There are 15 technicians and 11 chemists working in a research laboratory. In how many ways could they form a 5-member safety co

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

The correct answer is Option A. The calculation goes as follows: Number of millimoles of Na3PO4 = 1 × 100 = 100 Number of millimoles of AgNO3 = 1 × 100 = 100 Dissociating 1 mole of Na3PO4 yields 3 moles of sodium ions and 1 mole of phosphate ions, whereas 1 mole of AgNO3 releases 1 mole of Ag+ and 1 mole of NO3-. The Ag+ ion concentration becomes negligible since it forms a precipitate with the phosphate ion, indicating that the concentration of phosphate ions is also low. With 100 millimoles of Na3PO4, we get 300 millimoles of Na+ and 100 millimoles of PO43-, and with 100 millimoles of AgNO3 we have 100 millimoles of Ag+ and 100 millimoles of NO3-. Thus, the order of increasing concentration is: PO43- < NO3- < Na+.

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 %

Answer: The complete balanced chemical equation is,

Explanation:

A chemical equation represents the reactants on the left and the products on the right, separated by a right-pointing arrow indicating the reaction.

This representation includes the phases of the substances and uses subscripts and superscripts for numbers.

For the reaction given, the balanced chemical equation with phases is:

Answer:

The molality of the solution is 1.08 m.

Explanation:

First, determine the mass of the solvent.

A 13% solution by mass indicates that 13 grams are found in every 100 grams of solution.

Thus, solution mass = solute mass + solvent mass

100 g = 13 g + solvent mass

Therefore, solvent mass = 100 g - 13 g → 87 g

Next, we calculate the moles of solute (mass / molar mass):

13 g / 138.2 g/mol = 0.094 moles

Finally, to find the molality, which is the moles of solute per 1 kg of solvent (mol/kg), we convert the solvent mass to kg:

87 g. 1 kg / 1000 g = 0.087 kg

Then, molality → 0.094 mol / 0.087 kg = 1.08 m