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STALIN [3.7K]
3 years ago
10

Minerals transferred the same amont of energy into both minerals. Why did one mineral change and the other stayed the same?

Chemistry
1 answer:
IrinaK [193]3 years ago
8 0

Answer: Mineral A changed because Molecular energy transferred is equal or greater than than its Activation energy

Mineral B didn't change because Molecular energy transferred is less than its Activation energy.

Explanation:

The molecules of Mineral A has been disturbed by the addition of energy causing a change and the entropy is increased. The Molecules of Mineral B has not been disturbed because the residual energy has not been overcome and therefore yielding no visible change

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Marianna [84]

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Are there any vegetarian Italian Sausages?
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This is hard, isn't it? please help me though. PLEASE
Liono4ka [1.6K]
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2 years ago
A block of iron has a mass of 150 g and a volume of 20 ml. What is the density of iron? *
shtirl [24]

Answer:

<h3>The answer is 7.50 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question

mass of iron = 150 g

volume = 20 mL

We have

density =  \frac{150}{20}  =  \frac{15}{2}  \\

We have the final answer as

<h3>7.50 g/mL</h3>

Hope this helps you

4 0
3 years ago
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The pKa of lactic acid is 3.9. A lactate buffer will be useful from pH values ________. The pKa of lactic acid is 3.9. A lactate
Vedmedyk [2.9K]

Answer:

Explanation:

The usefulness of a buffer is its ability to resist changes in pH when small quantities of base or acid are added to it. This ability is the consequence of having both the conjugate base and the weak acid present in solution which will consume the added base or acid.  

This capacity is lost if the ratio of the concentration of conjugate base to the concentration of weak acid differ by an order of magnitude. Since  buffers having ratios differing by more will have their pH driven by either the weak acid or its conjugate base .

From the Henderson-Hasselbach equation we have that

pH = pKa + log [A⁻]/[HA]

thus

0.1 ≤  [A⁻]/[HA] ≤ 10

Therefore the log of this range is -1 to 1, and the pH will have a useful range of within +/- 1 the pKa of the buffer.

Now we are equipped to answer our question:

pH range = 3.9 +/- 1 = 2.9 through 4.9

7 0
3 years ago
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