Explain how a common housecat gets “worms.” <br><br> No google or else i report you.

Molodets [167]

Moves from the atmosphere to the oceans

3 0

2 years ago

What volume of 0.307 m naoh must be added to 200.0ml of 0.425m acetic acid (ka = 1.75 x 10-5 ) to produce a buffer of ph = 4.250

Blababa [14]

The buffer solution target has a pH value smaller than that of pKw (i.e., pH < 7.) The solution is therefore acidic. It contains significantly more protons $\text{H}^{+}$ than hydroxide ions $\text{OH}^{-}$ . The equilibrium equation shall thus contain protons rather than a combination of water and hydroxide ions as the reacting species.

Assuming that $x \; \text{L}$ of the 0.307 $\text{mol} \cdot \text{dm}^{-3}$ sodium hydroxide solution was added to the acetic acid. Based on previous reasoning, $x$ is sufficiently small that acetic acid was in excess, and no hydroxide ion has yet been produced in the solution. The solution would thus contain $0.2000 \times 0.425 - 0.307 \; x = 0.085 - 0.307 \; x$ moles of acetic acid and $0.307 \; x$ moles of acetate ions.

Let $\text{HAc}$ denotes an acetic acid molecule and $\text{Ac}^{-}$ denotes an acetate ion. The RICE table below resembles the hydrolysis equilibrium going on within the buffer solution.

$\begin{array}{lccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\end{array}$

The buffer shall have a pH of 4.250, meaning that it shall have an equilibrium proton concentration of $10^{4.250}\; \text{mol}\cdot \text{dm}^{-3}$ . There were no proton in the buffer solution before the hydrolysis of acetic acid. Therefore the table shall have an increase of $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ in proton concentration in the third row. Atoms conserve. Thus the concentration increase of protons by $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ would correspond to a decrease in acetic acid concentration and an increase in acetate ion concentration by the same amount. That is:

$\begin{array}{lcccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\text{C} & - 10^{-4.250} & & +10^{-4.250} & & +10^{-4.250} \\\text{E} & 0.085 - 10^{-4.250} - 0.307 \; x& & 10^{-4.250} & & 10^{-4.250} + 0.307 \; x\end{array}$

By definition:

$\text{K}_{a} = [\text{H}^{+}] \cdot [\text{Ac}^{-}] / [\text{HAc}]\\\phantom{\text{K}_{a}} = 10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x)$

The question states that

$\text{K}_{a} = 1.75 \times 10^{-5}$

such that

$10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x) = 1.75 \times 10^{-5}\\6.16 \times 10^{-5} \; x = 1.48 \times 10^{-6}\\x = 0.0241$

Thus it takes $0.0241 \; \text{L}$ of sodium hydroxide to produce this buffer solution.

6 0

3 years ago

Explain how dalton's theory of the atom and the conservation of mass are related

Scorpion4ik [409]

Answer:

Dalton's atomic theory states that atoms may be mixed in certain rations to produce compounds, while the law of conservation of mass states that the mass of reactants equals the mass of products. They are related because in the production of compounds, Dalton made it clear that mass can neither be destroyed or created, which supports the conservation of mass law.

Explanation:^^^

Hoped it helped...

8 0

2 years ago

Does the law of inertia apply to both moving and non-moving objects

zhuklara [117]

Yes becuase it says it maintains its velocity if it is moving or at rest

8 0

3 years ago

A buret is used to dispense standardized NaOH solution. The initial buret reading is 2.73 mL, and after dispensing a known volum

Novosadov [1.4K]

Answer:

1.62x10⁻³ moles of NaOH were dispensed

Explanation:

Molarity is an unit in chemistry defined as the ratio between moles of solute (In the problem, NaOH), per liter of solution.

The concentration of the solution is 0.125moles per liter. That means 1L of solution has 0.125 moles of NaOH.

The volume you dispensed in the buret was:

15.67mL - 2.73mL =

12.94mL of the 0.125M NaOH are:

12.94mL = 0.01294L * (0.125moles / L) =

<h3>1.62x10⁻³ moles of NaOH were dispensed</h3>

3 0

3 years ago

Explain how a common housecat gets “worms.” No google or else i report you.