The secondary structure of a protein results from _____. the secondary structure of a protein results from _____. bonds between

Answer:

Explanation:

This scenario is unrealistic since Al(OH)₃ is not soluble in water.

The question consists of two parts:

A. Stoichiometry — where we determine volumes, masses, and moles for the products

B. Calorimetry — where we assess the enthalpy of the reaction.

A. Stoichiometry

1. Determine the volume of Al(OH)₃

(a) The balanced chemical equation:

                 2Al(OH)₃ + 3H₂SO₄ ⟶ Al₂(SO₄)₃ + 6H₂O

M/V:            66.667

c/mol·L⁻¹:   4.000       3.000

(b) Moles of H₂SO₄

(c) Moles of Al(OH)₃

The molar ratio stands at 2 mmol Al(OH)₃: 3 mmol H₂SO₄

(d) Volume of Al(OH)₃

B. Calorimetry

This reaction has two energy exchanges.

q₁ = heat from the reaction

q₂ = heat used to heat the calorimeter

 q₁ + q₂ = 0

nΔH + mCΔT = 0

Data:

Moles of Al₂(SO₄)₃ = 0.066 667 mol

C = 1.10 J°C⁻¹g⁻¹

T_initial = 22.3 °C

T_final = 24.7 °C

Calculations

(a) Mass of solution

Assume solutions are as dense as water (though not realistic).

Mass of sulfuric acid solution            =   66.667 g 

Mass of aluminium hydroxide solution =  50.000    

                                             TOTAL =  116.667 g

(b) ΔT

ΔT = T_final - T_initial = 24.7 °C - 22.3 °C = 2.4°C

(c) ΔH

This result appears nonsensical, but it is derived from your given figures.

The issue with your setup is that the surface where you placed the drop is contaminated or dirty, preventing it from being wetted by the solution. To fix this, thoroughly clean the surface before applying the drop.

We assume that the stated 50% is measured by volume. Molarity defines the concentration in terms of moles of solute per volume of solution.

      To find the moles of NaOH, use: (0.1 moles / L)(0.4 L)
                      n = 0.04 moles of NaOH

Assuming we start with 1 mL of 50% NaOH solution, 
  
                        (1 mL solution)(1.525 g/mL)(0.50) = 0.7625 g
Then, the number of moles calculates as follows,[
                  0.7625 g NaOH x (1 mol / 40 g) = 0.01906 moles of NaOH

                The volume of solution required can be determined by:(0.04 moles of NaOH)(1 mL solution / 0.01906 moles of NaOH)
                 
                 Thus, the needed volume comes out to be 2.09 mL

Answer: 2.09 mL

Answer:

The temperature of a vessel at a pressure reading of 1.250 atm is 296.43 K.

Explanation:

At standard temperature and pressure (STP), the temperature is valued at 273.15 K with a pressure of 1.000 atm.

If the pressure is measured as 1.250 atm at temperature T.

Applying Gay Lussac's law:

(When volume is constant)

At a pressure of 1.250 atm, the temperature of the vessel is 296.43 K.

Answer:6.94

Explanation:Atomic mass of Li = 6.94

To find the number of atoms: 12.6 g Li x 1 mole/6.94 g x 6.02x10^23 atom/mole = 1.09x10^24 atoms