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3 years ago

What is the state of matter for 2NH3, 2NaCl and 2KNO3+PBI2.

Chemistry

1 answer:

Aloiza [94]3 years ago

7 0

Answer:

2NH3 - Gas Phase

2 NaCl - Solid Phase

2KNO3+PBI2 - Aqueous Phase

Explanation:

2NH3 is ammonia gas which is colorless and has a bad pungent smell.

NaCl is salt. It has crystalline structure and can exhibit both aqueous or a liquid phase. Since there is no H2O molecule in the formula of NaCl, hence it is not aqueous. It is not liquid as well because liquid NaCl is represented as NaCl (l) while aqueous NaCl is represented as NaCl (aq).

The reaction 2KNO3+PBI2 will take place in aqueous phase and hence these two exists in aqueous phase

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Does a physical change form a new substance

beks73 [17]

Answer:

No. No new substance is formed during Physical change.

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3 years ago

Why was it so important that the scientist investigating the atom be young and unattached to existing scientific thinking

tatiyna

Answer:

so they can find the aswer and not get hurt

Explanation:

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3 years ago

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

lisov135 [29]

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 $AlCl_3$

4 0

4 years ago

3. Indicate the type of reaction represented by each equation:

BartSMP [9]

3. Is double replacement

6 0

3 years ago

Strike anywhere matches contain the compound tetraphosphorus trisulfide, which burns to form tetraphosphorus decaoxide and sulfu

erica [24]

Answer:

194.6 mL of SO₂

Explanation:

The reaction that takes place is:

P₄S₃ + 6O₂(g) → P₄O₁₀ + 3SO₂(g)

To solve this problem we need to use PV=nRT, so first let's convert the given units:

23.8 °C → 23.8 + 273.15 = 296.95 K

747 torr → 747/760 = 0.983 atm

We need to calculate V, so in order to do that we calculate n, using the mass of the reactant (P₄S₃):

0.576 g P₄S₃ * $\frac{1molP_{4}S_{3}}{220gP_{4}S_{3}} *\frac{3molSO_{2}}{1molP_{4}S_{3}}$ = 7.85 * 10⁻³ mol SO₂ = n

Now we calculate V:

PV=nRT

0.983 atm * V = 7.85 * 10⁻³ mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 296.95 K

V = 0.1946 L

Finally we convert L into mL:

0.1946 * 1000 = 194.6 mL

8 0

3 years ago