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MaRussiya [10]
3 years ago
15

How many atoms are in 4.93 moles hcl?

Chemistry
1 answer:
trasher [3.6K]3 years ago
4 0
1 mole ------------- 6.02x10²³ atoms
4.93 moles ------- ??

4.93 x ( 6.02x10²³) / 1 = 

=> 2.96x10²⁴ atoms
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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
Which of the following is an example of oxidation? A. Charcoal is placed on a grill and lighting fluid poured over it. B. A car
Vikentia [17]

Answer:

<u>The Answer is (B)  A car get rusty over the course of few years</u>Explanation:

<u>Explanation:</u>

  • <u>Oxidation </u>refers to the process of loss of electrons by a molecule,atom or ion during a chemical reaction.The process which is just the opposite of oxidation is reduction,it occurs when their is gain of electrons .
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<u></u>

4 0
3 years ago
How do I convert 125000g to scientific notation?
lakkis [162]
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4 0
3 years ago
Salt is poured from a container at 10 cm³ s-¹ and it formed a conical pile whose height at any time is 1/5 the radius of the abo
Romashka-Z-Leto [24]

Answer:

\displaystyle \frac{dh}{dt} = \frac{1}{10 \pi}

Explanation:

Volume of a cone:

  • \displaystyle V=\frac{1}{3} \pi r^2 h

We have \displaystyle \frac{dV}{dt} = \frac{10 \ cm^3}{sec} and we want to find \displaystyle \frac{dh}{dt} \Biggr | _{h\ =\ 6}= \ ? when the height is 2 cm.

We can see in our equation for the volume of a cone that we have three variables: V, r, and h.

Since we only have dV/dt and dh/dt, we can rewrite the equation in terms of h only.

We are given that the height of the cone is 1/5 the radius at any given time, 1/5r, so we can write this as r = 5h.

Plug this value for r into the volume formula:

  • \displaystyle V =\frac{1}{3} \pi (5h)^2 h  
  • \displaystyle V =\frac{1}{3} \pi \ 25h^3

Differentiate this equation with respect to time t.

  • \displaystyle \frac{dV}{dt}  =\frac{25}{3} \pi \ 3h^2 \ \frac{dh}{dt}
  • \displaystyle \frac{dV}{dt}  =25 \pi h^2 \ \frac{dh}{dt}

Plug known values into the equation and solve for dh/dt.

  • \displaystyle 10 = 25 \pi (2)^2  \ \frac{dh}{dt}
  • \displaystyle 10 = 100 \pi  \ \frac{dh}{dt}  

Divide both sides by 100π to solve for dh/dt.

  • \displaystyle \frac{10}{100 \pi} = \frac{dh}{dt}
  • \displaystyle \frac{dh}{dt} = \frac{1}{10 \pi}

The height of the cone is increasing at a rate of 1/10π cm per second.

7 0
3 years ago
For every 4 hours that Alexa babysit her younger sister, her mom pays her $50
tiny-mole [99]

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7 0
3 years ago
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