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Misha Larkins [42]
3 years ago
11

Why does the temperature not change

Chemistry
1 answer:
S_A_V [24]3 years ago
6 0

Answer:

Option A

Explanation:

Temperature of a body is due to the heat gained or loss. During a phase change, the atoms or molecules of a substance are undergoing change is temperature due to which no temperature change is observed during phase change. The heat in the transition phase is used to break bonds and the change in temperature is felt when kinetic energy change is complete. During transition, the average kinetic energy of the molecules remains unchanged and hence during a phase change a temperature do not changes until unless the phase change is completed.

Hence, option A is correct

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What are three forms of the element carbon?
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Answer:

There are at least eight forms of carbon: amorphous, graphite, diamond, fullerenes, aggregated diamond nanorods, carbon nanofoam, glassy carbon, and linear acetylenic carbon.

Explanation:

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3 years ago
Trace amounts of sulfur (S) in coal are burned in the presence of diatomic oxygen (O2) to form sulfur dioxide (SO2). Determine t
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Answer:

0.99 kg O₂

1.9 kg SO₂

Explanation:

Let's consider the reaction between sulfur and oxygen to form sulfur dioxide.

S + O₂ → SO₂

The mass ratio of S to O₂ is 32.07:32.00. The mass of oxygen required to react with 1 kg of sulfur is:

1 kg S × (32.00 kg O₂/32.07 kg S) = 0.998 kg O₂

The mass ratio of S to SO₂ is 32.07:64.07. The mass of sulfur dioxide formed when 1 kg of sulfur is burned is:

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4 years ago
formic acid buffer containing 0.50 M HCOOH and 0.50 M HCOONa has a pH of 3.77. What will the pH be after 0.010 mol of NaOH has b
HACTEHA [7]

Answer:

pH = 3.95

Explanation:

It is possible to calculate the pH of a buffer using H-H equation.

pH = pka + log₁₀ [HCOONa] / [HCOOH]

If concentration of [HCOONa] = [HCOOH] = 0.50M and pH = 3.77:

3.77 = pka + log₁₀ [0.50] / [0.50]

<em>3.77 = pka</em>

<em />

Knowing pKa, the NaOH reacts with HCOOH, thus:

HCOOH + NaOH → HCOONa + H₂O

That means the NaOH you add reacts with HCOOH producing more HCOONa.

Initial moles of 100.0mL = 0.1000L:

[HCOOH] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOOH

[HCOONa] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOONa

After the reaction, moles of each species is:

0.0500moles HCOOH - 0.010 moles NaOH (Moles added of NaOH) = 0.0400 moles HCOOH

0.0500moles HCOONa + 0.010 moles NaOH (Moles added of NaOH) = 0.0600 moles HCOONa

With these moles of the buffer, you can calculate pH:

pH = 3.77 + log₁₀ [0.0600] / [0.0400]

<h3>pH = 3.95</h3>

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