a nurse practitioner orders Medrol to be given 1.5 mg/kg of body weight. if a child weighs 72.6lb and the available stock of Med

Explanation:

Initial moles of ethanoic acid = 0.020 mol

At equilibrium, half of the ethanoic acid molecules have reacted.

Thus, moles of ethanoic acid reacted = 0.020 mol * (50% / 100%)

                                                                     = 0.010 mol

Moles of ethanoic acid remaining = 0.020 mol - 0.010 mol = 0.010 mol

The moles of product gas formed are determined as follows:

0.010 mol CH3COOH * (1 mol / 2 mol CH3COOH)

= 0.005 mol

Consequently, the total moles of gas present in the vessel at equilibrium are 0.010 mol CH3COOH and 0.005 mol

Total gas moles at equilibrium = 0.010 mol + 0.005 mol = 0.015 mol

Next, let’s determine the pressure:

0.020 mol of gas has a pressure of 0.74 atm; so under the same conditions, we find the pressure exerted by 0.015 mol of gas:

P1/n1 = P2/n2

P2 = P1*(n2 / n1)

      = 0.74 atm * (0.015 mol / 0.020 mol)

     = 0.555 atm

Answer :

The percentage ionic character (%IC) equals 10%, indicating the bond is mostly covalent with slight polarity.

Percent Ionic Character:

This reflects the fraction of ionic nature within a polar covalent bond. The formula for %IC (% ionic character) is:

Here, Xa is the electronegativity of atom A and Xb is that of atom B.

Given: The compound is TiAl₃.

Electronegativity of Ti = 2.0

Electronegativity of Al = 1.6 (as shown in the provided image)

Substitute these values into the formula:

The value of e⁻¹ equals 0.90.

Therefore, percent ionic character = (1 - 0.90) × 100

Percent Ionic Character = 10%

Because the % IC is only 10%, which is relatively low, the bond is classified as covalent with minimal polarity.

The net ionic equation for the reaction between chromium (III) hydroxide and nitrous acid is:

Cr(OH)₃ (s) + 3H⁺ (aq) ⇒ Cr³⁺ (aq) + 3H₂O (l)

Additional Details

An electrolyte dissociates into ions in solution.

Chemical equations can also be represented with ionic species.

Strong electrolytes (fully ionized) are written as separate ions, whereas weak electrolytes (partially ionized) remain as intact molecules.

In ionic equations, spectator ions are those unchanged by the chemical process—they are present both before and after the reaction.

Removing these spectators results in the net ionic equation.

Gases, solids, and water (H₂O) are written as molecules, without ionization.

Therefore, only dissolved compounds are represented by their ions (aq).

The problem involves chromium (III) hydroxide reacting with nitrous acid.

The reaction occurring is:

Cr(OH)₃ (s) + 3HNO₂ (aq) ⇒ Cr(NO₂)₃ (aq) + 3H₂O (l)

Chromium (III) hydroxide is a solid and remains un-ionized, as does water.

Thus, the ionic equation is:

Cr(OH)₃ (s) + 3H⁺ (aq) + 3NO₂⁻ (aq) ⇒ Cr³⁺ (aq) + 3NO₂⁻ (aq) + 3H₂O (l)

The ion 3NO₂⁻ is a spectator ion; removing it yields the net ionic equation:

Cr(OH)₃ (s) + 3H⁺ (aq) ⇒ Cr³⁺ (aq) + 3H₂O (l)

Answer:

The new gas pressure within the chamber registers at 1,093.75 mmHg

Explanation:

The Gay-Lussac Law establishes a relationship between a gas's pressure and temperature when volume remains constant. This principle asserts that gas pressure is directly tied to its temperature: as temperature increases, pressure rises, and conversely, as temperature falls, pressure also diminishes. Therefore, the Gay-Lussac law can be depicted mathematically as:

Given an initial and final state of gas, we can apply the following formula:

In this scenario:

• P1= 1560 mmHg
• T1= 445 K
• P2=?
• T2= 312 K

Calculating:

P2=1,093.75 mmHg

The new gas pressure inside the chamber is 1,093.75 mmHg

Result:

94.7 %

Explanation:

The balanced reaction is:

2 S + 3 O₂ → 2 SO₃

The stoichiometric mole ratio is:

S: 2 moles

O₂: 3 moles

Moles are calculated as mass divided by molar mass:

n = w / m

where n = moles, w = mass, m = molar mass.

Given:

For sulfur: w = 6.0 g, molar mass = 32 g/mol, so n = 6 / 32 = 0.1871 mol

For oxygen: w = 5.0 g, molar mass = 32 g/mol, thus n = 5 / 32 = 0.15625 mol

Comparing to stoichiometric ratios, sulfur is in excess, so oxygen is the limiting reagent, controlling product formation.

Using proportions:

3 mol O₂ produce 2 mol SO₃, so 1 mol O₂ yields 2/3 mol SO₃.

Therefore, 0.15625 mol O₂ yields (2/3) × 0.15625 = 0.1042 mol SO₃.

Mass of SO₃ produced = n × molar mass = 0.1042 mol × 80 g/mol = 8.340 g

The percentage yield is actual yield divided by theoretical yield times 100:

Percent yield = (7.9 g / 8.340 g) × 100 = 94.7 %