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3 years ago

A 40.0-L sample of fluorine is heated from 90.0°C to 186.0°C. What volume will the sample occupy at the higher temperature?

Chemistry

1 answer:

jasenka [17]3 years ago

5 0

Answer:

The volume will be 82.67 L

Explanation:

Charles's Law is the relationship between the volume and temperature of a certain amount of ideal gas. In this way, Charles's law is a law that says that when the amount of gas and pressure are kept constant, the ratio between volume and temperature will always have the same value:

$\frac{V}{T} = k$

Having a certain volume of gas V1 that is at a temperature T1 at the beginning of the experiment, by varying the volume of gas to a new value V2, then the temperature will change to T2, and it will be true:

$\frac{V1}{T1} =\frac{V2}{T2}$

In this case, you know:

V1= 40 L
T1= 90 °C
V2= ?
T2= 186 °C

Replacing:

$\frac{40L}{90C} =\frac{V2}{186 C}$

Solving:

$V2=\frac{40L}{90C} *186 C$

V2= 82.67 L

The volume will be 82.67 L

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Give the name and formula of the compound formed from the following elements: (a) sodium and nitrogen; (b) oxygen and strontium;

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Answer:

For a: The chemical name and chemical formula formed is sodium nitride and $Na_3N$

For b: The chemical name and chemical formula formed is strontium oxide and $SrO$

For c: The chemical name and chemical formula formed is aluminium chloride and $AlCl_3$

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

The nomenclature of ionic compounds is given by:

1. Positive is written first.

2. The negative ion is written next and a suffix is added at the end of the negative ion. The suffix written is '-ide'.

For a: Sodium and nitrogen:

Sodium is the 11th element of periodic table having electronic configuration of $[Ne]3s^1$

To form $Na^{+}$ ion, this element will loose 1 electron.

Nitrogen is the 7th element of periodic table having electronic configuration of $[He]2s^22p^3$ .

To form $N^{3-}$ ion, this element will gain 3 electrons.

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for sodium nitride is $Na_3N$

For b: Oxygen and strontium

Strontium is the 38th element of periodic table having electronic configuration of $[Kr]5s^2$ .

To form $Sr^{2+}$ ion, this element will loose 2 electrons.

Oxygen is the 8th element of periodic table having electronic configuration of $[He]2s^22p^4$ .

To form $O^{2-}$ ion, this element will gain 2 electrons.

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for strontium oxide is $SrO$

For c: Aluminium and chlorine

Aluminium is the 13th element of periodic table having electronic configuration of $[Ne]3s^23p^1$

To form $Al^{3+}$ ion, this element will loose 3 electrons.

Chlorine is the 17th element of the periodic table having electronic configuration of $[Ne]3s^23p^5$

To form $Cl^{-}$ ion, this element will gain 1 electron.

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for aluminium chloride is $AlCl_3$

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3 years ago

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Answer:When you add baking powder to water or milk, the alkali and the acidreact with one another and produce carbon dioxide – the bubbles. Sodium bicarbonate is a weak base which is commonly known as baking soda and used in cooking. It weakly ionizes in water: NaHCO3 + H2O → H2CO3 + (OH-) + (Na+). u need to stop deleteing my answers ughh

Explanation:

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3 years ago

If the relative activities of two metals are known which metal is more easily oxidized

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Answer: The metal that has a greater reactivity is more easily oxidized.

Explanation:

Oxidation is when the elements lose electrons and increase their oxidation state.

The metals tend to react by losing electrons and form the corresponding cation.

For expample, sodium (an alkalyne metal) loses one elecron and form the cation Na¹⁺ , then this cation combine with an anion and form compounds like NaCl, NaOH. The same do the other alkalyne metals.

Magnesium (an alkalyne earth metal) loses two electrons and form the cation Mg²⁺, then it combines with some anions to form compounds, like MgSO₄, Mg(OH)₂.

So, the easier the metal gets oxidized the greater its reactivity.

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