(s)} \longrightarrow \text{N$_2$}_\text{(g)} + 3\text{Cu}_\text{(s)}+3\text{H$_2$O}_\text{(g)}$

And we want to determine the amount of products produced when 12.5 moles of NH₃ is reacted with excess CuO.

Compute using stoichiometry. From the equation, we can see the following stoichiometric ratios:

The ratio between NH₃ and N₂ is 2:1. (i.e. One mole of N₂ is produced from every two moles of NH₃.)

The ratio between NH₃ and Cu is 2:3.
The ratio between NH₃ and H₂O is 2:3. (i.e. Three moles of H₂O or Cu is produced frome every two moles of NH₃.)

Dimensional Analysis:

The amount of N₂ produced:

$\displaystyle 12.5\text{ mol NH$_3$} \cdot \frac{1\text{ mol N$_2$}}{2\text{ mol NH$_3$}} = 6.25\text{ mol N$_2$}$

The amount of Cu produced:

$\displaystyle 12.5\text{ mol NH$_3$} \cdot \frac{3\text{ mol Cu}}{2\text{ mol NH$_3$}} = 18.8\text{ mol Cu}$

And the amount of H₂O produced:

$\displaystyle 12.5\text{ mol NH$_3$} \cdot \frac{3\text{ mol H$_2$O}}{2\text{ mol NH$_3$}} = 18.8\text{ mol H$_2$O}$

In conclusion, 6.25 moles of N₂ is produced, and 18.8 moles of Cu and H₂O is produced.

8 0

2 years ago

You're legally obligated to yield the right-of-way:

kenny6666 [7]

It is most likely B.

3 0

3 years ago

How many atoms of oxygen are contained in 160 grams if N2O3?

Juli2301 [7.4K]

Answer:

3.80x1024 grams containing if not that then above 80 grams 6 grams

Explanation:

3 0

3 years ago

Read 2 more answers

When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced.

sp2606 [1]

Answer:

Mass of CaCl₂ = 20 g

CaCO is presewnt in excess.

Mass of of CaCO₃ remain unreacted = 7.007 g

Explanation:

Given data:

Mass of calcium carbonate = 25 g

Mass of hydrochloric acid = 13.0 g

Mass of calcium chloride produced = ?

Chemical equation:

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

Number of moles of CaCO₃:

Number of moles of CaCO₃ = Mass /molar mass

Number of moles of CaCO₃= 25.0 g / 100.1 g/mol

Number of moles of CaCO₃ = 0.25 mol

Number of moles of HCl:

Number of moles of HCl = Mass /molar mass

Number of moles of HCl = 13.0 g / 36.5 g/mol

Number of moles of HCl = 0.36 mol

Now we will compare the moles of CaCl₂ with HCl and CaCO₃ .

CaCO₃ : CaCl₂

1 : 1

0.25 : 0.25

HCl : CaCl₂

2 : 1

0.36 : 1/2 × 0.36 = 0.18 mol

The number of moles of CaCl₂ produced by HCl are less it will be limiting reactant.

Mass of CaCl₂ = moles × molar mass

Mass of CaCl₂ =0.18 mol × 110.98 g/mol

Mass of CaCl₂ = 20 g

The calcium carbonate is present in excess.

HCl : CaCO₃

2 : 1

0.36 : 1/2 × 0.36 = 0.18 mol

So, 0.18 moles react with 0.36 moles of HCl.

The moles of CaCO₃ remain unreacted = 0.25 -0.18

The moles of CaCO₃ remain unreacted = 0.07 mol

Mass of of CaCO₃ remain unreacted = Moles × molar mass

Mass of of CaCO₃ remain unreacted = 0.07 mol × 100.1 g/mol

Mass of of CaCO₃ remain unreacted = 7.007 g

7 0

3 years ago