Calculate the mass percent of sulfuric acid in a solid sample given a 0.8390g sample of that solid required 32.22 mL of 0.2012 M

Answer:

Explanation:

0.5678 G        X GRAMS

KHC8H4O4 + NaOH = NaKC8H4O4 + H2O

1 MOL               1 MOL

0.5678G X 204G/MOL = 0.00278 MOL KHC8H4O4

0.00278 MOL KHC8H4O4 X 1 MOLE NaOH/1 MOLE  KHC8H4O4=0.00278 MOL NaOH

0.00278 MOL NaOH/26.26ml=0.106 molar

Answer:

Explanation:

The atomic number refers to the total count of electrons or protons found in a neutral atom.

Nonetheless, when discussing the atomic number of an ion, it does not correspond to the electron count since electrons can be added or removed.

Therefore, it is more accurate to refer to the number of protons located in the atom's nucleus as the atomic number.

Consequently, the atomic number of phosphorus is 15

The mass number reflects the total count of particles in the nucleus, equating to the addition of protons and neutrons.

Given that the mass number is 32

Thus, the symbol for the isotope is:-

Explanation:

A mixture refers to a substance that contains two or more different kinds of substances that are combined physically.

For instance, air is a mixture that contains oxygen, nitrogen, and other gases.

A heterogeneous mixture is characterized by uneven distribution of solute particles in the solvent.

For example, sand suspended in water is a heterogeneous mixture.

Conversely, a homogeneous mixture is one where the solute particles are uniformly distributed in a solvent.

A homogeneous mixture appears as a clear solution.

For instance, when salt dissolves in water, it forms a homogeneous mixture.

A solution is defined as a mixture of two or more substances combined together.

A compound consists of two or more different elements chemically bonded together in a specific mass ratio.

An element is the simplest form of a substance that comprises only one type of atom.

For example, a piece of sodium is composed solely of sodium atoms.

Conversely, a pure substance refers to a material consisting of just one type of molecule or atom.

For example, , etc are considered pure substances.

Thus, it can be concluded that the air sample can be described using the terms:

• pure chemical substance.

• heterogeneous mixture.

• mixture.

Answer:

The rate law for the decomposition reaction is:

The unit for the rate constant will be

Explanation:

The rate law can be expressed as:

..[1]

When the drug concentration is tripled, the decomposition rate rises by a factor of nine.

...[2]

[1] ÷ [2]

Solving for x results in:

x = 2.

This indicates a second-order reaction.

The decomposition reaction's rate law is:

The unit for the rate constant will be:

The unit for the rate constant will be .

