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Alja [10]
3 years ago
11

If you drop a 50 gram piece of metal with a temperature of 125° Celsius into 1000 grams of water at 20° Celsius, what best descr

ibes what would occur?
Chemistry
1 answer:
Nostrana [21]3 years ago
3 0
Dropping a 50 gram piece of metal with a temperature of 125° Celsius into 1000 grams of water at 20° Celsius would lead to a heat loss by the metal and an heat gain by the water. These loss and gain should be equal since energy cannot be created nor destroyed.
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In dimensional analysis you should always have the same units adjacent to each other (top left to bottom right)
aleksklad [387]

Answer:

False

Explanation:

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Dehydration of 2-methyl-2-pentanol forms one major and one minor organic product. Draw the structures of the two organic product
Korvikt [17]

Answer:

The major product is 2-methyl-2-pentene [ CH₃-CH₂-CH=C(CH₃)₂ ] and a minor product 2-methyl-1-pentene [ CH₃-CH₂-CH₂-C(CH₃)=CH₂ ].

Explanation:

Dehydration reaction is a reaction in which a molecule loses a water molecule in the presence of a dehydrating agent like sulfuric acid (H₂SO₄).

<u>Dehydration reaction of 2-methyl-2-pentanol</u> gives a major product 2-methyl-2-pentene and a minor product 2-methyl-1-pentene.

CH₃-CH₂-CH₂-C(CH₃)₂-OH (2-methyl-2-pentanol)→ CH₃-CH₂-CH=C(CH₃)₂ (2-methyl-2-pentene, major) + CH₃-CH₂-CH₂-C(CH₃)=CH₂ (2-methyl-1-pentene, minor)

<u>Since more substituted alkene is more stable than the less substituted alkene. So, the trisubstituted alkene, 2-methyl-2-pentene is more stable than the disubstituted alkene, 2-methyl-1-pentene.</u>

<u>Therefore, the trisubstituted alkene, 2-methyl-2-pentene is the major product and the disubstituted alkene, 2-methyl-1-pentene is the minor product.</u>

7 0
2 years ago
An ethylene gas torch requires 300 L of gas at 0.8 atm. What will be the pressure of the gas if ethylene is supplied by a 200.0
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Answer:

1.2 atm

Explanation:

Given data

  • Volume of the gas in the tank (V₁): 200.0 L
  • Pressure of ethylene gas in the tank (P₁): ?
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If we consider ethylene gas to be an ideal gas, we can find the pressure of ethylene gas in the tank using Boyle's law.

P_1 \times V_1 = P_2 \times V_2\\P_1 = \frac{P_2 \times V_2}{V_1} = \frac{0.8atm \times 300L}{200.0L} = 1.2 atm

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I think the answer you are looking for is D.

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