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icang [17]
2 years ago
9

Which example is an endothermic process

Chemistry
1 answer:
Annette [7]2 years ago
4 0

Answer:

Endothermic reactions is the result of heat being absorbed.

Explanation:

Melting ice cubes. Melting solid salts. Evaporating liquid water. Converting frost to water vapor (melting, boiling, and evaporation, in general, are endothermic processes.

(These are all examples of endothermic reactions.)

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Graphics can sometimes convey more information in a brief amount of space than an author can explain in a paragraph.

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3 0
2 years ago
Give the electron confiuration of Chromium (Cr), atomic number 24.
nadezda [96]
<span>Chromium is a transition metal and it has 24 electrons and here is the orbital diagram. If we're going to make this short hand and make the electron configurationfor this we would make this 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d4 okay from now on every time you see 3d4 you're going to change it, we do not like 3d4.</span>
4 0
3 years ago
Read 2 more answers
Calculate the volume which 1.00 mole of a gas occupies at 1 atm and 298K?
elena-14-01-66 [18.8K]

Answer:

25.45 Liters

Explanation:

Using Ideal Gas Law PV = nRT => V = nRT/P

V = (1mole)(0.08206Latm/molK)(298K)/(1atm) = 25.45 Liters

7 0
2 years ago
The following reaction shows the products when sulfuric acid and aluminum hydroxide react.
Elden [556K]

The correct answer is approximately 11.73 grams of sulfuric acid.

The theoretical yield of water from Al(OH)3 is lower than that of H₂SO₄. As a consequence, Al(OH)3 is the limiting reactant, H₂SO₄ is in excess.

The balanced equation is:

2Al(OH)₃ + 3H₂SO₄ ⇒ Al₂(SO₄)₃ + 6H₂O

Each mole of Al(OH)3 corresponds to 3/2 moles of H₂SO₄. The molecular mass of Al(OH)3 is 78.003 g/mol. There are 15/78.003 = 0.19230 moles of Al(OH)3 in the five grams of Al(OH)3 available. Al(OH)3 is in limiting, which means that all 0.19230 moles will be consumed. Accordingly, 0.19230 × 3/2 = 0.28845 moles of H₂SO₄ will be consumed.

The molar mass of H₂SO₄ is 98.706 g/mol. The mass of 0.28845 moles of H₂SO₄ is 0.28845 × 98.706 = 28.289 g

40 grams of sulfuric acid is available, out of which 28.289 grams is consumed. The remaining 40-28.289 = 11.711 g is in excess, which is closest to the first option, that is, 11.73 grams of H₂SO₄.

6 0
2 years ago
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