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34kurt
3 years ago
5

A solution of the ionic salt CaSO3 would have blank pH

Chemistry
1 answer:
stepladder [879]3 years ago
8 0

Answer: basic

Explanation:

I just answered it on acellus it's correct ;)

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Iontophoresis is a noninvasive process that transports drugs through the skin without needles. In the photo, the red electrode i
anyanavicka [17]

Answer:

Follows are the explanation to this question:

Explanation:

When the drug is negatively charged, its negative electrolyte is annihilated to just the positive electrode. It is enticed, and it may not have a picture showing the electrode, however, We suppose that electrodes from either side of a skin slice. Its negative electrode will bypass or push thru the skin if in front of the counter terminal this becomes a red-positive electrode.

8 0
3 years ago
An ant carries a morsel of food 4.26 meters along a straight path to his nest. he then turns around and follows the path back to
SVEN [57.7K]
<span>Answer: 0.00 meters

  Solution:

    Step 1: Define displacement DISPLACEMENT = a vector quantity that describes "linear or angular distance in a given direction between a body or point and a reference position."

  Step 2: Understand the question

 Assumption 1: Assume that when the ant moves 4.25 meters from its origin to its nest, it is moving in a positive direction (on a graph you would draw a line along the x-axis from its origin to +4.25).

  Assumption 2: Assume that when the ant "turns around...back to the source of food", it is moving back in the negative direction (towards the origin).

   Step 3: Analyze the question

  What is the distance between where the ant originally started and where it ended its journey?

  The ant started and ended its journey in the same place.

 While it traveled a distance of 8.52 meters (2 * 4.26 = 8.52), it's displacement is actually 0.00 meters (4.26 + (-4.26) = 0.00)

  Therefore, the answer is 0.00 meters</span>
5 0
3 years ago
Mendeleev created this table as he noticed that a
Nimfa-mama [501]

Answer:

Mendeleev had left the noble gases out of his periodic table.

Explanation:

Mendeleev's periodic table is pictured in the image attached to the question.

Mendeleev's table obviously lacked the noble gases. The reason for this grave omission is simple; the noble gases were not known as at the time when he formulated his periodic table. There weren't any known elements  whose properties were similar to the  properties of the noble gases. This would have lead him to suspect their existence.

4 0
3 years ago
Which pair of aqueous solutions can create a buffer solution if present in the appropriate concentrations?.
Julli [10]

HF and NaF - If the right concentrations of aqueous solutions are present, they can produce a buffer solution.

<h3>What are buffer solutions and how do they differ?</h3>
  • The two main categories of buffers are acidic buffer solutions and alkaline buffer solutions.
  • Acidic buffers are solutions that contain a weak acid and one of its salts and have a pH below 7.
  • For instance, a buffer solution with a pH of roughly 4.75 is made of acetic acid and sodium acetate.
<h3>Describe buffer solution via an example.</h3>
  • When a weak acid or a weak base is applied in modest amounts, buffer solutions withstand the pH shift.
  • A buffer made of a weak acid and its salt is an example.
  • It is a solution of acetic acid and sodium acetate CH3COOH + CH3COONa.

learn more about buffer solutions here

<u>brainly.com/question/8676275</u>

#SPJ4

8 0
1 year ago
A buffer solution contains 0.345 M acetic acid and 0.377 M sodium acetate . If 0.0613 moles of potassium hydroxide are added to
melamori03 [73]

Answer:

pH = 5.54

Explanation:

The pH of a buffer solution is given by the <em>Henderson-Hasselbach (H-H) equation</em>:

  • pH = pKa + log\frac{[CH_3COO^-]}{[CH_3COOH]}

For acetic acid, pKa = 4.75.

We <u>calculate the original number of moles for acetic acid and acetate</u>, using the <em>given concentrations and volume</em>:

  • CH₃COO⁻ ⇒ 0.377 M * 0.250 L = 0.0942 mol CH₃COO⁻
  • CH₃COOH ⇒ 0.345 M * 0.250 L = 0.0862 mol CH₃COOH

The number of CH₃COO⁻ moles will increase with the added moles of KOH while the number of CH₃COOH moles will decrease by the same amount.

Now we use the H-H equation to <u>calculate the new pH</u>, by using the <em>new concentrations</em>:

  • pH = 4.75 + log\frac{(0.0942+0.0613)mol/0.250L}{(0.0862-0.0613)mol/0.250L} = 5.54
6 0
3 years ago
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