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julia-pushkina [17]
3 years ago
13

210.0 °C a gas has a volume of 8.00 L. What is the volume of this gas at -23.0°C?

Chemistry
1 answer:
Firlakuza [10]3 years ago
5 0
You will need to use Charles Law. Volume1/Temperature1 = Volume2/Temperature2 Note all temperatures must be in Kelvin. You can convert Celsius to Kelvin by adding 273. Then plug in and solve . You should get 4.14L
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The answer to this question would be C. I hope this helps!

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3 years ago
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The following diagrams represent mixtures of NO(g) and O2(g). These two substances react as follows: 2NO(g)+O2(g)→2NO2(g) It has
Alja [10]

This is an incomplete question, here is a complete question and an image is attached below.

The following diagrams represent mixtures of NO(g) and O₂(g). These two substances react as follows:

2NO(g)+O_2(g)\rightarrow 2NO_2(g)

It has been determined experimentally that the rate is second order in NO and first order in O₂.

Based on this fact, which of the following mixtures will have the fastest initial rate?

The mixture (1). The mixture (2). The mixture (3).

Answer : The mixture 1 has the fastest initial rate.

Explanation :

The given chemical reaction is:

2NO(g)+O_2(g)\rightarrow 2NO_2(g)

The rate law expression is:

Rate=k[NO]^2[O_2]

Now we have to determine the number of molecules of NO\text{ and }O_2

In mixture 1 : There are 5 NO and 4 O_2 molecules.

In mixture 2 : There are 7 NO and 2 O_2 molecules.

In mixture 3 : There are 3 NO and 5 O_2 molecules.

Now we have to determine the rate law expression for mixture 1, 2 and 3.

The rate law expression for mixture 1 is:

Rate=k[NO]^2[O_2]

Rate=k(5)^2\times (4)

Rate=k(100)

The rate law expression for mixture 2 is:

Rate=k[NO]^2[O_2]

Rate=k(7)^2\times (2)

Rate=k(98)

The rate law expression for mixture 3 is:

Rate=k[NO]^2[O_2]

Rate=k(3)^2\times (5)

Rate=k(45)

Hence, the mixture 1 has the fastest initial rate.

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Explanation:

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Explanation:

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3 years ago
Calculate the number of grams of oxygen required to convert 48.0 g of glucose to co2 and h2o.
worty [1.4K]

I believe that the balanced chemical reaction is:

C6H12O6 + 6 O2 → 6 CO2 + 6 H2O 

 

So the number of grams of oxygen required is:

mass O2 required = 48 g C6H12O6 * (1 mole C6H12O6 / 180.16 g) * (6 mole O2 / 1 mole C6H12O6) * (32 grams O2 / 1 mole)

<span>mass O2 required = 51.15 grams</span>

4 0
3 years ago
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