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OlgaM077 [116]
2 years ago
10

To demonstrate the formation of iron (iii) chloride from iron fillings

Chemistry
1 answer:
Pavel [41]2 years ago
5 0

Iron (iii) chloride is obtained by vapor condensation from the reaction between chlorine gas and iron fillings.

<h3>How can iron (iii) chloride be formed from iron fillings?</h3>

Iron (ii) chloride can be formed from iron fillings in the laboratory as follows:

  • Iron fillings + Cl₂ → FeCl₃

Chlorine gas is introduced into a reaction vessel containing iron fillings and the iron (iii) chloride vapor formed is obtained by condensation.

In conclusion, iron (iii) chloride is formed by the the direct combination of iron fillings and chlorine gas.

Learn more about iron (iii) chloride at: brainly.com/question/14653649

#SPJ1

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During _____, bonds between monomers are broken by adding water.
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During _<span>A.   hydrolysis</span>, bonds between monomers are broken by adding water.
Prefix "hydro-" means water.
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3 years ago
What is the density of object A? Does it sink<br> or float in water?
Oksana_A [137]

Answer:

~1.5 g/cm3 and it does NOT float in water.

Explanation:

If you look at the graph, Object A weighs ~6 grams and is ~4 cm3 in volume

Density = Mass/Volume

So 6 grams/4 cm3 = 1.5 g/cm3

Water has a density of 1 g/cm3 and because Object A density is higher than that of water, it sinks.

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3 0
3 years ago
What is the molar mass of nitrogen?
sukhopar [10]
The molar mass of nitrogen is 14.0067amu or just 14amu.
4 0
3 years ago
In a lab experiment 80.0 g of ammonia [NH3] and 120 g of oxygen are placed in a reaction vessel. At the end of the reaction 72.2
valentinak56 [21]

The percent yield of the reaction : 89.14%

<h3>Further explanation</h3>

Reaction of Ammonia and Oxygen in a lab :

<em>4 NH₃ (g) + 5 O₂ (g) ⇒ 4 NO(g)+ 6 H₂O(g)</em>

mass NH₃ = 80 g

mol NH₃ (MW=17 g/mol):

\dfrac{80}{17}=4.706

mass O₂ = 120 g

mol O₂(MW=32 g/mol) :

\tt \dfrac{120}{32}=3.75

Mol ratio of reactants(to find limiting reatants) :

\tt \dfrac{4.706}{4}\div \dfrac{3.75}{5}=1.1765\div 0.75\rightarrow O_2~limiting~reactant(smaller~ratio)

mol of H₂O based on O₂ as limiting reactants :

mol H₂O :

\tt \dfrac{6}{5}\times 3.75=4.5

mass H₂O :

4.5 x 18 g/mol = 81 g

The percent yield :

\tt \%yield=\dfrac{actual}{theoretical}\times 100\%\\\\\%yield=\dfrac{72.2}{81}\times 100\%=89.14\%

6 0
3 years ago
How many grams are in 11.9 moles of chromium?
raketka [301]

Explanation:

It is known that one mole of chromium or molar mass of chromium is 51.99 g/mol.

It is given that number of moles is 11.9 moles.

Therefore, calculate the mass of chromium in grams as follows.

     No. of moles = \frac{mass in grams}{Molar mass}

    mass in grams = No. of moles × Molar mass

                             = 11.9 moles × 51.99 g/mol

                             = 618.68 g

Thus, we can conclude that there are 618.68 g in 11.9 moles of chromium.

7 0
3 years ago
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