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Lemur [1.5K]
2 years ago
11

Please do not write answer in a zip file

Chemistry
1 answer:
Contact [7]2 years ago
5 0

Answer:

Explanation:

Part A

The pH of a solution is given by the negative concentration of hydrogen ions in the solution

2.0 mL = 0.002 L

The number of moles of HCl in 2.0 mL of 4.0 M HCl is given as follows;

1 Liter of 4.0 M HCl contains 4.0 moles of HCl

2.0 mL = 0.002 L 4.0 M HCl contains 0.002 L/(1 L) × 4.0 M = 0.008 moles of HCl

The concentration of 0.008 moles in 1.50 L is given as follows;

Concentration = The number of moles/(The volume in liters)

∴ The concentration of 0.008 moles in 1.50 L, C = 0.008 moles/(1.5 L + 0.002 L)

∴ The concentration of 0.008 moles in 1.50 L, C ≈ 0.00533 moles/liter = 0.00533 M HCl

Given that HCl is a strong acid, we have that HCl dissociates completely to give equal number of H⁺ and Cl⁻ ions;

The number of moles of [H⁺] in the solution = 0.00533 moles

The pH of the solution = -log[H⁺]

∴ pH = -log[5.33 × 10⁻³] ≈ 2.273

The pH of the 1.5 L of pure water will be approximately 2.273

Part B

The pH of the pure water has changed from neutral (pH = 7) tp pH = 2.273

The change in pH is ΔpH = 2.274 - 7 = -4.726

ΔpH ≈ -4.726

Part C

When 2.0 mL of the 4.0 M HCl is added, the solution above, we have;

C = (0.008 + 0.008)/(1.5 + 0.002 + 0.002) ≈ 1.06383 × 10⁻²

The concentration of the solution becomes, C ≈ 1.06383 × 10⁻² mole/liter

The pH becomes, pH = -log(1.06383 × 10⁻²) ≈ 1.973

Part D

The amount by which the pH has changed, ΔpH ≈ 1.973 - 2.274 = -0.301.

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False

Explanation:

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List two detection (i.e. visualisation) techniques commonly used to visualise<br> compounds in TLC.
olya-2409 [2.1K]

Answer:

The most common non-destructive visualization method for TLC plates is ultraviolet (UV) light. A UV lamp can be used to shine either short-waved (254nm) or long-waved (365nm) ultraviolet light on a TLC plate with the touch of a button

Explanation:

hope this helps

7 0
3 years ago
a book with a mass of 1 kg is dropped from a height of 3 m. what is the potential energy of the book when it reaches the floor?
mafiozo [28]
Well, first of all, the formula for finding potential energy is;
PE=mgh
  Where; m is the mass
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                h is the height. 
   Anyway, according to the question, the mass is 1kg, the acceleration due to gravity has a constant value of 10ms² . And the height is 3m. Now you just have to use all these in the formula. So;
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5 0
3 years ago
Water is poured into a conical container at the rate of 10 cm3/sec. The cone points directly down, and it has a height of 20 cm
8090 [49]

Answer:

\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{2}

Explanation:

Hello,

The suitable differential equation for this case is:

\frac{dV}{dt}=10\frac{cm^3}{s}

As we're looking for the change in height with respect to the time, we need a relationship to achieve such as:

\frac{dh}{dt} = ?*\frac{dV}{dt}

Of course, ?=\frac{dh}{dV}.

Now, since the volume of a cone is V=\pi r^2h/3 and the ratio r/h=15/20=3/4 or r=3/4h, the volume becomes:

V=\pi (\frac{3}{4} h)^2h/3= \frac{3}{16}\pi h^3

We proceed to its differentiation:

\frac{dV}{dh} =\frac{9}{16} \pi h^2\\\frac{dh}{dV} =\frac{16}{9 \pi h^2}

Then, we compute \frac{dh}{dt}

\frac{dh}{dt} = \frac{16}{9 \pi h^2}*\frac{dV}{dt}\\\frac{dh}{dt} = \frac{16}{9\pi h^2}*10\frac{cm^3}{s} =\frac{160}{9 \pi h^2}

Finally, at h=2:

\frac{dh}{dt}_{h=2cm} =\frac{160}{9\pi 2^2}\\\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{s}

Best regards.

4 0
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Answer:

I would say, what helps me is really paying attention in class and asking questions, also making sure you study for upcoming test's and quizzes and completely assingments on time

Explanation:

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