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sdas [7]
3 years ago
13

A volume of 5.0 L of gas at a temperature of 25°C is cooled to a temperature of -120°C at constant pressure. What is the new vol

ume of the gas?
Chemistry
1 answer:
lorasvet [3.4K]3 years ago
5 0
V1 = 5L = 5000ml V2 = ?
T1 = 25°C = 25+273 T2 = -120°C
= 297K = -120+273
= 153K
By Charles Law,
V1/T1 = V2/T2
5000/297 = V2/153
By solving,
V2 = 2.576L
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Which method is used to separate the following mixture (a) water + kerosene​
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Answer:

The separation of kerosene, oil, and water are immiscible liquids, so they can be separated funnel.

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3 years ago
Help please
Mekhanik [1.2K]

Answer:

it's the regolith

hope this helps you

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7 0
3 years ago
A six-carbon organic compound containing oxygen is suspected of being either a secondary alcohol or a ketone. which chemical or
White raven [17]
Following are the possible isomers of secondary alcohol and ketones for six carbon molecules. In order to distinguish between sec. alcohol and ketone we can simply treat the unknown compound with acidified Potassium Dichromate (VI) in the presence of acid. If with treatment with unknown compound the colour of K2Cr2O7 (potassium dichromate VI) changes from orange to green then it is confirmed that the unknown compound is sec. alcohol, or if no change in colour is detected then ketone is confirmed. This is because ketone can not be further oxidized while, sec. alcohol can be oxidized to ketones as shown below,

6 0
3 years ago
How would having too much sample in the melting point tube most likely affect the melting point measurement? Select the correct
oksano4ka [1.4K]

Answer:

2-4 mm height of capillary tube.

Explanation:

Sample should be around 2-4 mm in height.

It should be packed well so that it does not have air packets that caues the lowering of melting point.

If you take greater amount, then there will be needed more heat, resulting a wide range of melting point.

7 0
3 years ago
If 15 g of C₂H₆ reacts with 60.0 g of O₂, how many moles of water (H₂O) will be produced?
IceJOKER [234]

Answer:

n_{H_2O}=1.5molH_2O

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O

Next, we identify the limiting reactant by computing the available moles of ethane and the moles of ethane consumed by 60.0 grams of oxygen:

n_{C_2H_6}^{available}=15g*\frac{1mol}{30g} =0.50molC_2H_6\\n_{C_2H_6}^{reacted}=60.0gO_2*\frac{1molO_2}{32gO_2}*\frac{2molC_2H_6}{7molO_2} =0.536molC_2H_6

Thus, we notice there are less available moles, for that reason, the ethane is the limiting reactant. Finally, we can compute the produced moles of water by:

n_{H_2O}=0.50molC_2H_6*\frac{6molH_2O}{2molC_2H_6}\\\\n_{H_2O}=1.5molH_2O

Best regards.

5 0
3 years ago
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