How many mL of a 0.63 M solution would contain 12g of Al(NO3)3

1 atomic mass unit (amu) represents the mass of an atom or is used to measure mass on an atomic scale. It is also referred to as a dalton, abbreviated as Da, while atomic mass unit is indicated as amu.

1 amu can be translated into grams as follows:

For conversion into grams:

Therefore, the mass of Te is 204.16 * 10^-2^4 g

Answer:

a. H₂O (conjugate acid); b. OH⁻ (conjugate base), H₃O⁺ (conjugate acid); c. H₂CO₃ (conjugate acid), CO₃⁻² (conjugate base); d. NH₄⁺ (conjugate strong acid) e. H₂SO₄ (conjugate acid), SO₄⁻² (conjugate base); f. No conjugate acid or base exists; g. H₂S (conjugate acid), S⁻² (conjugate base);

h. H₄N₂ (conjugate base)

Explanation:

a. OH⁻ + H⁺ ⇄ H₂O

The hydroxide functions as a Bronsted-Lowry base, allowing it to capture a proton, thus water serves as the conjugate acid.

b. H₂O is amphoteric, capable of acting as either an acid or a base. As a base, its conjugate acid is H₃O⁺, whereas as an acid, its conjugate base is OH⁻.

c. HCO₃⁻ + H⁺ ⇄ H₂CO₃

HCO₃⁻ + H₂O ⇄ CO₃⁻² + H₃O⁺

Bicarbonate is also amphoteric. When it captures a proton, it forms carbonic acid as the conjugate acid when acting as a base. When HCO₃⁻ acts as an acid and releases a proton, carbonate becomes the conjugate base.

d. Ammonia functions as a weak base, with ammonium being the conjugate strong acid.

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻

e. Another amphoteric compound. Acid sulfate can function as both an acid and a base.

(similar to bicarbonate). Acting as a base yields sulfuric acid as the conjugate acid, while acting as an acid leads to sulfate as the conjugate base.

HSO₄⁻ + H₂O ⇄ SO₄⁻² + H₃O⁺

HSO₄⁻ + H⁺ ⇄ H₂SO₄

f. H₂O₂ does not accept H⁺ or OH⁻ nor does it expel H⁺. It’s neutral and does not function as an acid or base.

g. HS⁻ is amphoteric.

HS⁻ + H⁺ ⇄ H₂S

HS⁻ + H₂O ⇄ S⁻² + H₃O⁺

This is similar to the case of bicarbonate or acid sulfate.

h. H₅N₂⁺ + H₂O ⇄ H₄N₂ + H₃O⁺

Hydrazinium acts as an acid, making hydrazine its conjugate base.

The answer is C: hydrogen bonds. Explanation: The surface tension and the capacity for heat storage in water are due to its hydrogen bonds. Water molecules have a strong attraction to one another through hydrogen bonding. These bonds are continuously forming and breaking within water molecules. The result of this hydrogen bonding is surface tension, which allows water to have a greater capacity for heat retention. Consequently, during the night, the temperature on Earth drops much faster than it does for water, as water gradually releases heat, helping maintain a moderate atmospheric temperature at night.

<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>

The question lacks completeness; the full question is:

Determine the theoretical yield:

When excess aqueous copper(II) nitrate reacts with aqueous sodium sulfide, sodium nitrate and copper(II) sulfide precipitate. For this reaction, 469 grams of copper(II) nitrate was combined with 156 grams of sodium sulfide yielding 272 grams of sodium nitrate.

Answer:

The theoretical yield of sodium nitrate is 340 grams.

Explanation:

Calculating moles of copper(II) nitrate =

Moles of sodium sulfide =

Based on the reaction, 1 mole of copper(II) nitrate reacts with 1 mole of sodium sulfide.

Thus, 2 moles of sodium sulfide will react with:

of copper(II) nitrate

Since sodium sulfide is in limiting quantities, the amount of sodium nitrate produced will depend on the moles of sodium sulfide available.

According to the reaction, 1 mole of sodium sulfide generates 2 moles of sodium nitrate; thus, 2 moles of sodium sulfide will yield:

sodium nitrate

The total mass of 4 moles of sodium nitrate is:

85 g/mol × 4 mol = 340 g

The theoretical yield of sodium nitrate amounts to 340 g.

The theoretical yield of sodium nitrate is 340 grams.