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oksano4ka [1.4K]
2 years ago
9

Molecules have Question 8 options: A) only kinetic energy. B) neither kinetic nor potential energy. C) only potential energy. D)

both potential and kinetic energy.
Chemistry
1 answer:
emmasim [6.3K]2 years ago
6 0
The answer is C. only potential energy
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Most chemical reactions involve the breaking and formation of chemical bonds. It takes energy to break a chemical bond but energy is released when chemical bonds are formed. If more energy is released than consumed, then the chemical reaction evolves heat and is said to be exothermic.

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Who is your favorite scientist?
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2 years ago
Select the missing words to complete the definition of buffer capacity. Buffer capacity is the _____________ of acid or base a b
horsena [70]

Answer:

amount, pH value.

Explanation:

The buffer range is the pH range in which the buffer performs optimally, i.e., neutralizes even when a strong acid or base is introduced to it and resists any major change in its pH value.

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Thus, the final statement becomes,

Buffer capacity is the amount of acid or base a buffer can handle before pushing the pH value outside of the buffer range.

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2 years ago
Hypothesis and Data Collection Part A The image shows three different liquids: water, water with salt, and vinegar. The pH of ea
Norma-Jean [14]

Answer:

The water would be neutral, (usually 7). The salt water would be the same (7) and the vinegar would be very acidic. (probably 2).

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7 0
1 year ago
In Lab 10 you make a stock solution of salicylic acid, and then four dilutions. The stock solution is made by diluting 5.00 ml o
Zigmanuir [339]

Answer:

Stock  solution =  1.25 *10^-3 M

Dilution 1 = 5*10^-4 M

Dilution 2= 3.75 * 10^-4 M

Dilution 3 = 2.5 *10^-4 M

Dilution 4 = 1.25 *10^-4 M

Explanation:

<u>Step 1:</u> Data given

The stock solution is made by diluting 5.00 ml of 1.250 x 10-2 M salicylic acid in 50.00 mL of solution.

<u>Step 2</u>: Calculate the concentration of the stock solution:

M1*V1 = M2*V2

⇒ with M1 = the initial concentration = 1.250 *10^-2 M

⇒ with V1 = 5 mL = 5*10^-3 L

⇒ with M2 = TO BE DETERMINED

⇒ with V2 = 50 mL = 50 *10^-3 L

M2 = (M1*V1)/V2

M2 = (1.250 *10^-2 * 5*10^-3 L) / 50 *10^-3

M2 = 0.00125 M = 1.25 *10^-3 M

<u>Step 3:</u> Calculate dilution 1

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 10 *10^-3)/(25*10^-3L)

M2 = 0.0005 M = 5*10^-4 M

<u>Step 4</u>: Calculate dilution 2

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 10 * 7.5*10^-3)/(25*10^-3)

M2 = 0.000375 M = 3.75 * 10^-4 M

<u>Step 5:</u> Calculate dilution 3

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 5*10^-3) /(25*10^-3)

M2 = 0.00025 M = 2.5 *10^-4 M

<u>Step 6</u>: Calculate dilution 4

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 2.5*10^-3)/(25*10^-3)

M2 = 0.000125 M = 1.25 *10^-4 M

5 0
3 years ago
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