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marin [14]
3 years ago
5

gas has an initial volume of 2.4 L at a pressure of 1.5 atm and a temperature of 273 K. The pressure of the gas increases to 4.5

atm, and the temperature of the gas increases to 313 K.
Chemistry
2 answers:
QveST [7]3 years ago
8 0
We are given with two set of conditions. To calculate the final volume, we get first the number of moles under the first condition. Using PV=nRT, the number of moles is equal to 0.16 moles. We substitute this together with the other conditions in PV=nRT, the final volume is 0.92 liters.
JulsSmile [24]3 years ago
5 0

Answer:

B. 0.92 L

Explanation:

Edge2020

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How many calories of heat are needed to raise the temperature of 33.0 g of diethyl ether from 14.0 ∘C to 25.0 ∘C ?
Xelga [282]

The amount of heat (in calories) needed to raise the temperature is 324.885 calories

<h3>How to determine the temperature change </h3>

We'll begin by obtaining the temperature change. This can be obtained as followed:

  • Initial temperature (T₁) = 14 °C
  • Final temperature (T₂) = 25 °C
  • Change in temperature (ΔT) = ?

ΔT = T₂ - T₁

ΔT = 25 - 14

ΔT = 11 °C

<h3>How to determine the heat (in Calories)</h3>

The amount of heat needed to raise the temperature can bee obtaimedals follow:

  • Mass (M) = 33 g
  • Change in temperature (ΔT) = 11 °C
  • Specific heat capacity (C) of diethyl ether = 0.895 cal/g°C
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Q = MCΔT

Q = 33 × 0.895 × 11

Q = 324.885 calories

Thus, the amount of heat needed is 324.885 calories

Learn more about heat transfer:

brainly.com/question/10286596

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3 0
1 year ago
(CH3)2-CH-CH2-O(CH3)3IUPAC NAME
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Answer:

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Note: there is a mistake in formula, the correct formula is (CH₃)₂-CH-CH₂-O-C(CH₃)₃ not (CH₃)₂-CH-CH₂-O(CH₃)₃, because oxygen is a divalent compound.

Explanation:

<em>Structural formula is attached</em>

IUPAC naming rules

1. start numbering the chain from the functional group. In this compound we        start from oxygen side.

2. Here we can see that at position 1 there is an oxy group along with a tertiary carbon having three methyl groups. So we write it as 1-tert-butoxy. Which means that there is a methoxy group at position 1 along with a tertiary carbon.

3. At position 2 we can see that there is a methyl group attached to the main chain, so we write it as 2-methyl.

4. Now we count the total number of carbons in the main chain. As we can see that there are 3 carbons in the remaining or parent chain, so we write it as propane

5. So the IUPAC name of the compound will be 1-(tert-butoxy)-2-methylpropane

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