1) The ionic compound present in solution b is K₂CrO₄ (potassium chromate). This compound contains two potassiums (oxidation state +1), a single chromium (oxidation state +6), and four oxygen atoms. The oxidation state of oxygen is -2, resulting in a neutral compound: 2 · (+1) + 6 + x · (-2) = 0. Hence, x = 4, denoting the count of oxygen atoms. 2) The ionic compound in solution a is AgNO₃ (silver nitrate). ω(N) = 8.246% ÷ 100%. Thus, ω(N) = 0.08246, indicating the mass percentage of nitrogen. M(MNO₃) = M(N) ÷ ω(N). It follows that M(MNO₃) = 14 g/mol ÷ 0.08246, leading to M(MNO₃) = 169.8 g/mol; the molar mass of the metal nitrate. M(M) = M(MNO₃) - M(N) - 3 · M(O). Consequently, M(M) = 169.8 g/mol - 14 g/mol - 3 · 16 g/mol, resulting in M(M) = 107.8 g/mol which is the atomic mass of silver (Ag). 3) The balanced chemical equation is: 2AgNO₃(aq) + K₂CrO₄(aq) → Ag₂CrO₄(s) + 2KNO₃(aq). In ionic form: 2Ag⁺(aq) + 2NO₃⁻(aq) + 2K⁺(aq) + CrO₄²⁻(aq) → Ag₂CrO₄(s) + 2K⁺(aq) + 2NO₃⁻(aq). The net ionic equation is: 2Ag⁺(aq) + CrO₄²⁻(aq) → Ag₂CrO₄(s). Thus, the red precipitate is identified as silver chromate (Ag₂CrO₄). 4) The mass of solid silver chromate created is m(Ag₂CrO₄) = 331.8 g. The amount is determined by n(Ag₂CrO₄) = m(Ag₂CrO₄) ÷ M(Ag₂CrO₄). Therefore, n(Ag₂CrO₄) = 331.8 g ÷ 331.8 g/mol yields n(Ag₂CrO₄) = 1 mol. From the balanced equation, n(Ag₂CrO₄): n(AgNO₃) = 1: 2, it follows n(AgNO₃) = 2 · 1 mol, which means n(AgNO₃) = 2 mol. Then, the mass of silver nitrate is computed as m(AgNO₃) = n(AgNO₃) · M(AgNO₃). Hence, m(AgNO₃) = 2 mol · 169.8 g/mol gives m(AgNO₃) = 339.6 g; thus, m(AgNO₄) equals m(K₂CrO₄). Therefore, m(K₂CrO₄) = 339.6 g; amount of potassium chromate is n(K₂CrO₄) = m(K₂CrO₄) ÷ M(K₂CrO₄). Thus, n(K₂CrO₄) = 339.6 g ÷ 194.2 g/mol thus arrives at n(K₂CrO₄) = 1.75 mol. 5) The dissociation of silver nitrate in water is expressed as: AgNO₃(aq) → Ag⁺(aq) + NO₃⁻(aq). Volume of solution a = 500 mL ÷ 1000 mL/L results in V(solution a) = 0.5 L. Concentration equation c(AgNO₃) = n(AgNO₃) ÷ V(solution a), thus c(AgNO₃) = 2 mol ÷ 0.5 L, yielding c(AgNO₃) = 4 mol/L = 4 M. As a result: c(AgNO₃) = c(Ag⁺) = c(NO₃⁻). Thus, c(Ag⁺) = 4 M; the concentration of silver ions in the initial solution a. 6) The dissociation of potassium chromate in water is represented as: K₂CrO₄(aq) → 2K⁺(aq) + CrO₄²⁻(aq). Volume of solution b = 500 mL ÷ 1000 mL/L results in V(solution b) = 0.5 L. Following, c(K₂CrO₄) is calculated as n(K₂CrO₄) ÷ V(solution b). So c(AgNO₃) = 1.75 mol ÷ 0.5 L gives c(AgNO₃) = 3.5 mol/L = 3.5 M. Consequently: c(K⁺) = 7 M; the concentration of potassium ions in solution b. Therefore, c(CrO₄²⁻) = 3.5 M; the concentration of chromium ions in the same solution. 7) The total final volume is V(final solution) = V(solution a) + V(solution b). Thus, V(final solution) = 500.0 mL + 500.0 mL leads to V(final solution) = 1000 mL ÷ 1000 mL/L results in V(final solution) = 1 L. Then n(NO₃⁻) = 2 mol. Therefore, c(NO₃⁻) = n(NO₃⁻) ÷ V(final solution) finds c(NO₃⁻) = 2 mol ÷ 1 L and results in c(NO₃⁻) = 2 M; the concentration of nitrate anions in the final solution. 8) In solution b, there are 3.5 mol of potassium cations while part of that combines with 2 moles of nitrate anions: K⁺(aq) + NO₃⁻(aq) → KNO₃(aq). From the reaction: n(K⁺): n(NO₃⁻) = 1: 1. Thus, Δn(K⁺) = 3.5 mol - 2 mol results in Δn(K⁺) = 1.5 mol, signifying the remaining potassium anions in the final solution. Thus, c(K⁺) = Δn(K⁺) ÷ V(final solution) yields c(K⁺) = 1.5 mol ÷ 1 L, leading to c(K⁺) = 1.5 M; the final concentration of potassium cations.