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Allushta [10]
3 years ago
5

At which temperature would a can of soda be most fizzy when it is opened? 30°C 40°C 20°C 10°C

Chemistry
2 answers:
scoray [572]3 years ago
5 0

I think the answer is 40⁰C

I hope this helps you!

Helga [31]3 years ago
4 0

The answer to your question is 40° because freezing temp for a liquid is 32°

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Place the following substances in the following order: soluble, somewhat soluble, and insoluble.
dimulka [17.4K]
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3 years ago
What’s the formula equation for
jeyben [28]
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7 0
3 years ago
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An "empty" container is not really empty if it contains air. How may moles of nitrogen are in an "empty" two-liter cola bottle a
Lisa [10]

Answer:

1. 0.0637 moles of nitrogen.

2. The partial pressure of oxygen is 0.21 atm.  

Explanation:

1. If we assume ideal behaviour, we can use the Law of ideal gases to find the moles of nitrogen, considering that air composition is mainly nitrogen (78%), oxygen (21%) and argon (1%):  

V_{N_2}=V_{T}\times 0.78=2L \times 0.78 =1.56 L\\PV=nRT\\n_{N_2}=\frac{PV}{RT}=\frac{1 atm\times 1.56 L}{0.0821\frac{atmL}{molK}\times 298 K}\\n_{N_2}= 0.0637 mol

2. Now, in order to find he partial pressure of oxygen we need to find the total moles of air, and then the moles of oxygen. Then, we use these results to determine the molar fraction of oxygen, to multiply it with total pressure and get the partial pressure of oxygen as follows:

n_{total}=\frac{1 atm \times 2L}{0.0821 \frac{atmL}{molK}298K}=0.0817 mol

V_{O_2}=2L \times 0.21 = 0.42 L\\n_{O_2}=\frac {1atm \times 0.42 L}{0.0821 \frac{atm L}{mol K}298 K}=0.0172 mol\\X_{O_2}=\frac{n_{O_2}}{n_{total}}=\frac{0.0172 mol}{0.0817 mol}= 0.21

P_{O_2}=X_{O_2} \times P = 1 atm \times 0.21 = 0.21 atm

As you see, the molar fraction and volume fraction are the same because of the assumption of ideal behaviour.  

3 0
3 years ago
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algol13

the cathode is where reduction occurs!

5 0
3 years ago
Water has a density of 1g/ml. what is the mass of the water if it fills a 10ml container?
notka56 [123]
It's 10.
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M = 1g/ml(10ml) = 10g
8 0
3 years ago
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