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Andrew [12]
3 years ago
15

A sample of argon initially has a volume of 5.0 L and the pressure is 2 atm. If the final temperature is 30° C, the final volume

is 6 L, and the final pressure is 8atm, what was the initial temperature of the argon? Please give answer WITHOUT the unit of KELVIN and round to the tenth position (two places to the right of the decimal). Example: 25.14 not 25.4K
Chemistry
1 answer:
jenyasd209 [6]3 years ago
4 0
Volume of Argon V1 = 5.0 L  
Pressure of Argon P1 = 2 atm 
Final temperature T2 = 30 C = 30 + 273 = 303 K 
Volume at final temperature V2= 6 L 
Pressure at final temperature P2 = 8 atm  
We know that (P1 x V1) / T1 = (P2 x V2) / T2  
(2 x 5)/ T1 = (8 x 6)/ 303 => T1 = (10 x 303) / 48 
Initial Temperature T1 = 3030 / 48 = 63.12 
Initial Temperature = -209. 8 C
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Explanation:

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Explanation:

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Compute the sugar content in an 8 oz sample of a soft drink. If the sugar content as per label on the product =10g per 100ml.​
Nataly_w [17]

Answer:

m_{sugar}=23.7g\ sugar

Explanation:

Hello,

In this case, we can first compute the volume of the sample in mL from the ounces:

8oz*\frac{29.5735mL}{1oz} =236.6mL

Thus, with the volume of the sample, we can compute the amount of sugar given the 10 g of sugar per 100 mL of soft drink as shown below:

m_{sugar}=236.6mL*\frac{10g\ sugar}{100mL}\\ \\m_{sugar}=23.7g\ sugar

Best regards.

6 0
3 years ago
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