A 2.25-g sample of magnesium nitrate, mg(no3)2, contains __________ mol of this compound.

1) reacting hydrochloric acid with nickel:

Balanced molecular equation: Ni(s) + 2HCl(aq) → NiCl₂(aq) + H₂(g).

Ionic equation: Ni(s) + 2H⁺(aq) + 2Cl⁻(aq) → Ni²⁺(aq) + 2Cl⁻(aq) + H₂(g).

Net ionic equation: Ni(s) + 2H⁺(aq) → Ni²⁺(aq) + H₂(g).

In this reaction, nickel undergoes oxidation, changing from an oxidation state of 0 to +2, while hydrogen is reduced from +1 to 0 (H₂).

2) reacting sulfuric acid with iron:

Balanced molecular equation: Fe(s) + H₂SO₄(aq) → FeSO₄(aq) + H₂(g).

Ionic equation: Fe(s) + 2H⁺(aq) + SO₄²⁻(aq) → Fe²⁺(aq) + SO₄²⁻(aq) + H₂(g).

Net ionic equation: Fe(s) + 2H⁺(aq) → Fe²⁺(aq) + H₂(g).

In this scenario, iron is oxidized from an oxidation state of 0 to +2, while hydrogen experiences reduction from +1 to 0 (H₂).

3) hydrobromic acid reacting with magnesium:

Balanced molecular equation: Mg(s) + 2HBr(aq) → MgBr₂(aq) + H₂(g).

Ionic equation: Mg(s) + 2H⁺(aq) + 2Br⁻(aq) → Mg²⁺(aq) + 2Br⁻(aq) + H₂(g).

Net ionic equation: Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g).

This reaction sees magnesium oxidized from 0 to +2, and hydrogen reduced from +1 to 0 (H₂).

4) acetic acid reacting with zinc:

Balanced molecular equation: Zn(s) + 2CH₃COOH(aq) → (CH₃COO)₂Zn(aq) + H₂(g).

Ionic equation: Zn(s) + 2H⁺(aq) + 2CH₃COO⁻(aq) → Zn²⁺(aq) + 2CH₃COO⁻(aq) + H₂(g).

Net ionic equation: Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g).

Here, zinc gets oxidized from 0 to +2 (Zn²⁺), while hydrogen is reduced from +1 to 0 (H₂).

I believe the answer is D, though I'm not entirely certain unfortunately.

The result is 200 g. Given that the molar mass of CaCl2 is 110.98 g/mol, this indicates that there are 110.98 g in 1 L of a 1 M solution. Let's calculate the amount of CaCl2 in 0.720 M. Using the proportion 110.98 g: 1 M = x: 0.720 M, we find x to be 79.90 g. Therefore, in 1 L of a 0.720 M solution, there is 79.90 g. Next, we need to create ten beakers with 250 mL each, totaling 10 * 250 mL = 2500 mL or 2.5 L. Then, using the equation 79.90 g: 1 L = x: 2.5 L, we calculate x = 79.90 g * 2.5 L: 1 L, resulting in x = 199.75 g, approximately 200 g.

Answer:

C₂H₅O₂

Explanation:

From the information provided in the question, we have the following details:

Mass of compound = 0.25 g

Mass of CO₂ = 0.3664 g

Mass of H₂O = 0.15 g

Empirical formula =?

Now, let’s calculate the masses of carbon, hydrogen, and oxygen within the compound as follows:

For Carbon (C):

Mass of CO₂ = 0.3664 g

Molar mass of CO₂ = 12 + (2×16) = 44 g/mol

Mass of C = 12/44 × 0.3664

Mass of C = 0.1

For Hydrogen (H):

Mass of H₂O = 0.15 g

Molar mass of H₂O = (2×1) + 16 = 18 g/mol

Mass of H = 2/18 × 0.15

Mass of H = 0.02 g

For Oxygen (O):

Mass of C = 0.1 g

Mass of H = 0.02 g

Mass of compound = 0.25 g

Mass of O =?

Mass of O = (Mass of compound) – (Mass of C + Mass of H)

Mass of O = 0.25 – (0.1 + 0.02)

Mass of O = 0.25 –0.12

Mass of O = 0.13 g

In conclusion, we will now find the empirical formula for the compound:

C = 0.1

H = 0.02

O = 0.13

Next, we divide by their respective molar mass:

C = 0.1 / 12 = 0.0083

H = 0.02 / 1 = 0.02

O = 0.13 / 16 = 0.0081

Then we divide by the smallest value:

C = 0.0083 / 0.0081 = 1

H = 0.02 / 0.0081 = 2.47

O = 0.008 / 0.008 = 1

Finally, we multiply by 2 to present in whole numbers:

C = 1 × 2 = 2

H = 2.47 × 2 = 5

O = 1 × 2 = 2

Therefore, the empirical formula for the compound is C₂H₅O₂

The molar masses for H2S and NH3 are 34 and 17 g/mol, respectively. The appropriate equation to represent this is,
                                  Rate A/Rate B = √(molar mass B/molar mass A)
Substituting values,
                               x/77 = √(17 /34 )
                                 x = 54.4
This means NH3 will take 54.4 seconds to escape through the container.