You are trying to make balloon sculptures. You twist the balloons gently, but they keep popping. Besides trimming your nails, ho

The balanced equation is:

NaCH₃COO + HCl → NaCl + HCH₃COO

Make 

Response:

The mass percentage of a solution comprising 7.6 grams of sucrose and 83.4 grams of water equals 8.351 %.

Details:

Provided data:

Sucrose mass = 7.6 grams

Water mass = 83.4 grams

In this scenario, sucrose acts as the solute, while water is the solvent.

The calculation for mass percent of a solution is done using the following formula:

Mass percent = (Mass of Solute/Mass of Solution)(100)

As sucrose is the solute, the mass equals 7.6 grams.

The total mass of the solution, which includes both sucrose and water, comes out to:

Total mass = 7.6 grams + 83.4 grams = 91 grams

Therefore, applying the values gives mass percent = (7.6/91)(100) = 8.351 %.

<span>BaCl2 + Na2SO4 --> BaSO4 + 2NaCl In this reaction, 1.0 g of BaCl2 and 1.0 g of Na2SO4 are present. We need to identify the limiting reactant. "First, convert grams to moles" 1.0 g BaCl2 * (1 mol BaCl2 / 208.2 g BaCl2) = 4.8 x 10^-3 mol BaCl2 1.0 g Na2SO4 * (1 mol Na2SO4 / 142.04 g Na2SO4) = 7.0 x 10^-3 mol Na2SO4 (7.0 x 10^-3 mol Na2SO4 / 4.8 x 10^-3 mol BaCl2) = 1.5 mol Na2SO4 per mol BaCl2 "Using this ratio to compare with the balanced equation, BaCl2 + Na2SO4 --> BaSO4 + 2NaCl" The balanced equation indicates that 1 mol of BaCl2 reacts with 1 mol of Na2SO4. However, we found that 1.5 mol of Na2SO4 is available for each mol of BaCl2. Therefore, BaCl2 is the limiting reagent.</span>

Answer:

Explanation:

The percent composition indicates the mass percentage of a specific element within the compound.

The chemical formula for chromium(III) nitrate is .

The molar mass for chromium(III) nitrate is calculated at 238.011 g/mol.

Each mole of chromium(III) nitrate includes 9 moles of oxygen.

The molar mass of oxygen is 16 g/mol.

Hence, Mass = Molar mass * Moles = 16 * 9 g = 144 g.

Answer:

7.3

Explanation:

Using the Henderson Hasselbalch equation, one can determine the pH or pOH of a solution via its pKa. Remember, , and pKa = -logKa, where Ka denotes the acid's equilibrium constant.

The Henderson Hasselbalch formula:

In this context, acid X possesses two ionic forms: the carboxyl group and an alternative form. Initially, we have 0.1 mol/L of acid in 100 mL, which gives:

n1 = (0.1 mol/L)×(0.1 L) = 0.01 mol

Upon dissociation, it yields 0.005 mol of the carboxyl form and 0.005 mol of the other form with stoichiometry assumed constant.

Introducing NaOH at a concentration of 0.1 mol/L and 75 mL, the moles of become:

n2 = (0.1 mol/L)×(0.075 L) = 0.0075 mol

Thus, 0.0075 mol of reacts with 0.005 mol of the carboxyl form, leading to 0.0025 mol of , which in turn reacts with 0.005 mol of the alternating group, leaving 0.0025 mol of the latter.

The new solution’s volume is 175 mL, but the concentrations of both forms remain unchanged in volume, so we can utilize the moles in the equation.

<pNote, the moles of the acid form remain 0.01 mol as it doesn’t undergo reaction!

Thus, we arrive at:

6.72 = pKa - log 4

pKa - log 4 = 6.72

pKa = 6.72 + log 4

pKa = 6.72 + 0.6

pKa = 7.3