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3 years ago

WHY ARE FROGS SAID to have "two lives?'

Chemistry

2 answers:

kompoz [17]3 years ago

3 0

Because they live in water half of their lives and land most of their lives

stiks02 [169]3 years ago

3 0

Because they live in water half of their lives and land most of their lives

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A neutral atom possesses an atomic number of 15 and an atomic mass of 31. Three electrons are gained. What is the result of this

12345 [234]

On a neutral state, the number of proton of a substance will be equal to the number of its electron (in this case, it has 15 electrons). When three electrons are gained, the total number of the electron will change to 18.

Therefore this result in the substance becomes - D. A negatively charged ion of 3-

Hope this helps!

5 0

3 years ago

Read 2 more answers

What is the formula for the compound that forms between barium and oxygen?

Dmitry [639]

solution:

Barium is a chemical element with symbol Ba and atomic number 56. It is the fifth element in group 2 and is a soft, silvery alkaline earth metal. Because of its high chemical reactivity, barium is never found in nature as a free element

Oxygen is a chemical element with symbol O and atomic number 8. It is a member of the chalcogen group on the periodic table, a highly reactive nonmetal, and an oxidizing agent that readily forms oxides with most elements as well as with other compounds.

Barium can form two distinct compounds with oxygen as the only other element in the compound: barium oxide with formula BaO and barium peroxide with formula BaO 2 . The first of these compounds is more common and more stable.

8 0

3 years ago

The standard enthalpy of reaction for the dissolution of silica in aqueous HF is 4.6 kJ mol–1 . What is the standard enthalpy of

Stels [109]

Answer:

B) ) –1615.1 kJ mol^–1

Explanation:

since

SiO2(s) + 4 HF(aq) → SiF4(g) + 2 H2O(l) ∆Hºrxn = 4.6 kJ mol–1

the enhalpy of reaction will be

∆Hºrxn = ∑νp*∆Hºfp - ∑νr*∆Hºfr

where ∆Hºrxn= enthalpy of reaction , ∆Hºfp= standard enthalpy of formation of products , ∆Hºfr = standard enthalpy of formation of reactants , νp=stoichiometric coffficient of products, νr=stoichiometric coffficient of reactants

therefore

∆Hºrxn = ∑νp*∆Hºfp - ∑νr*∆Hºfr

4.6 kJ/mol = [1*∆HºfX + 2*(–285.8 kJ/mol)] - [1*(–910.9kJ/mol) + 4*(–320.1 kJ/mol)]

4.6 kJ/mol =∆HºfX -571.6 kJ/mol + 2191.3 kJ/mol

∆HºfX = 4.6 kJ/mol + 571.6 kJ/mol - 2191.3 kJ/mol = -1615.1 kJ/mol

therefore ∆HºfX (unknown standard enthalpy of formation = standard enthalpy of formation of SiF4(g) ) = -1615.1 kJ/mol

8 0

3 years ago

[Standard Enthalpy of Formation]

Schach [20]

Answer:

3. ΔH = 0.30 kJ; 4. ΔH = -84.6 kJ

Step-by-step explanation:

Question 3:

We have three equations:

(I) S(r) + O₂ → SO₂; ΔH = -296.06 kJ

(II) S(m) + O₂ ⟶ SO₂; ΔH = -296.36 kJ

From these, we must devise the target equation:

(III) S(r) ⟶ S(m); ΔH = ?

The target equation has 1S(r) on the left, so you rewrite Equation(I).

(IV) S(r) + O₂ ⟶ SO₂; ΔH = -296.06 kJ

Equation (IV) has 1SO₂ on the right, and that is not in the target equation.

You need an equation with 1SO₂ on the left, so you reverse Equation (II).

When you reverse an equation, you change the sign of its ΔH.

(V) SO₂ ⟶ S(m) + O₂ ; ΔH = 296.36 kJ

Now, you add equations (IV) and (V), cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you add their ΔH values.

We get the target equation (III)

(IV) S(r) + O₂ ⟶ SO₂; ΔH = -296.06 kJ

(V) SO₂ ⟶ S(m) + O₂; ΔH = 296.36 kJ

(III) S(r) ⟶ S(m); ΔH = 0.30 kJ

Question 4

We have three equations:

(I) C + O₂ ⟶ CO₂; ΔH = -393.5 kJ

(II) H₂ + ½O₂ ⟶ H₂O; ΔH = -285.8 kJ

(III) 2C₂H₆ + 7O₂ ⟶ 4CO₂ + 6H₂O; ΔH = -296.8 kJ

From these, we must devise the target equation:

(IV) 2C + 3H₂ → C₂H₆; ΔH = ?

The target equation has 2C on the left, so you double Equation(I).

When you double an equation, you double its ΔH.

(V) 2C + 2O₂ ⟶ 2CO₂; ΔH = -787.0 kJ

Equation (V) has 2CO₂ on the right, and that is not in the target equation.

You need an equation with 2CO₂ on the left, so you reverse Equation (III) and divide it by 2.

(VI) 2CO₂ + 3H₂O ⟶ C₂H₆ + ⁷/₂O₂; ΔH = 148.4 kJ

Equation (V) has 3H₂O on the left, and that is not in the target equation.

You need an equation with 3H₂O on the right. Triple Equation (III).

(VII) 3H₂ + ³/₂O₂ ⟶ 3H₂O; ΔH = -857.4 kJ

Now, you add equations (V), (VI), and (VII), cancelling species that appear on opposite sides of the reaction arrows.

We get the target equation (IV).

(V) 2C + 2O₂ ⟶ 2CO₂; ΔH = -787.0 kJ

(VI) 2CO₂ + 3H₂O ⟶ C₂H₆ + ⁷/₂O₂; ΔH = 1559.8 kJ

(VII) 3H₂ + ³/₂O₂ ⟶ 3H₂O; ΔH = - 857.4 kJ

(IV) 2C + 3H₂ → C₂H₆; ΔH = -84.6 kJ

4 0

4 years ago

The strength of an acid or a base is determined by the

Tresset [83]

Ionization/dissociation, Ka, larger, concentration/molarity, mostly/completely, strong, weak, base, water, acid, strong

3 0

3 years ago